# How to integrate $\int \frac{x}{\sqrt{x^4+10x^2-96x-71}}dx$?

I read about $$\int \dfrac{x}{\sqrt{x^4+10x^2-96x-71}}dx$$ on the Wikipedia Risch algorithm page. They gave an answer but I don't understand how they got it.

• Plus or minus $96$?
– 5xum
Feb 19, 2014 at 9:16
• @5xum Can you share a link to its solution? Feb 19, 2014 at 9:18
• @5xum this
– Mike
Feb 19, 2014 at 9:57
• You are at a loss, eh? An explanation of the Risch algorithm is probably a whole book, so unless you are a specialist, leave it to the computer programs made by the specialists. Feb 19, 2014 at 15:39
• Don't look to deep into the machine eh.. just be glad it works! Feb 20, 2014 at 3:07

I originally intended a briefer version of this to be a comment, but because of the length of the briefer version and because it might spur someone to take up the challenge, I’m giving it as an answer. Also, I didn’t realize this question was over 4 years old until I had written my comments below, so perhaps my comments will generate some interest in this old question.

Let $$P(x) = x^4 + 10x^2 – 96x – 71.$$ The integral $$\int \frac{x}{P(x)}dx$$ is a pseudo-elliptic integral (but apparently not when $$71$$ is replaced by $$72),$$ and information on how to evaluate such integrals is given in Article 113, pp. 234-236 in Goursat/Hedrick’s A Course in Mathematical Analysis, Volume I (1904). The term involutory relation, which is mentioned near the beginning of Article 113, is discussed on pp. 231-233 in the first half of Article 112.

Regarding the roots of $$P(x),$$ which the method described by Goursat/Hedrick requires, these can be found using the quartic formula. Also possibly useful for the integration are the factorizations below (first factorization and second factorization), which can be obtained by expanding appropriate products of linear factors once you have the roots. Possibly Descartes' method for solving quartic equations will lead to the quadratic factors quicker, but I have not looked into this. The "difference of squares method" I described here for factoring $$x^4 + 1$$ doesn't seem to work for factoring $$P(x)$$ into a pair of quadratics.

$$x^4 \; + \; 10x^2 \; – \; 96x \; – \; 71 \;\; = \;\; \left(x^2 \; – \; \sqrt{12}\,x \; + \; 11 \; – \; \sqrt{192}\right)\left(x^2 \; + \; \sqrt{12}\,x \; + \; 11 \; + \; \sqrt{192}\right)$$

$$x^2 \; – \; \sqrt{12}\,x \; + \; 11 \; – \; \sqrt{192} \;\; = \;\; \left(x \; + \; \sqrt{\sqrt{192} \; – \; 8\;} \; - \; \sqrt{3}\right)\left(x \; - \; \sqrt{\sqrt{192} \; – \; 8\;} \; - \;\sqrt{3}\right)$$

$$= \;\; \left[\left(x - \sqrt{3}\right) \: + \; \sqrt{8\sqrt{3} \; – \; 8\;}\;\right]\left[\left(x - \sqrt{3}\right) \: - \; \sqrt{8\sqrt{3} \; – \; 8\;}\;\right]$$

From the last factorization above we see that the two real roots are symmetric about $$\sqrt{3},$$ which suggests the substitution $$u = x - \sqrt{3}$$ could be algebraically useful. However, I don't know how one would know a priori to make this substitution in $$P(x).$$

A few days after I posted the above remarks, no one had yet added anything. Therefore, since I had a little free time (read: I put some other things aside and made some free time), I decided to attack this integral myself. In what follows, essentially everything was done by hand --- only rarely did I even use a basic calculator for the multiplication of integers. However, I later conducted extensive algebraic and numeric checks with a computer algebra system. I’ve included most of my hand-calculation details --- so many details, in fact, that with only a few exceptions (these being where I felt there would be simply too much clutter to include them), the reader should not even need to resort to paper and pencil to completely follow what I’ve done.

# Contents

2. Summary of What Follows

3. Quadratic Factors of $$\;x^4 + 10x^2 - \; 96x - 71$$

4. Simultaneously Eliminate First-Degree Terms in the Two Quadratic Factors

5. Applying the Change of Variable $$\;x = \frac{at+b}{t+1}\;$$ to the Integrand

6. Decluttering the Coefficients of the Two Quadratics

7. Writing $$R(t)$$ as the Sum of an Even Function and an Odd Function

8. Evaluating the Integral Involving $$R_o(t)$$

9. Reduction of the Integral Involving $$R_e(t)$$ to Jacobi Standard Forms

10. Some Additional References (Not Cited Above)

My original goal was to give all the details for at least one method of indefinite integral evaluation and subsequent simplification that would result in the specific antiderivative given in the Wikipedia page that Hashir Omer cited in his question. I did not achieve this goal. In fact, I did not even obtain an antiderivative explicitly in terms of elementary functions. Instead, I have only carried out a relatively standard reduction process that gives, for general integrals of this type (rational functions of $$x$$ and $$y,$$ where $$y$$ is the square root of a quartic function of $$x),$$ the elementary function component of the antiderivative and the elliptic integral component of the antiderivative, with the latter reduced to Jacobi standard forms.

Because I'll be involved with an out of town work-related trip for the next several days, I can’t spend any more time on this at present. However, perhaps someone can use what I’ve done to complete the task of obtaining the precise form of the antiderivative given in the Wikipedia page. Incidentally, while researching some background information about the “standard reduction process” that I use here, I never saw a fully worked out example, either in print somewhere or on the internet. Thus, the following fully worked out example is likely to be of use to others from time to time.

When I decided to work on this integral, I spent the first several days trying to evaluate it by following the procedure given in Goursat/Hedrick, the book I cited in my original comments above. I finally gave up, and followed the procedure given in Appendix. Elliptic Integrals (pp. 124-137) in The Indefinite Integral by Grigorii Mikhailovich Fichtenholz (1971). However, for those who might be interested in following the Goursat/Hedrick’s book procedure, I found some helpful background information in Introduction to the Theory of Analytic Functions by James Harkness and Frank Morley (1898; see Chapter III on pp. 27-45 and Chapter V on pp. 57-66) and in Introduction to the Geometry of Complex Numbers by Roland Deaux (1957/2008). For what it’s worth, I have two other books that are marginally similar to Deaux’s book, namely Geometry of Complex Numbers by Hans Schwerdtfeger (1962/1979) and Complex Numbers by Liang-shin Hahn (1994). However, I found Deaux’s book much more useful than either of these two other books in trying to follow the discussion in Goursat/Hedrick.

In what follows, I have mostly avoided discussing all the various special cases that arise in general, such as issues relating to the nature of the roots of the quartic (4 distinct real roots, 2 distinct real roots and 2 complex conjugate non-real roots, etc.) and the change of variable from $$t$$ to $$z$$ in Section 9 (the change of variable that converts $$\;(1 - mt^2)(1 - nt^2)\;$$ to $$\;(1 - z^2)(1 - kz^2)\;$$ for $$0 < k < 1$$ has several cases, depending on the signs and magnitudes of the constants $$m$$ and $$n).$$ I have also tried to avoid mentioning or using any mathematical tools, concepts, or terms that would be unfamiliar to a student with a moderately strong U.S. Calculus 2 (Integral Calculus) background. Finally, the focus here is entirely on formal analytical calculations, without paying attention to restrictions that might need to be imposed on some of the variables and unspecified constants.

# 2. Summary of What Follows

In Section 3 we show how to factor $$\;x^4+10x^2-96x-71\;$$ into two quadratic polynomials with real coefficients. Then, in Section 4 we find real constants $$a$$ and $$b$$ such that the change of variable given by $$\;x = \frac{at+b}{t+1}\;$$ allows us to rewrite each of the quadratic polynomials in $$x$$ as $$(t+1)^{-2}$$ times a quadratic polynomial in $$t$$ having no first-degree-$$t$$ term. After this, in Section 5 we’ll go through all the details of rewriting the $$x$$-integral as a $$t$$-integral.

At this point, the irrational term in the integrand has the form $$\;\sqrt{({\gamma}'t^2\;+\;\delta)({\delta}'t^2\;+\; \gamma)\;}\,.\;$$ In Section 6 we show how, by factoring out an appropriate constant factor, we can replace the use of the four constant parameters $${\gamma},$$ $${\gamma}',$$ $${\delta},$$ $${\delta}'$$ with the use of one constant parameter, getting $$\;\sqrt{\left(1+{\alpha}t^2\right)\left(1-\frac{1}{8}{\alpha}t^2\right)}\;$$ where $$\;\alpha = 6\sqrt{3}-10.$$

We now have an integral with respect to $$t$$ that is $$\left(2 \sqrt{30 + 18\sqrt{3}}\right)^{-1}$$ times a rational function $$R(t)$$ divided by $$\;\sqrt{\left(1+{\alpha}t^2\right)\left(1-\frac{1}{8}{\alpha}t^2\right)}\,,\;$$ where $$\;\alpha = 6\sqrt{3}-10.$$ In Section 7 we write $$\;R(t)=R_e(t)+R_o(t),\;$$ where $$\;R_e(t)=\frac{1}{2}\left[R(t)+R(-t)\right]\;$$ is an even function and $$\;R_o(t)=\frac{1}{2}\left[R(t)-R(-t)\right]\;$$ is an odd function. After doing this, the integral we began with will be expressed as an integral involving $$R_e(t),$$ where the rational function part of the integrand winds up being a rational function of $$t^{2},$$ and an integral involving $$R_o(t),$$ where the rational function part of the integrand winds up being $$t$$ times a rational function of $$t^{2}.$$

In Section 8 we give all the details for evaluating the integral involving $$R_o(t)$$ in terms of elementary functions. In general, if the present method (Sections 4-9) is used to evaluate an integral of a rational function of $$x$$ and $$y,$$ where $$y$$ is the square root of a quartic function of $$x,$$ then the integral involving $$R_o(t)$$ can be expressed in terms of elementary functions, and the “essential elliptical’ness aspect of the integral” belongs to the integral involving $$R_e(t).$$

In Section 9 we give all the details for reducing the integral involving $$R_e(t)$$ to Jacobi (sometimes called Legendre-Jacobi) standard forms. In our particular case, there will be an elliptic integral of the first kind and an elliptic integral of the third kind. Apparently, in our particular case, the integral involving $$R_e(t)$$ can also be expressed in terms of elementary functions, but I have not done this. Those who wish to try doing this will want to investigate pseudo-elliptic integrals, perhaps beginning with the Stack Exchange question/answers When is an elliptic integral expressible in terms of elementary functions?. Finally, Section 10 gives a list of some relevant books and papers that I came across while working on this. These references, at least the specific sections and chapters and pages I mention in these references, were chosen for their usefulness to my intended audience and for their mostly free availability.

# 3. Quadratic Factors of $$\;x^4 + 10x^2 - 96x - 71$$

Despite what I said in my initial brief answer --- The "difference of squares method" $$[\cdots]$$ doesn't seem to work for factoring $$P(x)$$ into a pair of quadratics --- the difference of squares method wound up being the ONLY method that led me to being able to factor $$P(x)$$ without a great deal of difficulty.

For a simple example of this method, consider $$x^4 + 1.$$ If we add and subtract $$2x^2,$$ then we get $$x^4+2x^2+1-2x^2,$$ which can be written as the difference of squares $$(x^2 +1)^2 -(\sqrt{2}\,x)^2,$$ and therefore $$x^4 +1$$ can be factored as $$(x^2 +1+\sqrt{2}\,x)(x^2 +1-\sqrt{2}\,x).$$ The idea is to add a term that completes the square, then subtract that term, then apply the difference of squares factorization, using radicals if necessary.

For another example of this method, consider $$x^2+xy+y^2.$$ If the middle term had been $$2xy$$ instead of $$xy,$$ then this would have been a perfect square. So we add and subtract $$xy,$$ getting $$x^2+2xy+y^2-xy,$$ which can be written as the difference of squares $$(x+y)^2-(\sqrt{xy})^{2},$$ and therefore $$x^2 + xy + y^2$$ can be factored as $$(x+y+\sqrt{xy})(x+y-\sqrt{xy}).$$

This method can be found in a large number of older algebra textbooks. For example, see Article 172 on pp. 83-84 and Article 459 on p. 281 in Advanced Course in Algebra by Webster Wells (1904) AND Article 111 on p. 134 in Academic Algebra by George Wentworth and David Eugene Smith (1913) AND Chapter VII, Article 10 on pp. 132-133 in Algebra. An Elementary Text-Book (Part I) by George Chrystal (1886).

A generalization of this method can be used to solve quartic equations --- a method used by Ferrari and Simpson. See Ferrari’s Solution of a Quartic Equation (this is especially recommended) AND Article 63 in The Theory of Equations by William Snow Burnside and Arthur William Panton (1912 7th edition) AND Encyclopedia of Mathematics: Ferrari method.

The generalization of this method is to find constants $$a,$$ $$b,$$ $$A,$$ $$B$$ such that

$$(x^2 + ax + b)^2 \;-\;(Ax + B)^2 \;\;=\;\; x^4 + 10x^2 -96x-71$$

Expanding the left side of the above equation and collecting like powers of $$x$$ gives the following. Incidentally, the square of a trinomial can be expanded by adding the squares of each of the 3 terms to double the product of each of the 3 pairs of distinct terms.

$$x^4 + (2a)x^3+(a^2+2b-A^2)x^2 +(2ab-2AB)x+(b^2 - B^2)\;\;=\;\;x^4 +10x^2 -96x-71$$

Equating the $$x^3$$ coefficients gives $$a = 0.$$ Equating the other coefficients and using $$a = 0$$ gives

$$\begin{matrix} \text{(a)} & & 2b - A^2 & = & 10 \\ \text{(b)} & & -2AB & = & -96 \\ \text{(c)} & & b^2 - B^2 & = & -71 \\ \end{matrix}$$

From (b) we have $$A = \frac{48}{B}.$$ From (c) we have $$B^2 = b^2 + 71.$$ Applying these last two equalities successively to (a) gives

$$2b \;\;=\;\; A^2 +10 \;\;=\;\; \left(\frac{48}{B}\right)^2 +10 \;\;=\;\; \frac{{48}^2}{b^2 +71}+10$$

Now multiply both sides of the above (far left side and far right side) by $$b^2 + 71$$ to get

$$2b(b^2 + 71)\;\;=\;\;{48}^2 +10(b^2 +71)$$

$$b^3 \; - \; 5b^2 \; + \; 71b \; - \; 1507\;=\; 0$$

Since $$1507 = 11 \cdot 137,$$ the rational root test tells us that any rational solution of this equation must be among the 8 numbers $$\pm \,1,\, \pm\, 11,\, \pm\, 137,\, \pm\, 1507.$$ A numerical check shows that $$b = 11$$ is a solution. Note that only $$\pm \, 1$$ and $$\pm \, 11$$ need to be checked, since a unit-digit analysis shows that the other possibilities result in evaluations whose unit digits are either $$2$$ or $$8.$$ Using this value of $$b,$$ we get $$A^2 = 12$$ from equation (a) and $$B^2 = 192$$ from equation (c). We only need one choice of values for $$a,$$ $$b,$$ $$A,$$ $$B$$ so we use $$a=0$$ and $$b=11$$ and $$A = \sqrt{12} = 2\sqrt{3}$$ and $$B = \sqrt{192} = 8\sqrt{3},$$ which gives us

$$x^4 + 10x^2 - 96x - 71 \;\; = \;\; (x^2 + 11)^2 \; - \; (2\sqrt{3}\,x + 8\sqrt{3})^2$$

$$=\;\;\left[(x^2 + 11)\;+\;(2\sqrt{3}\,x + 8\sqrt{3})\right]\left[(x^2 + 11) \;-\;(2\sqrt{3}\,x + 8\sqrt{3})\right]$$

$$= \;\; \left(x^2 \; + \; 2\sqrt{3}\,x\;+\;11\;+\; 8\sqrt{3}\right)\left(x^2 \; -\; 2\sqrt{3}\,x \;+\; 11\;-\; 8\sqrt{3}\right)$$

An alternate method of factoring $$x^4 + 10x^2 - 96x - 71$$ is to seek a factorization of the form $$(x^2 + ax + b)(x^2 + Ax + B)$$ --- see Article 63 in The Theory of Equations by William Snow Burnside and Arthur William Panton (1912 7th edition). This leads to $$A + a = 0$$ and then, after replacing $$A$$ with $$-a,$$ to the equations (1) $$B - a^2 + b = 10$$ and (2) $$ab - aB = 96$$ and (3) $$bB = -71.$$ The seemingly most promising reduction I was able to obtain was the pair of equations (4) $$2ab = a^3 + 10a + 96$$ and (5) $$ab^2 + 71a = 96b,$$ where equation (4) can be obtained from $$a$$ times equation (1) and equation (2), and equation (5) can be obtained by replacing $$B$$ with $$-\frac{71}{b}$$ in equation (2). By rewriting equation (5) as $$(ab)b + 71a = 96b$$ and using equation (4) to replace the factor $$ab$$ and then the two remaining appearances of $$b$$ (all three being replaced with expressions only involving $$a),$$ leads to the equation $$a^6 + 20a^4 + 384a^2 - 9216 = 0,$$ or equivalently, $$m^3 + 20m^2 + 384m - 9216 = 0$$ where $$m = a^{2}.$$ A solution could be discovered for $$m,$$ namely $$m = 12,$$ by applying the rational root theorem. However, because $$9216 = 2^{10}3^{2},$$ which has $$(10 + 1)(2 + 1) = 33$$ many positive integer factors, applying the rational root theorem involves considering $$66$$ many possibilities, a task far more difficult than before.

# 4. Simultaneously Eliminate First-Degree Terms in the Two Quadratic Factors

Recall that the first-degree term of a quadratic can be removed by an appropriate change of variable. Specifically, there exists a constant $$a$$ such that the change of variable $$x = t+a$$ allows us to rewrite $$Ax^2+Bx+C$$ in the form $$A't^2+C'.$$ To see this, substitute $$x=t+a$$ into $$Ax^2+Bx+C,$$ which gives $$At^2+(2Aa+B)t+(Aa^2+Ba+C)=0,$$ then set the coefficient of $$t$$ equal to zero and solve for $$a$$ to get $$a=-\frac{B}{2A}.$$ This is essentially the method of completing the square.

It is possible to simultaneously "essentially eliminate" the first-degree terms in two quadratics by a less simple change of variable. Specifically, one can show that there exist constants $$a$$ and $$b$$ such that the change of variable $$x=\frac{at+b}{t+1}$$ allows us to rewrite both $$Ax^2 + Bx + C$$ and $$Dx^2+Ex+F$$ into the forms $$\frac{A't^2+C'}{(t+1)^2}$$ and $$\frac{D't^2+F'}{(t+1)^2}.$$ I will not prove this in general. Instead, I will find such a change of variable for the specific quadratic factors we found earlier. Incidentally, my answers to Proving a quadratic polynomial has no real roots without using derivatives or any formulas and to The sum - product problem give a geometric way of understanding why the change of variable $$x = t+a$$ works for one quadratic. There is surely a geometric way of understanding why the change of variable $$x = \frac{at+b}{t+1}$$ works for two quadratics (e.g. certain geometric properties of homographic/linear-fractional transformations), but I have not investigated this.

Substituting $$\;x=\frac{at+b}{t+1}\;$$ into $$\;x^2+2\sqrt{3}\,x+11+8\sqrt{3}\;$$ gives

$$\left(\frac{at+b}{t+1}\right)^2\;+\;\;2\sqrt{3}\left(\frac{at+b}{t+1}\right) \;+\;(11+8\sqrt{3})$$

$$=\;\;\frac{a^2t^2+2abt+b^2}{(t+1)^2}\;+\;\frac{2\sqrt{3}(at+b)(t+1)}{(t+1)^2}\;+\;\frac{(11+8\sqrt{3})(t+1)^2}{(t+1)^2}$$

$$=\;\;\frac{A't^2\;+\;\left[2ab+2\sqrt{3}(a+b)+22+16\sqrt{3}\right]t\;+\;C'}{(t+1)^2},$$

where for later use we note that

$$A'=a^2+2\sqrt{3}\,a+11+8\sqrt{3}\;\;\text{and}\;\; C'=b^2+2\sqrt{3}\,b+11+8\sqrt{3}.$$

Substituting $$\;x=\frac{at+b}{t+1}\;$$ into $$\;x^2- 2\sqrt{3}\,x+11-8\sqrt{3}\;$$ gives

$$\left(\frac{at+b}{t+1}\right)^2 \;-\;\;2\sqrt{3}\left(\frac{at+b}{t+1}\right)\;+\;(11-8\sqrt{3})$$

$$= \;\;\frac{a^2t^2+2abt+b^2}{(t+1)^2}\;-\;\frac{2\sqrt{3}(at+b)(t+1)}{(t+1)^2}\;+\;\frac{(11-8\sqrt{3})(t+1)^2}{(t+1)^2}$$

$$= \;\;\frac{D't^2\;+\;\left[2ab-2\sqrt{3}(a+b)+ 22 - 16\sqrt{3}\right]t\; +\;F'}{(t+1)^2},$$

where for later use we note that

$$D'=a^2-2\sqrt{3}\,a+11-8\sqrt{3}\;\;\text{and}\;\;F'=b^2-2\sqrt{3}\,b+11-8\sqrt{3}.$$

The first-degree terms in the denominators will vanish when both of the following equations are satisfied.

$$\begin{matrix} \text{(d)} & & 2ab+2\sqrt{3}(a+b)+22+16\sqrt{3} & = & 0 \\ \text{(e)} & & 2ab-2\sqrt{3}(a+b)+22-16\sqrt{3} & = & 0 \\ \end{matrix}$$

Adding equations (d) and (e) gives $$4ab + 44 = 0,$$ and hence $$ab = -11.$$ Subtracting equation (e) from equation (d) gives $$4\sqrt{3}(a+b) + 32\sqrt{3} = 0,$$ and hence $$a+b = \frac{-32\sqrt{3}}{4\sqrt{3}}=-8.$$ Thus, we have reduced the solution to that of solving the system $$ab=-11$$ and $$a+b=-8.$$ This system can be solved by standard methods (e.g. $$ab=-11$$ becomes $$a(-8-a) = -11,$$ etc.), but interestingly this system can also be solved by using Vieta's formulas to obtain the equation $$Q^2+8Q-11=0.$$ There are two solutions for the pair $$(a,b),$$ namely $$\;(a,b)=(-4+3\sqrt{3},\;-4 -3\sqrt{3})\;$$ and $$\;(a,b)=(-4-3\sqrt{3}, \;-4+3\sqrt{3}).$$ Each solution gives an appropriate change of variable that can be used. I'll use the first solution for our quadratics.

# 5. Applying the Change of Variable $$\;x = \frac{at+b}{t+1}\;$$ to the Integrand

The specific change of variable we will use is

$$x \;\; = \;\; \frac{at+b}{t+1}\;\;=\;\; \frac{(-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})}{t+1}$$

We now apply this change of variable to

$$\int \frac{x \, dx}{\sqrt{x^4+10x^2-96x-71}} \;\; = \;\; \int \frac{x \, dx}{\sqrt{Ax^2 + Bx + C}\;\sqrt{Dx^2 + Ex + F}}$$

$$= \;\; \int \frac{x \, dx} {\sqrt{x^2 \; + \; 2\sqrt{3}\,x \; + \; 11 \; + \; 8\sqrt{3}\;} \; \sqrt{x^2 \; - \; 2\sqrt{3}\,x \; + \; 11 \; - \; 8\sqrt{3}\;}}$$

to obtain the integral

$$\int \frac{R(t)\,dt}{\sqrt{A't^2 + C'}\,\sqrt{D't^2 + F'}}$$

where $$R(t)$$ is a rational function and $$A',$$ $$C',$$ $$D',$$ $$F'$$ are as given above.

Applying this change of variable and using the results above where the first boldface usage of "Substituting" appears, the quadratic $$\; Ax^2 + Bx + C\;$$ becomes $$\frac{A't^2 + C'}{(t+1)^2},\;$$ where

$$A' \;\; = \;\; a^2 + 2\sqrt{3}\,a + (11 + 8\sqrt{3}) \;\; = \;\; (-4 + 3\sqrt{3})^2 + 2\sqrt{3}(-4 + 3\sqrt{3}) + (11 + 8\sqrt{3})$$

$$=\;\;16 - 24\sqrt{3} + 27 - 8\sqrt{3} + 18 + 11 + 8\sqrt{3} \;\; = \;\; 72 - 24\sqrt{3}$$

and

$$C' \;\; = \;\; b^2 + 2\sqrt{3}\,b + (11 + 8\sqrt{3}) \;\; = \;\; (-4 - 3\sqrt{3})^2 + 2\sqrt{3}(-4 - 3\sqrt{3}) + (11 + 8\sqrt{3})$$

$$= \;\; 16 + 24\sqrt{3} + 27 - 8\sqrt{3} - 18 + 11 + 8\sqrt{3} \;\; = \;\; 36 + 24\sqrt{3}$$

Therefore,

$$\sqrt{x^2 \; + \; 2\sqrt{3}\,x \; + \; 11 \; + \; 8\sqrt{3}\;} \;\; = \;\; \sqrt{ \frac {(72 - 24\sqrt{3})t^2 \; + \; (36 + 24\sqrt{3})}{(t+1)^2} }$$

$$= \;\; (t+1)^{-1} \sqrt{(72 - 24\sqrt{3})t^2 \; + \; (36 + 24\sqrt{3})}$$

$$= \;\; 2\sqrt{3}\,(t+1)^{-1} \sqrt{(6 - 2\sqrt{3})t^2 \; + \; (3 + 2\sqrt{3})}$$

Applying this change of variable and using the results above where the second boldface usage of "Substituting" appears, the quadratic $$\; Dx^2 + Ex + F\;$$ becomes $$\frac{D't^2 + F'}{(t+1)^2},\;$$ where

$$D' \;\; = \;\; a^2 - 2\sqrt{3}\,a + (11 - 8\sqrt{3}) \;\; = \;\; (-4 + 3\sqrt{3})^2 - 2\sqrt{3}(-4 + 3\sqrt{3}) + (11 - 8\sqrt{3})$$

$$= \;\; 16 - 24\sqrt{3} + 27 + 8\sqrt{3} - 18 + 11 - 8\sqrt{3} \;\; = \;\; 36 - 24\sqrt{3}$$

and

$$F' \;\; = \;\; b^2 - 2\sqrt{3}\,b + (11 - 8\sqrt{3}) \;\; = \;\; (-4 - 3\sqrt{3})^2 - 2\sqrt{3}(-4 - 3\sqrt{3}) + (11 - 8\sqrt{3})$$

$$= \;\; 16 + 24\sqrt{3} + 27 + 8\sqrt{3} + 18 + 11 - 8\sqrt{3} \;\; = \;\; 72 + 24\sqrt{3}$$

Therefore,

$$\sqrt{x^2 \; - \; 2\sqrt{3}\,x \; + \; 11 \; - \; 8\sqrt{3}\;} \;\; = \;\; \sqrt{ \frac {(36 - 24\sqrt{3})t^2 \; + \; (72 + 24\sqrt{3})}{(t+1)^2} }$$

$$= \;\; (t+1)^{-1} \sqrt{(36 - 24\sqrt{3})t^2 \; + \; (72 + 24\sqrt{3})}$$

$$= \;\; 2\sqrt{3}\,(t+1)^{-1} \sqrt{(3 - 2\sqrt{3})t^2 \; + \; (6 + 2\sqrt{3})}$$

Combining these results for the two quadratic factors under the square root gives

$$\sqrt{x^4 + 10x^2 - 96x - 71} \;\; = \;\; \sqrt{x^2 \; + \; 2\sqrt{3}\,x \; + \; 11 \; + \; 8\sqrt{3}\;} \; \sqrt{x^2 \; - \; 2\sqrt{3}\,x \; + \; 11 \; - \; 8\sqrt{3}\;}$$

$$= \;\; 2\sqrt{3}\,(t+1)^{-1} \sqrt{(6 - 2\sqrt{3})t^2 \; + \; (3 + 2\sqrt{3})} \; \cdot \; 2\sqrt{3}\,(t+1)^{-1} \sqrt{(3 - 2\sqrt{3})t^2 \; + \; (6 + 2\sqrt{3})}$$

$$= \;\; 12\,(t+1)^{-2} \sqrt{(6 - 2\sqrt{3})t^2 \; + \; (3 + 2\sqrt{3})} \; \sqrt{(3 - 2\sqrt{3})t^2 \; + \; (6 + 2\sqrt{3})}$$

Incorporating the change of variable made to the radical term of the integrand gives

$$\int \frac{x \, dx}{\sqrt{x^4+10x^2-96x-71}}$$

$$= \;\; \int \frac {(t+1)^2}{ 12 \sqrt{(6 - 2\sqrt{3})t^2 \; + \; (3 + 2\sqrt{3})} \; \sqrt{(3 - 2\sqrt{3})t^2 \; + \; (6 + 2\sqrt{3})} } \; \cdot \; x\,dx$$

$$= \;\; \int \frac {(t+1)^2}{12} \cdot \frac{1}{\sqrt{{\gamma}'t^2 \; + \; \delta} \; \sqrt{{\delta}'t^2 \; + \; \gamma} } \; \cdot \; x\,dx,$$

where $$\;\gamma = 6 + 2\sqrt{3}\;$$ and $$\;{\gamma}' = 6 - 2\sqrt{3}\;$$ and $$\;\delta = 3 + 2\sqrt{3}\;$$ and $$\;{\delta}' = 3 - 2\sqrt{3}.$$

We now apply the change of variable to $$x \, dx.$$

$$dx \; = \; d\left(\frac{U}{V}\right) \; = \; \frac{dU \cdot V \; - \; U \cdot dV}{V^2},$$

where $$\;U \; = \; (-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})\;$$ and $$\;V = t+1.\;$$ Thus,

$$dx \;\; = \;\; d \left( \frac{(-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})}{t + 1} \right)$$

$$= \;\; \frac{[(-4 + 3\sqrt{3})\,dt]\cdot(t+1) \; - \; [(-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})]\cdot dt }{(t+1)^2}$$

$$= \;\; \left[ \frac{-4t + 3\sqrt{3}\,t - 4 + 3\sqrt{3} + 4t - 3\sqrt{3}\,t + 4 + 3\sqrt{3}}{(t+1)^2} \right] \, dt \;\; = \;\; \frac{6\sqrt{3}}{(t+1)^2}\,dt$$

Hence,

$$x\cdot dx \;\; = \;\; \frac{(-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})}{t + 1} \cdot \frac{6\sqrt{3}}{(t+1)^2}\,dt$$

$$= \;\; \frac{6}{(t+1)^3} \cdot \left[ (-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9) \right] \, dt$$

Therefore,

$$\int \frac{x \, dx}{\sqrt{x^4+10x^2-96x-71}}$$

$$= \;\; \int \; \frac {(t+1)^2}{12} \cdot \frac{1}{\sqrt{{\gamma}'t^2 \; + \; \delta} \; \sqrt{{\delta}'t^2 \; + \; \gamma}} \; \cdot \; \frac{6}{(t+1)^3} \cdot \left[ (-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9) \right] \, dt$$

$$= \;\; \int \frac{R(t)\,dt}{2\sqrt{{\gamma}'t^2 \; + \; \delta} \; \sqrt{{\delta}'t^2 \; + \; \gamma}}\, ,$$

where

$$R(t) \;\; = \;\; \frac{(-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9)}{t+1}$$

# 6. Decluttering the Coefficients of the Two Quadratics

By factoring out an appropriate constant factor, which can be taken outside the integral sign, we can describe the coefficients of

$$\sqrt{{\gamma}'t^2 \; + \; \delta} \; \sqrt{{\delta}'t^2 \; + \; \gamma}$$

with the use of one constant parameter, where $$\;\gamma = 6 + 2\sqrt{3}\;$$ and $$\;{\gamma}' = 6 - 2\sqrt{3}\;$$ and $$\;\delta = 3 + 2\sqrt{3}\;$$ and $$\;{\delta}' = 3 - 2\sqrt{3}.$$

First, factor out appropriate constants so that each of the quadratics has constant coefficient $$1.$$

$$\sqrt{{\delta}\left(1 \; + \; \frac{{\gamma}'}{\delta}t^2\right)} \; \sqrt{{\gamma}\left(1 \; + \; \frac{{\delta}'} {\gamma}t^2 \right)} \;\; = \;\; \sqrt{{\delta}{\gamma}} \; \sqrt{1 \; + \; \frac{{\gamma}'}{\delta}t^2} \; \sqrt{1 \; + \; \frac{{\delta}'} {\gamma}t^2}$$

where

$${\delta}{\gamma} \;\; = \;\; (3 + 2\sqrt{3})(6 + 2\sqrt{3}) \;\; = \;\; 30 + 18\sqrt{3}$$

$$\frac{{\gamma}'}{\delta} \;\; = \;\; \frac{6 - 2\sqrt{3}}{3 + 2\sqrt{3}} \;\; = \;\; \frac{6 - 2\sqrt{3}}{3 + 2\sqrt{3}} \cdot \frac{3 - 2\sqrt{3}}{3 - 2\sqrt{3}} \;\; = \;\; \frac{30 - 18\sqrt{3}}{-3} \;\; = \;\; 6\sqrt{3} - 10$$

$$\frac{{\delta}'}{\gamma} \;\; = \;\; \frac{3 - 2\sqrt{3}}{6 + 2\sqrt{3}} \;\; = \;\; \frac{3 - 2\sqrt{3}}{6 + 2\sqrt{3}} \cdot \frac{6 - 2\sqrt{3}}{6 - 2\sqrt{3}} \;\; = \;\; \frac{30 - 18\sqrt{3}}{24} \;\; = \;\; \frac{5 - 3\sqrt{3}}{4}$$

If we let $$\;\alpha = \frac{{\gamma}'}{\delta} = 6\sqrt{3} - 10,\;$$ then $$\;\frac{{\delta}'}{\gamma} = -\frac{1}{8}\alpha.\;$$ In general, note that we can always express both $$\;\frac{{\gamma}'}{\delta}\;$$ and $$\;\frac{{\delta}'}{\gamma}\;$$ in terms of a single constant parameter $$\;\alpha = \frac{{\gamma}'}{\delta},\;$$ since $$\;\frac{{\delta}'}{\gamma} \; = \; \left(\frac{ {\delta}'/\gamma} { {\gamma}'/\delta}\right) \frac{{\gamma}'}{\delta} \; = \; \left(\frac{ {\delta}'/\gamma} { {\gamma}'/\delta}\right) \alpha.$$

Therefore,

$$\int \frac{x \, dx}{\sqrt{x^4+10x^2-96x-71}}\;\;=\;\;\int\frac{R(t)\,dt}{2\sqrt{{\gamma}'t^2 \; + \; \delta} \; \sqrt{{\delta}'t^2 \; + \; \gamma}}$$

$$=\;\; \frac{1}{2 \sqrt{30 + 18\sqrt{3}}} \cdot \int \frac{R(t)\,dt}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}}$$

where

$$R(t)\;\;=\;\; \frac{(-4\sqrt{3}+9)t\;+\;(-4\sqrt{3} - 9)}{t+1} \;\;\; \text{and} \;\;\; \alpha = 6\sqrt{3} - 10$$

• It only took four years! Hahah :) Apr 12, 2018 at 17:20
• FYI, the reason I made two answers is that there is a 30,000 character limit on single answer submissions, and my "total answer" is about 55,000 characters. Apr 30, 2018 at 9:22
• I've never seen such detailed answer... It justifies the answer in en.wikipedia.org/wiki/Risch_algorithm#Problem_examples right? May 2, 2018 at 2:04
• @Tony Ma: No, to do that I'd need to trace back through all the substitutions and show that you get the integer-coefficient polynomial stuff in the Wikipedia example. I tried that with the integral involving $R_o(t)$ (end of Section 8; my attempts are not included in the write-up), and didn't get anything close to it, but I really doubt I made any mistakes because I was VERY CAREFUL with the computations, proof-reading, and safety checks. Of course, I didn't even get an explicit form for the other part, which involves $R_e(t)$ (what's dealt with in Section 9), but I expected (continued) May 2, 2018 at 23:09
• to get something that at least resembled parts of the Wikipedia answer with the integral involving $R_o(t),$ and I didn't. It might be because there's some really non-trivial simplification and rewriting that needs to take place, or it might be that I need the $R_e(t)$ integral part in order to cancel out the "non-Wikipedia looking stuff" in the $R_o(t)$ integral. For what it's worth, I have an idea now of how to evaluate the $R_e(t)$ integral in terms of elementary functions, but I don't know when I'll have a chance to get to it. The work-related trip I was on involves (continued) May 2, 2018 at 23:13

THIS CONTINUES MY ANSWER THAT BEGINS HERE

# 7. Writing $$R(t)$$ as the Sum of an Even Function and an Odd Function

Recall that every function can be written as the sum of an even function and an odd function. For discussions of how such a decomposition can be discovered, see How do I divide a function into even and odd sections?. We want an even function $$R_e(t),$$ hence $$R_e(-t) = R_e(t),$$ and an odd function $$R_o(t),$$ hence $$R_o(-t) = -R_o(t),$$ such that $$R(t) = R_e(t) + R_o(t).$$ Thus, we want to express $$R(t)$$ as

$$R(t) \;\; = \;\; \underbrace{\frac{R(t) + R(-t)}{2}}_{R_e(t)} \;\; + \;\; \underbrace{\frac{R(t) - R(-t)}{2}}_{R_o(t)}$$

Carrying out the details, we get

$$R_e(t) \;\; = \;\; \frac{(-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9)}{t+1} \;\; + \;\; \frac{(-4\sqrt{3} + 9)(-t) \; + \; (-4\sqrt{3} - 9)}{-t + 1}$$

$$= \;\; \frac{ \left[(-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9)\right](-t + 1) \;\; + \;\; (t+1)\left[(-4\sqrt{3} + 9)(-t) \; + \; (-4\sqrt{3} - 9)\right]}{- t^2 + 1}$$

$$= \;\; \frac{(18 - 8\sqrt{3})t^2 \; + \; (18 + 8\sqrt{3})}{t^2 - 1}$$

and

$$R_o(t) \;\; = \;\; \frac{(-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9)}{t+1} \;\; - \;\; \frac{(-4\sqrt{3} + 9)(-t) \; + \; (-4\sqrt{3} - 9)}{-t + 1}$$

$$= \;\; \frac{ \left[(-4\sqrt{3} + 9)t \; + \; (-4\sqrt{3} - 9)\right](-t + 1) \;\; - \;\; (t+1)\left[(-4\sqrt{3} + 9)(-t) \; + \; (-4\sqrt{3} - 9)\right]}{- t^2 + 1}$$

$$= \;\; \frac{-18t}{t^2 - 1}$$

Therefore,

$$\int \frac{x \, dx}{\sqrt{x^4+10x^2-96x-71}}$$

$$= \;\;\;\; \frac{1}{2 \sqrt{30 + 18\sqrt{3}}} \cdot \left[ \int \frac{R_e(t)\,dt}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}} \;\; + \;\; \int \frac{R_o(t)\,dt}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}}\right]$$

where $$\;\alpha = 6\sqrt{3} - 10\;$$ and $$\;R_e(t),\;R_o(t)\;$$ are given above.

# 8. Evaluating the Integral Involving $$R_o(t)$$

Applying the substitution $$\;u = t^2\;$$ and $$\;du = 2t\,dt\;$$ and $$\;dt = \frac{du}{2t},\;$$ we get

$$\int \frac{R_o(t)\,dt}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}} \;\; = \;\; \int \frac{-18t\,dt}{t^2 - 1} \cdot \frac{1}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}}$$

$$= \;\; \int \frac{-18t \left(\frac{du}{2t}\right)}{u - 1} \cdot \frac{1}{ \sqrt{ \left( 1 + {\alpha}u \right) \left( 1 - \frac{1}{8}{\alpha}u \right)}} \;\; = \;\; \int \frac{-9\,du}{u - 1} \cdot \frac{1}{ \sqrt{ \left( 1 + {\alpha}u \right) \left( 1 - \frac{1}{8}{\alpha}u \right)}}$$

This integral belongs to a standard class of integrals that can be expressed using elementary functions, namely those whose integrands have the form $$R(x,y)\,dx$$ where $$y^2$$ is a quadratic function of $$x$$ and $$R(x,y)$$ is a rational function of $$x$$ and $$y.$$ In general, such an integral is often first reduced to sums and differences of a handful of special forms, and each of these forms has its own method of evaluation. Fortunately, the integral we have is already in one of these special forms.

Since this class of integrals is rarely covered in modern elementary calculus textbooks, I’ll mention a few freely available references that discuss both the more general class and the specific form we have. Elements of the Integral Calculus by William Elwood Byerly (1888; see Article 68 on pp. 59-61) AND Infinitesimal Analysis by William Benjamin Smith (1898; see Article 124 on pp. 99-101, especially the last complete sentence on p. 99) AND The Integration of Functions of a Single Variable by Godfrey Harold Hardy (1916; see Chapter V, especially Section V.5 on pp. 26-27) AND A Treatise on the Integral Calculus, Volume I, by Joseph Edwards (1921; see Chapter VIII on pp. 275-323) AND A Course of Pure Mathematics by Godfrey Harold Hardy 1921; see Article 139 on pp. 242-245; or see Article 142 on pp. 261-264 of any of the recent reprints of the 1952 10th edition) AND An Elementary Course of Infinitesimal Calculus by Horace Lamb (1942; see Article 86.3 on pp. 190-191).

Let $$\;v = \frac{1}{u-1}.\;$$ Then $$\;u = 1 + \frac{1}{v} = \frac{v+1}{v}\;$$ and $$\;dv = -\frac{1}{(u-1)^2}\,du = -v^{2}du.\;$$ Therefore, $$\;du = -\frac{dv}{v^2},\;$$ and hence $$\;\frac{du}{u-1} = du \cdot v = \left(-\frac{dv}{v^2}\right) \cdot v = -\frac{dv}{v}.\;$$ Using this substitution, the integral becomes

$$-9 \int \frac{du}{u-1} \cdot \frac{1}{ \sqrt{ \left[ 1 + {\alpha}u \right] \left[ 1 - \frac{1}{8}{\alpha}u \right]}}$$

$$= \;\; -9 \int \left(-\frac{dv}{v}\right) \cdot \frac{1}{ \sqrt{ \left[ 1 + {\alpha}\left(\frac{v+1}{v}\right) \right] \left[ 1 - \frac{1}{8}{\alpha}\left(\frac{v+1}{v}\right) \right]}}$$

$$= \;\; 9 \int \left(\frac{dv}{v}\right) \cdot \frac{1}{ \sqrt{ \left(\frac{1}{v}\right) \left( v + {\alpha}v + \alpha \right) \left(\frac{1}{v}\right) \left( v - \frac{1}{8}{\alpha}v - \frac{1}{8}{\alpha} \right)}}$$

$$= \;\; 9 \int \frac{dv}{v \left(\frac{1}{v}\right) \sqrt{( v + {\alpha}v + \alpha) \left(\frac{1}{8}\right)(8v - {\alpha}v - {\alpha})}}$$

$$= \;\; 9 \int \frac{dv}{\sqrt{\frac{1}{8}} \cdot \sqrt{( v + {\alpha}v + \alpha)(8v - {\alpha}v - {\alpha})}}$$

$$= \;\; 9\sqrt{8} \int \frac{dv}{ \sqrt{( v + {\alpha}v + \alpha)(8v - {\alpha}v - {\alpha})}} \;\; = \;\; 18\sqrt{2} \int \frac{dv}{ \sqrt{( v + {\alpha}v + \alpha)(8v - {\alpha}v - {\alpha})}}$$

Using $$\; \alpha = 6\sqrt{3} - 10,\;$$ and a computer algebra system (only for the next few numerical computations, ending with completing the square), we get

$$( v + {\alpha}v + \alpha)(8v - {\alpha}v - {\alpha}) \;\; = \;\; 8v^2- {\alpha}v^2 - {\alpha}v + 8{\alpha}v^2 - {\alpha}^2v^2 - {\alpha}^2v + 8{\alpha}v - {\alpha}^2v - {\alpha}^2$$

$$= \;\; (-{\alpha}^2 + 7{\alpha} + 8)v^2 \; + \; (-2{\alpha}^2 + 7{\alpha})v \; - \; {\alpha}^2$$

$$= \;\; (162\sqrt{3} - 270)v^2 \; + \; (282\sqrt{3} - 486)v \; + \; (120\sqrt{3} - 208)$$

$$= \;\; (162\sqrt{3} - 270)\left[ v^2 \; + \; \left(\frac{9 - 4\sqrt{3}}{9}\right)v \; + \; \underline{\;\;\;\;\;\;\;\;\;\;\;} \; \right] \;\; + \;\; (120\sqrt{3} - 208) \;\; - \;\; (162\sqrt{3} - 270)(\; \underline{\;\;\;\;\;\;\;\;\;\;\;} \;)$$

$$= \;\; (162\sqrt{3} - 270)\left[ v^2 \; + \; \left(\frac{9 - 4\sqrt{3}}{9}\right)v \; + \; \left(\frac{9 - 4\sqrt{3}}{18}\right)^2 \right] \;\; + \;\; (120\sqrt{3} - 208) \;\; - \;\; (162\sqrt{3} - 270) \left(\frac{9 - 4\sqrt{3}}{18}\right)^2$$

$$= \;\; (162\sqrt{3} - 270)\left(v \; + \; \frac{9 - 4\sqrt{3}}{18}\right)^2 \; + \; \frac{15 - 9\sqrt{3}}{2}$$

Using this, the integral now becomes

$$\frac{18\sqrt{2}}{\sqrt{162\sqrt{3} - 270}} \int \frac{dv}{\sqrt{\left(v \; + \; \frac{9 - 4\sqrt{3}}{18}\right)^2 \; + \; \frac{15 - 9\sqrt{3}}{2}}}$$

Letting $$\;w = v + \frac{9 - 4\sqrt{3}}{18},\;$$ we get

$$\frac{18\sqrt{2}}{\sqrt{162\sqrt{3} - 270}} \int \frac{dw}{\sqrt{w^2 \; + \; \frac{15 - 9\sqrt{3}}{2}}}$$

Before continuing, we'll clean up the radical expression $$\;\frac{18\sqrt{2}}{\sqrt{162\sqrt{3} - 270}}\;$$ a little.

$$\frac{18\sqrt{2}}{\sqrt{162\sqrt{3} - 270}} \cdot \frac{\sqrt{162\sqrt{3} + 270}}{\sqrt{162\sqrt{3} + 270}} \;\; = \;\; \frac{18\sqrt{2} \cdot \sqrt{162\sqrt{3} + 270}}{\sqrt{78732 - 72900}}$$

$$= \;\; \frac{18\sqrt{2} \cdot 3\sqrt{18\sqrt{3} + 30}}{\sqrt{5832}} \;\; = \;\; \frac{54\sqrt{2} \cdot \sqrt{18\sqrt{3} + 30}}{54\sqrt{2}} \;\; = \;\; \sqrt{18\sqrt{3} + 30}$$

Thus, we have reduced the integral to

$$\sqrt{18\sqrt{3} + 30} \int \frac{dw}{\sqrt{w^2 \; + \; \frac{15 - 9\sqrt{3}}{2}}} \;\; = \;\; \sqrt{18\sqrt{3} + 30} \int \frac{dw}{\sqrt{w^2 \; - \; \beta}}$$

where $$\; \beta = \frac{9\sqrt{3} - 15}{2} > 0.\;$$ Since the integral involving $$R_o(t)$$ has a constant factor $$\;\frac{1}{2 \sqrt{30 + 18\sqrt{3}}}\;$$ (see the end of Section 7), the integral involving $$R_o(t)$$ becomes

$$\frac{1}{2} \int \frac{dw}{\sqrt{w^2 \; - \; \beta}}$$

This last integral is a standard form involving $$\cosh$$ (hyperbolic cosine), but let's evaluate it without the use of hyperbolic functions. Since the trigonometric substitution $$x = a\sec \theta$$ is used for "$$x^2 - a^2$$ forms", we let $$\;w = \sqrt{\beta}\sec {\theta}.\;$$ Then $$\;w^2 - \beta = \beta({\sec}^2 \theta - 1) = \beta\,{\tan}^2 \theta \;$$ and $$\;dw = \sqrt{\beta}\sec \theta \tan \theta \, d\theta.\;$$ Therefore we get

$$\frac{1}{2} \int \frac{dw}{\sqrt{w^2 \; - \; \beta}} \;\; = \;\; \frac{1}{2} \int \frac{\sqrt{\beta}\sec \theta \tan \theta \, d\theta}{\sqrt{\beta \, {\tan}^2 \theta} } \;\; = \;\; \frac{1}{2} \int \sec \theta \,d\theta \;\; = \;\; \frac{1}{2} \ln|\sec \theta + \tan \theta|$$

I’m omitting the details for integrating $$\sec {\theta}$$ because this is so well known. See Ways to evaluate $$\int \sec \theta \, \mathrm d \theta$$ several such methods. By drawing an appropriately labeled right triangle, we find that $$\;\sec \theta = \frac{w}{\sqrt{\beta}}\;$$ and $$\;\tan \theta = \frac{\sqrt{w^2 - \beta}}{\sqrt{\beta}}.\;$$ Thus, we get

$$\frac{1}{2} \int \frac{dw}{\sqrt{w^2 \; - \; \beta}} \;\; = \;\; \frac{1}{2} \ln\left[ \frac{w}{\sqrt{\beta}} + \frac{\sqrt{w^2 - \beta}}{\sqrt{\beta}} \right] \;\; = \;\; \frac{1}{2} \ln \left[ \left(\frac{1}{ \sqrt{\beta} }\right)\left(w + \sqrt{w^2 - \beta}\right) \right]$$

$$= \;\; \frac{1}{2} \ln \left( \frac{1}{\sqrt{\beta}}\right) \;\; + \;\; \frac{1}{2} \ln \left(w + \sqrt{w^2 - \beta}\right)$$

The constant $$\;\frac{1}{2} \ln \left( \frac{1}{\sqrt{\beta}}\right)\;$$ can be absorbed into the integration constant, and hence the integral involving $$R_o(t)$$ evaluates to

$$\frac{1}{2} \int \frac{dw}{\sqrt{w^2 \; - \; \beta}} \;\; = \;\; \frac{1}{2} \, \ln \left(w + \sqrt{w^2 - \beta}\right)$$

where

$$\beta = \frac{9\sqrt{3} - 15}{2} \;\;\; \text{and} \;\;\; w = v + \frac{9 - 4\sqrt{3}}{18} \;\;\; \text{and} \;\;\; v = \frac{1}{u-1} \;\;\; \text{and} \;\;\; u = t^2$$

$$\text{and} \;\;\;\;\;\; t \;\; = \;\; -\left(\frac{x + 4 + 3\sqrt{3}}{x + 4 – 3\sqrt{3}}\right)$$

The last equation is obtained by solving $$\;x \; = \; \frac{(-4 + 3\sqrt{3})t \; + \; (-4 - 3\sqrt{3})}{t+1}\;$$ for $$t$$ in terms of $$x.$$

# 9. Reduction of the Integral Involving $$R_e(t)$$ to Jacobi Standard Forms

The integral involving $$R_e(t)$$ is

$$\frac{1}{2 \sqrt{30 + 18\sqrt{3}}} \cdot \int \frac{R_e(t)\,dt}{ \sqrt{ \left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}}$$

where

$$\alpha \; = \; 6\sqrt{3} - 10 \;\;\;\; \text{and} \;\;\;\; R_e(t) \;\; = \;\; \frac{(18 - 8\sqrt{3})t^2 \; + \; (18 + 8\sqrt{3})}{t^2 - 1}$$

The integral involving $$R_e(t)$$ is the "elliptic function" part of the original integral, in the sense that for an arbitrary quartic polynomial under the square root, the integral involving $$R_o(t)$$ can always be evaluated in terms of elementary functions and the integral involving $$R_e(t)$$ cannot always be evaluated in terms of elementary functions.

Let $$\;t = \sqrt{\frac{8}{\alpha}}\cdot\sqrt{1 - z^2}.\;$$ Then

$$dt \;\; = \;\; \sqrt{\frac{8}{\alpha}} \cdot \frac{1}{2\sqrt{1 - z^2}} \cdot (-2z\,dz) \;\; = \;\; -\sqrt{\frac{8}{\alpha}} \cdot \frac{z \, dz}{\sqrt{1 - z^2}}$$

$$1 + {\alpha}t^2 \;\; = \;\; 1 \; + \; \alpha \left[ \frac{8}{\alpha}(1-z^2) \right] \;\; = \;\; 9 - 8z^2$$

$$1 - \frac{1}{8}\alpha t^2 \;\; = \;\; 1 \; - \; \frac{1}{8}\alpha \left[ \frac{8}{\alpha}(1-z^2) \right] \;\; = \;\; z^2$$

Therefore, we get

$$\frac{dt}{\sqrt{\left( 1 + {\alpha}t^2 \right) \left( 1 - \frac{1}{8}{\alpha}t^2 \right)}} \;\; = \;\; \frac{-\sqrt{\frac{8}{\alpha}} \cdot \frac{z \, dz}{\sqrt{1 - z^2}}}{\sqrt{(9 - 8z^2)(z^2)\;}} \;\; = \;\; -\sqrt{\frac{8}{\alpha}} \cdot \frac{z \, dz}{\sqrt{1 - z^2}} \cdot \frac{1}{z\sqrt{9 - 8z^2}}$$

$$= \;\; -\sqrt{\frac{8}{\alpha}} \cdot \frac{dz}{\sqrt{(1 - z^2)(9 - 8z^2)}} \;\; = \;\; -\sqrt{\frac{8}{\alpha}} \cdot \frac{dz}{\sqrt{(1 - z^2)(9)(1 - \frac{8}{9}z^2)}}$$

$$= \;\; -\sqrt{\frac{8}{\alpha}} \cdot \frac{dz}{3\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}} \;\; = \;\; -\frac{2}{3}\sqrt{\frac{2}{\alpha}} \cdot \frac{dz}{\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

Before continuing, we'll clean up the radical expression $$\;\frac{1}{2 \sqrt{30 + 18\sqrt{3}}} \cdot \frac{2}{3}\sqrt{\frac{2}{\alpha}}\;$$ a little.

$$\frac{\frac{2}{3}\sqrt{\frac{2}{\alpha}}}{2 \sqrt{30 + 18\sqrt{3}}} \;\; = \;\; \frac{\frac{1}{3} \sqrt{\frac{2}{6\sqrt{3} - 10}}}{\sqrt{30 + 18\sqrt{3}}} \;\; = \;\; \frac{1}{3} \sqrt{\frac{1}{(3\sqrt{3} - 5)(30 + 18\sqrt{3})} }$$

$$\frac{1}{3} \sqrt{\frac{1}{90\sqrt{3} + 162 - 150 - 90\sqrt{3}}} \;\; = \;\; \frac{1}{3}\sqrt{\frac{1}{12}} \;\; = \;\; \frac{\sqrt{3}}{18}$$

Thus, the integral involving $$R_e(t)$$ becomes

$$-\frac{\sqrt{3}}{18} \cdot \int \frac{R_e(t) \, dz}{\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

We now rewrite $$\;-\frac{\sqrt{3}}{18} \cdot R_e(t)\;$$ in terms of $$z,$$ which will complete the change of variable from $$t$$ to $$z.$$

Since $$\;t = \sqrt{\frac{8}{\alpha}}\cdot\sqrt{1 - z^2},\;$$ it follows that

$$t^2 \;\; = \;\; \frac{8}{\alpha}(1-z^2) \;\; = \;\; \frac{8}{6\sqrt{3} - 10}(1-z^2) \;\; = \;\; \frac{8(6\sqrt{3} + 10)(1-z^2)}{(6\sqrt{3} - 10)(6\sqrt{3} + 10)}$$

$$= \;\; \frac{8(6\sqrt{3} + 10)(1-z^2)}{108 - 100} \;\; = \;\; (6\sqrt{3} + 10)(1-z^2)$$

Therefore,

$$-\frac{\sqrt{3}}{18} \cdot R_e(t) \;\; = \;\; -\frac{\sqrt{3}}{18} \cdot \left[ \frac{(18 - 8\sqrt{3})t^2 \; + \; (18 + 8\sqrt{3})}{t^2 - 1} \right]$$

$$= \;\; \frac{1}{3} \cdot \left[ \frac{(- 3\sqrt{3} + 4)t^2 - 3\sqrt{3} - 4}{t^2 - 1}\right]$$

$$= \;\; \frac{1}{3} \cdot \left[ \frac{(- 3\sqrt{3} + 4)(6\sqrt{3} + 10)(1-z^2) \; - 3\sqrt{3} - 4}{(6\sqrt{3} + 10)(1-z^2) \; - 1}\right]$$

$$= \;\; \frac{1}{3} \cdot \left[ \frac{(14 + 6\sqrt{3})z^2 \; - \; (18 + 9\sqrt{3})}{-(10 + 6\sqrt{3})z^2 \; + \; (9 + 6\sqrt{3})} \right] \;\; = \;\; \frac{1}{3} \cdot \left[ \frac{\left(\frac{14 + 6\sqrt{3}}{10 + 6\sqrt{3}}\right)z^2 \; - \; \frac{18 + 9\sqrt{3}}{10 + 6\sqrt{3}}}{-z^2 \; + \; \frac{9 + 6\sqrt{3}}{10 + 6\sqrt{3}}} \right]$$

$$= \;\; \frac{1}{3} \cdot \left[ \frac{(3\sqrt{3} - 4)z^2 \; - \; \frac{9\sqrt{3} - 9}{4}}{-z^2 \; + \; \frac{9 - 3\sqrt{3}}{4}} \right] \;\; = \;\; \frac{\left(\frac{3\sqrt{3} - 4}{3}\right)z^2 \; - \; \frac{3\sqrt{3} - 3}{4}}{-z^2 \; + \; \frac{9 - 3\sqrt{3}}{4}}$$

$$= \;\; \frac{\left(\frac{4 - 3\sqrt{3}}{3}\right)z^2 \; + \; \frac{3\sqrt{3} - 3}{4}}{z^2 \; - \; \frac{9 - 3\sqrt{3}}{4}} \;\; = \;\; \frac{4 - 3\sqrt{3}}{3} \;\; + \;\; \frac{\frac{9 - 5\sqrt{3}}{2}}{z^2 \; - \; \frac{9 - 3\sqrt{3}}{4}\;\;}$$

The last equality follows by a short polynomial long division that I'm not going to try to format visually, but I will mention that the remainder term winds up being $$\;\frac{3\sqrt{3} - 3}{4} \; - \; \left(\frac{4 - 3\sqrt{3}}{3}\right) \left(-\frac{9 - 3\sqrt{3}}{4}\right).\;$$

Putting everything together, the integral involving $$R_e(t)$$ becomes

$$\int \frac{-\frac{\sqrt{3}}{18} \cdot R_e(t) \, dz}{\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

$$= \;\;\;\; \frac{4 - 3\sqrt{3}}{3} \cdot \int \frac{dz}{\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

$$+ \;\;\;\; \frac{9 - 5\sqrt{3}}{2} \cdot \int \frac{dz}{\left(z^2 - \frac{9 - 3\sqrt{3}}{4}\right)\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

On the right side of this last equality, the two integrals are in Jacobi (sometimes called Legendre-Jacobi) standard form --- the first integral being of the first kind and the second integral being of the third kind.

A Jacobi third kind integral is sometimes reduced further using a partial fraction expansion.

$$\frac{1}{z^2 \; - \; \frac{9 - 3\sqrt{3}}{4}\;} \;\; = \;\; \frac{A}{z \; + \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}}\;} \;\; + \;\; \frac{B}{z \; - \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}}\;}$$

$$1 \;\; = \;\; A\left(z \; - \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}})\right) \;\; + \;\; B\left(z \; + \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}}\right)$$

In this last equation, substituting $$\;z = \frac{1}{2}\sqrt{9 - 3\sqrt{3}}\;$$ gives

$$1 \;\; = \;\; A\cdot 0 \;\; + \;\; B \left( \sqrt{9 - 3\sqrt{3}} \right) \;\;\; \implies \;\;\; B \; = \; \frac{1}{\sqrt{9 - 3\sqrt{3}}}$$

and substituting $$\;z = -\frac{1}{2}\sqrt{9 - 3\sqrt{3}}\;$$ gives

$$1 \;\; = \;\; A \left(- \sqrt{9 - 3\sqrt{3}} \right) \;\; + \;\; B \cdot 0 \;\;\; \implies \;\;\; A \; = \; \frac{-1}{\sqrt{9 - 3\sqrt{3}}}$$

Now that we have found the constants $$A$$ and $$B$$ for the partial fraction expansion, we incorporate $$\;\frac{9 - 5\sqrt{3}}{2}\;$$ --- the coefficient in front the Jacobi third kind integral --- into these constants. Doing this leads to the radical expression $$\;\frac{9 - 5\sqrt{3}}{2} \cdot \frac{1}{\sqrt{9 - 3\sqrt{3}}},\;$$ which we will now clean up a little.

$$\frac{9 - 5\sqrt{3}}{2} \cdot \frac{1}{\sqrt{9 - 3\sqrt{3}}} \;\; = \;\; \frac{9 - 5\sqrt{3}}{2} \cdot \frac{\sqrt{9 + 3\sqrt{3}}}{\sqrt{(9 - 3\sqrt{3})(9 + 3\sqrt{3})}} \;\; = \;\; \frac{9 - 5\sqrt{3}}{2} \cdot \frac{\sqrt{9 + 3\sqrt{3}}}{\sqrt{81 - 27}}$$

$$= \;\; \frac{(9 - 5\sqrt{3})\sqrt{9 + 3\sqrt{3}}}{6\sqrt{6}} \;\; = \;\; \frac{(9 - 5\sqrt{3})\sqrt{3}\sqrt{3 + \sqrt{3}}}{6\sqrt{2}\sqrt{3}} \;\; = \;\; \frac{(9 - 5\sqrt{3})\sqrt{3 + \sqrt{3}}}{6\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$

$$= \;\; \frac{(9 - 5\sqrt{3})\sqrt{6 + 2\sqrt{3}}}{12}$$

Therefore,

$$\frac{\frac{9 - 5\sqrt{3}}{2}}{\;\;z^2 \; - \; \frac{9 - 3\sqrt{3}}{4}\;\;} \;\; = \;\; \frac{-\frac{1}{12} (9 - 5\sqrt{3})\sqrt{6 + 2\sqrt{3}}} {z \; + \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}}} \;\; + \;\; \frac{\frac{1}{12} (9 - 5\sqrt{3})\sqrt{6 + 2\sqrt{3}}} {z \; - \; \frac{1}{2}\sqrt{9 - 3\sqrt{3}}}$$

Putting everything together, the integral involving $$R_e(t)$$ is equal to $$\;I_{e}^{(1)} + I_{e}^{(2)} + I_{e}^{(3)},\;$$ where

$$I_{e}^{(1)} \;\; = \;\; \frac{4 - 3\sqrt{3}}{3} \cdot \int \frac{dz}{\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

$$I_{e}^{(2)} \;\; = \;\; -\,\frac{(9 - 5\sqrt{3})\sqrt{6 + 2\sqrt{3}}}{12} \cdot \int \frac{dz}{\left(z + \frac{1}{2}\sqrt{9 - 3\sqrt{3}} \right)\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

$$I_{e}^{(3)} \;\; = \;\; \frac{(9 - 5\sqrt{3})\sqrt{6 + 2\sqrt{3}}}{12} \cdot \int \frac{dz}{\left(z - \frac{1}{2}\sqrt{9 - 3\sqrt{3}} \right)\sqrt{(1 - z^2)(1 - \frac{8}{9}z^2)}}$$

$$\text{and} \;\;\;\; z \; = \; \sqrt{1 \; + \; \left(\frac{5 - 3\sqrt{3}}{4}\right)t^2\;} \;\;\;\; \text{and} \;\;\;\;\;\; t \;\; = \;\; -\left(\frac{x + 4 + 3\sqrt{3}}{x + 4 – 3\sqrt{3}}\right)$$

Regarding the expression for $$z$$ in terms of $$t,$$ recall that $$\;t = \sqrt{\frac{8}{\alpha}}\,\sqrt{1 - z^2}.\;$$ Thus, we have $$\;1 - z^2 \; = \; \left(\frac{\alpha}{8}\right)t^2 \; = \; \left(\frac{6\sqrt{3} - 10}{8}\right)t^2 \; = \; \left(\frac{3\sqrt{3} - 5}{4}\right)t^2,\;$$ and hence $$\;z^2 \; = \; 1 - \left(\frac{3\sqrt{3} - 5}{4}\right)t^2 \; = \; 1 + \left(\frac{5 - 3\sqrt{3}}{4}\right)t^2.$$

# 10. Some Additional References (Not Cited Above)

 Elliptic Integral at Wolfram MathWorld

 Traité des Fonctions Elliptiques et des Intégrales Eulériennes [Treatise of Elliptic Functions and Eulerian Integrals], Volume 1, by Adrien Marie Legendre (1825; see Chapter II on pp. 6-8 and Article 143 on pp. 175-176)

 Pafnuty Lvovich Chebyshev, Sur l'Intégration de la différentielle $$\frac{x + A}{\sqrt{x^4 + {\alpha}x^3 + {\beta}x^2 + {\gamma}x + \delta}}\,dx$$ [On the integration of the differential $$\frac{x + A}{\sqrt{x^4 + {\alpha}x^3 + {\beta}x^2 + {\gamma}x + \delta}}\,dx$$], Bulletin de l'Académie Impériale des Sciences de St.-Pétersbourg 3 (1861), 1-12.

For a discussion in English about this paper, see top of p. 15 in On the present state and prospects of some branches of pure mathematics by Henry John Stephen Smith [Proceedings of the London Mathematical Society (1) 8 (1876-1877), pp. 6-29] (reprinted on pp. 175-176 in The Collected Mathematical Papers of Henry John Stephen Smith, Volume II, 1894).

 Elliptic Functions by Arthur Latham Baker (1890; see Chapter I on pp. 4-15)

 An Introduction to the Study of the Elements of the Differential and Integral Calculus by Carl Gustav Axel Harnack (1891 translation of 1881 German edition; see Third Chapter on pp. 197-218, especially Article 125 on pp. 212-218)

 The Applications of Elliptic Functions by George Greenhill (1892; see Articles 70-75 on pp. 59-62)

 The Elementary Properties of the Elliptic Functions by Alfred Cardew Dixon (1894; see parts of Chapter X, which begins on p. 82)

 An Elementary Treatise on Elliptic Functions by Arthur Cayley (1895; see Chapter XII (pp. 311-325)

 Lectures on the Theory of Elliptic Functions by Harris Hancock (1910; see Chapter VIII on 180-205)

 Elliptic Integrals by Harris Hancock (1917; see Articles 1-3 on pp. 9-13 and Article 32 on pp. 52-54)

 Gosta Mittag-Leffler, An introduction to the theory of elliptic functions, Annals of Mathematics (2) 24 #4 (June 1923), 271-351.

See especially approximately the first half, which gives a very nice historical survey. Incidentally, for a very thorough historical treatment, see 1. Elliptic Functions (pp. 15-79) and 4. Complex Functions and Elliptic Integrals (pp. 217-258) in Hidden Harmony---Geometric Fantasies by Umberto Bottazzini and Jeremy John Gray (2013; MAA Review).

 An Introduction to the Theory of Elliptic Functions and Higher Transcendentals by Ganesh Prasad (1928; see First Lecture on pp. 1-12)

 Homer Benjamin Scretchings, Investigations in Theory of Elliptic Integrals, Master of Science thesis, Atlanta University (Atlanta, Georgia), July 1936, ii + 32 pages.

 Howard William McElrath, The elliptic integral and some applications, Master of Arts thesis, Atlanta University (Atlanta, Georgia), January 1932, ii + 58 pages.

 Introduction to Elliptic Functions With Applications by Frank Bowman (1961; see Chapter IX: Reduction to the Standard Form on pp. 86-98)

 Algebraic Technique of Integration by Harris Franklin MacNeish (1952; see Chapter VI. The Integration of Irrational Functions and Chapter VIII. Elliptic Integrals)

 Special Functions, Lecture notes for Math 483 at Missouri University of Science and Technology, by Leon Morris Hall (1995; see 3.3. Evaluation of Elliptic Integrals on pp. 26-29)

 Athanase Papadopoulos, Looking backward: from Euler to Riemann, pp. 1-94 in From Riemann to Differential Geometry and Relativity (2017; see 3. Elliptic Integrals, also on pp. 21-30 of the arXiv:1710.03982 version, for an informative historical overview)

• "I originally intended a briefer version of this to be a comment" and then proceeds to write a paper Oct 24, 2018 at 16:10
• @user438666: FYI, the "... briefer version..." comment refers to the stuff above "(ADDED ABOUT 2 WEEKS LATER)", which was my original answer. The "write a paper" occurred a couple of weeks later as an extensive addition to my original sketchy comments. I spent those two weeks working on it, however. At the time I was between contract jobs and had about 2 weeks free from work, so I decided to do something I've been wanting to do for at least 20 years, which is to carefully go through how one reduces elliptic integrals to elliptic standard forms. Oct 24, 2018 at 16:21
• Don't get me wrong, for what my opinion is worth this is an incredible answer and you putting all of this time and effort out here for free is awesome, thank you Oct 24, 2018 at 21:24
• OP left the chat Dec 3, 2020 at 18:51

The problem is $$\int\frac{x\,\mathrm{d}x}{y}$$ where $$y^2=x^4+10x^2-96x-71$$ is a polynomial with integer coefficients and no double roots. There is a rarely-known “Chebyshev substitution” which decides if integrals like this one are elementary functions.

His criterion is the following: the $$\mathbb{Z}[x]$$-continued fraction expansion of $$y$$ contains a quadratic coefficient after the constant (e.g. $$y=[c_0;c_1,\ldots,c_n,c_{n+1},\ldots]$$ where $$c_0,c_{n+1}$$ are quadratic and $$c_1,\ldots,c_n$$ are linear) iff the integral is an elementary function. In this case “$$w=$$ the numerator of $$[c_0;c_1,\ldots,c_n]$$” is a good substitution.

So we first work out the continued fraction expansion of $$y$$: $$y=x^2+5-\frac{48}{x-1}\mathbin{\raise{-11mu}{+}}\frac{6}{x+5}\mathbin{\raise{-11mu}{+}}\frac{24}{x-4}\mathbin{\raise{-11mu}{-}}\frac{3}{x-4}\mathbin{\raise{-11mu}{+}}\frac{24}{x+5}\mathbin{\raise{-11mu}{+}}\frac{12}{x-1}\mathbin{\raise{-11mu}{-}}\frac{48}{x^2+5+y}.$$ Now $$c_1,\ldots,c_6$$ are linear, but $$c_7$$ is quadratic, so the integral is an elementary function (or, a pseudoelliptic integral). We have $$[c_0;c_1,\ldots,c_6]=\frac{x^8+20x^6-128x^5+54x^4-1408x^3+3124x^2+10001}{x^6+15x^4-80x^3+27x^2-528x+781}=\frac{w}{z}.$$ For these polynomials we observe that $$w^2-y^2z^2=2^{16}\cdot3^7$$ and $$w'=8xz$$, so $$\int\frac{x\,\mathrm{d}x}{y}=\int\frac{xz\,\mathrm{d}x}{yz}=\frac18\int\frac{\mathrm{d}w}{\sqrt{w^2-2^{16}\cdot3^7}}=\frac18\log\left(w+\sqrt{w^2-2^{16}\cdot3^7}\right)+C.$$

This method is found in Chebyshev's paper “On the integration of differentials that contain a square root of a third or fourth degree polynomial”, available in French here.

• Fantastic explanation of an otherwise obscure result. May 16 at 2:09