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One learns evaluating indefinite integrals of the form $$ \int \frac{f(x)}{p(x)\sqrt{q(x)}} \ dx $$ where $p(x)$ and $q(x)$ are linear polynomials.

How to evaluate these integrals if $p(x)$ and $q(x)$ are quadratic or cubic in nature. Is there any method for evaluating them.

Examples of such integrals are: $\displaystyle \int \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx$ or $\displaystyle \int\frac{1}{(x^{3}+a^{3})\sqrt{x^{3}+b^{3}}} \ dx$

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5 Answers

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If $q$ is quadratic, then these can be done by trigonometric/hyperbolic substiuitions. If $q$ is cubic then they are are elliptic integrals, not in general expressible in elementary terms, but expressible in terms of elementary functions and the standard examples of elliptic integrals.

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For Chandru: As a matter of fact, as long as $q$ has degree $\leq 4$, the integral is an elliptic integral; of course, it can happen due to degeneracy that your "elliptic integral" becomes expressible as a logarithm (that may or may not have square roots). –  J. M. Aug 28 '10 at 13:55
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There are algorithms due to Abel, Liouville, Risch, Davenport, Trager et al. for solving the general problem of the algebraic (vs. transcendental) case of integration in finite terms (elementary and algebraic functions). Here's an illuminating excerpt from the introduction of Barry Trager's 1984 MIT Thesis (coincidentally Barry was working right next door to me disproving Hardy's statement that no algorithm exists at the same time I was disproving an analogous Hardy statement while inventing algorithms for computing limits / asymptotics - while we were both student members of the MIT Mathlab Group - the research group that developed the Macsyma computer algebra system).

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I've been meaning to look for a copy of Trager's thesis, so thanks for pointing it out! Too bad it is not a printable version, but I can still use this. –  J. M. Aug 29 '10 at 0:19
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For the quadratic, first put $x = a \tan \theta$ then put $t = \sin \theta$

(First write $p(t)$ in the form $x^2 + a^2$ where $x = t+r$, $a$ could possibly be complex)

Don't know about the cubic.

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Moron: the cubic requires a whole 'nother new set of functions: dlmf.nist.gov/19 ; unless of course your cubic has for instance multiple roots. –  J. M. Aug 28 '10 at 13:58
@J.Mangaldan: I guessed so! Thanks. –  Aryabhata Aug 28 '10 at 14:08
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Here i have found one more method for evaluvating integrals of the form $$ I= \frac{1}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}} \ dx $$

Let $y = \displaystyle \sqrt{\frac{x^{2}+b^{2}}{x^{2}+a^{2}}}$, then by differentiating one can see that $$ \frac{dy}{dx} = \frac{(a^{2}-b^{2})x}{(x^{2}+a^{2})^{3/2} (x^{2}+b^{2})^{1/2}}$$

Thus $I$ becomes $$\frac{(x^{2}+a^{2})^{1/2}}{(a^{2}-b^{2})x} \ dy$$

Also $(x^{2}+a^{2})y^{2}=x^{2}+b^{2}$ so that we have $\displaystyle x^{2}= \frac{b^{2}-a^{2}y^{2}}{y^{2}-1}$ and $$x^{2}+a^{2} = \frac{b^{2}-a^{2}}{y^{2}-1}$$ So $I$ further reduces to $$ I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{b^{2}-a^{2}}}{\sqrt{b^{2}-a^{2}y^{2}}} \ dy = \frac{1}{a\sqrt{b^{2}-a^{2}}} \cos^{-1}\frac{ay}{b} \ (a < b)$$

For $a>b$ one can rearrange $$I = \frac{1}{a^{2}-b^{2}} \int \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{a^{2}y^{2}-b^{2}}} \ dy$$

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@J.M: It should be fine now, i believe. I have altered $y$, where there was a mistake at the beginning –  anonymous Aug 31 '10 at 17:11
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Gradshteyn & Rizhik makes a good discussion about that in chapter 2 ( Indefinite Integrals ).

In particular, they discuss an important method: Ostrogradskiy–Hermite.

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