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I can't figure out how to integrate

$$\int \sqrt{1+x^{\frac{3}{2}}} \operatorname d x$$

I've tried substitution by letting $u = x^3$, but it didn't go anywhere. I also tried to integrate using a trigonometric substitution, but that also got me nowhere. Then I tried Wolfram Alpha, and it got me even nowhere-er!

If you could give me a hint as to where to go, I'll try to answer this question at a later time. Thanks!

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@cable What happened to the limits on the integration? – Srivatsan Sep 20 '11 at 4:44
I took them out of the paragraph... – The Chaz 2.0 Sep 20 '11 at 4:49
Possibly relevant: If you make the substitution $u=\sqrt{x}$ you get an elliptic integral, and there is another question on M.SE on when they have closed forms in terms of elementary functions. – anon Sep 20 '11 at 4:50
We haven't learned what an elliptic integral is yet. I'm sure, from what you say, I could solve it using that method, but there has to be a way to solve it using other methods. Otherwise, why assign it? :/ – Caleb Jares Sep 20 '11 at 5:06
@cable729: My point in posting the links is that often integrals just like these don't have closed forms. It's possible your teacher made a typographical error, it's possible your teacher is some kind of sadist that thinks trick questions are a learning experience, and it's also possible (haven't really thought about it much) that there's some substitutions that take us to elementary functions but no one's posted them yet. – anon Sep 20 '11 at 6:29
up vote 16 down vote accepted

The preliminary substitution $x=u^2$ yields the integral

$$2\int u\sqrt{1+u^3}\mathrm du$$

This is indeed in the form of an elliptic integral, where the cubic under the square root factorizes as $(u+1)(u^2-u+1)$.

For handling integrals like these, one first performs a preliminary Möbius substitution. In this case, we let


to give

$$12\int\frac{(1-\sqrt{3}+(1+\sqrt{3})v)}{(1+v)^5}\sqrt{(1-v^2)\left(2\sqrt{3}-3+(2\sqrt{3}+3\right) v^2)}\mathrm dv$$

We can now use the Jacobian elliptic functions. Letting $v=\mathrm{cn}(w|m)$, $\mathrm dv=-\mathrm{sn}(w|m)\mathrm{dn}(w|m)\mathrm dw$ and using the Pythagorean formula $\mathrm{sn}^2(w|m)+\mathrm{cn}^2(w|m)=1$ yields

$$-12\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\sqrt{2\sqrt{3}-3+(2\sqrt{3}+3)\mathrm{cn}^2(w|m)}\mathrm{sn}^2(w|m)\mathrm{dn}(w|m)\mathrm dw$$

We perform a further application of the Pythagorean formula, and factor out a constant:

$$-24\sqrt[4]{3}\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\sqrt{1-\frac{2+\sqrt{3}}{4}\mathrm{sn}^2(w|m)}\mathrm{sn}^2(w|m)\mathrm{dn}(w|m)\mathrm dw$$

From that, if we let $m=\dfrac{2+\sqrt{3}}{4}$, we can then exploit the identity $\mathrm{dn}^2(w|m)+m\,\mathrm{sn}^2(w|m)=1$:

$$-24\sqrt[4]{3}\int\frac{(1-\sqrt{3}+(1+\sqrt{3})\mathrm{cn}(w|m))}{(1+\mathrm{cn}(w|m))^5}\mathrm{sn}^2(w|m)\mathrm{dn}^2(w|m)\;\mathrm dw$$

We can express everything in terms of $\mathrm{cn}$; using the Pythagorean relations and splitting into partial fractions yields


where for brevity

$$\varrho_k=\int\frac{\mathrm dw}{(1+\mathrm{cn}(w\mid m))^k}$$

Evaluating $\varrho_k$ is algebraically a rather complicated affair; for brevity, I will instead refer you to formulae 341.52-55 in Byrd and Friedman, where a recursion relation is listed. The required members are:

$$\begin{align*}\varrho_1&=w-\varepsilon\left(w \mid \frac{2+\sqrt{3}}{4}\right)+\frac{\mathrm{sn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)\mathrm{dn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)}{1+\mathrm{cn}\left(w \mid \frac{2+\sqrt{3}}{4}\right)}\\\varrho_2&=\frac13\left((3+\sqrt{3})\varrho_1-\frac12(2+\sqrt{3})w+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^2}\right)\\\varrho_3&=\frac15\left(2\left(3+\sqrt{3}\right)\varrho_2-\frac32(2+\sqrt{3})\varrho_1+\frac{2+\sqrt{3}}{4}w+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^3}\right)\\\varrho_4&=\frac17\left(3\left(3+\sqrt{3}\right)\varrho_3-\frac52\left(2+\sqrt{3}\right)\varrho_2+\frac{2+\sqrt{3}}{2}\varrho_1+\frac{\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^4}\right)\end{align*}$$

where $\varepsilon(w|m)=E(\mathrm{am}(w|m)|m)$ is the Jacobi epsilon function. After much algebra and tears, we end up with

$$\begin{split}\frac{\sqrt[4]{3}}{7} \left((3-\sqrt{3})w-6\,\varepsilon\left(w \mid \frac{2+\sqrt{3}}{4}\right)+\frac{2\,\mathrm{sn}\left(w\mid \frac{2+\sqrt{3}}{4}\right) \mathrm{dn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)}{\left(1+\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)\right)^4}\left(3\mathrm{cn}^3\left(w\mid \frac{2+\sqrt{3}}{4}\right)+\right.\right. \\ \left.\left.(21+8\sqrt{3})\mathrm{cn}^2\left(w\mid \frac{2+\sqrt{3}}{4}\right)+(9-8\sqrt{3})\mathrm{cn}\left(w\mid \frac{2+\sqrt{3}}{4}\right)+8 \sqrt{3}-9\right)\right)\end{split}$$

Undoing the substitutions yields

$$\begin{split}\frac2{7\sqrt[4]{3}}\left(\frac{2\sqrt[4]{3}\sqrt{1+x^{3/2}}(3+x(1+\sqrt{3}+\sqrt{x}))}{1+\sqrt{3}+\sqrt{x}}-6\sqrt{3} E\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt{x}}-1\right)\mid \frac{2+\sqrt{3}}{4}\right)\right. \\ \left.+3(\sqrt{3}-1)F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\sqrt{x}}-1\right)\mid \frac{2+\sqrt{3}}{4}\right)\right)\end{split}$$

You can check yourself that the derivative of this expression is $\sqrt{1+x^\frac32}$.

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Holy smoke! That was a nasty homework assignment... ;-) – Hans Lundmark Sep 20 '11 at 17:03
Indeed, I'm not at my Mathematica computer, so I needed four letter-size sheets and about two hours of writing, me being a slow writer and all... :D at least an explicit elliptic representation is now here for posterity. – J. M. Sep 20 '11 at 17:07

Let $I = \int \sqrt{1 + x^{3/2}} \, \mathrm{d} x$. Integrating by parts: $$ \begin{eqnarray} I &=& x \sqrt{1 + x^{3/2}} - \int x \frac{3}{4} \, \frac{\sqrt{x}}{\sqrt{1+x^{3/2}}} \, \mathrm{d} x = x \sqrt{1 + x^{3/2}} - \frac{3}{4} \int \sqrt{1+x^{3/2}} \, \mathrm{d} x + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}} \\ &=& -\frac{3}{4} I + x \sqrt{1 + x^{3/2}} + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}} \end{eqnarray} $$ Thus $$ I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3}{7} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}} $$ The latter integral is not elementary and can be evaluated in terms of Gauss hypergeometric function: $$ I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3 x}{7} \, {}_2F_1\left(\frac{1}{2}, \frac{2}{3} ; \frac{5}{3} ; -x^{3/2} \right) $$

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Let's rearrange the integral:

$I=\int \sqrt{1^2+(x^\frac{3}{4})^2} dx$ , and make the substitution $x^\frac{3}{4}=t \Rightarrow \frac{3}{4}x^\frac{-1}{4} dx=dt$ ,

Since $x=t^\frac{4}{3} \Rightarrow dx=\frac{4}{3}t^\frac{1}{3} dt$; if we substitute this into integral we get the following:

$I=\frac{4}{3}\int \sqrt {(1^2+t^2)} t^\frac{1}{3} dt$,

This integral can be solved using integration by parts where

$u=t^\frac{1}{3}$ and $dv=\sqrt {1^2+t^2} dt$, and the integral $v=\int\sqrt {1^2+t^2} dt$ can be solved by applying the first formula from this list.

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This doesn't exactly lead to anything simpler... – Hans Lundmark Sep 20 '11 at 8:16
@HansLundmark,for the last integral one can use substitution $t=\tan{\alpha}$ and identity $1+(\tan{\alpha})^2=(\sec{\alpha})^2$ , in order to make simpler expression... – pedja Sep 20 '11 at 9:03
@HansLundmark,I have visited your you were in Novi Sad last year...that's town where I studied calculus :) – pedja Sep 20 '11 at 9:29
Dobar dan! It's a small world. :-) What I meant with my comment was that $3I/4=\int u \, dv = uv - \int v \, du$, and I don't see how to get anywhere with $\int v \, du$, since $v$ will be a fairly complicated expression and $du$ will still involve the cube root of $t$. – Hans Lundmark Sep 20 '11 at 11:37
It's not computing $v$ that's worrying me. It's the integral $\int v \, du$ in the next step that looks hopeless. – Hans Lundmark Sep 20 '11 at 13:06

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