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When I encounter an integral being a rational function of x and a radical $\sqrt {ax^2+bx+c}$, I am used to employ the Euler substitution:

https://en.wikipedia.org/wiki/Euler_substitution

However, today I have encountered the following integral

$\displaystyle I = \int \frac{xdx}{\sqrt{1-x^2}(1-a^2x^2)^{3/2}} $

that contains two different radicals. At first I have searched in my books, and found it looks like an elliptic integral of the third kind, except for the $x$ at the numerator. As I almost know nothing about elliptic integrals, I have entered this integral in Wolfram Alpha, that finds:

$\displaystyle I = \frac {\sqrt{1-x^2}} {(a^2-1)\sqrt{1-a^2x^2}} $

Differentiation shows that this is the correct result. There is no elliptic function inside. Is this simple result the outcome of many calculations with many simplifications, or is it the consequence of some simple manipulations I have not seen (I have tried integration by parts, without much success)?

Thanks in advance for your help.

Edition: Solution following Lucian indications

  • At least I should have seen that noting $X=x^2$ we get a form containing a single radical $\sqrt {aX^2+bX+c}$ allowing to use Euler substitution.
  • But thanks to Lucian I have found another easier method: simply note $X=1/\sqrt{1-x^2}$. It leads to:

$\displaystyle I=\int \frac {XdX} {(a^2+(1-a^2)X^2)^{3/2}}$

which is easily integrated.

1 Answers 1

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It looks like an elliptic integral of the third kind, except for the x at the numerator.

Before jumping to any conclusions about encounters of the third kind, please take a deep breath, and a step back, and recall the fact that the expression of $\displaystyle\int xe^{x^2}~dx$ also looks suspiciously close to that of an error function, but, following a simple substitution, it can be shown to be elementary. Now, can you tell me what that simple substitution is? And, more importantly, can something similar $($or perhaps even identical$)$ also be employed here ? ;-$)$

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    Thanks for your help. I have edited my question to give the solution. I looked for something too much difficult. In fact I did not examine the integral enough close (so it was not a close encounter of the third kind). Thanks for the suggestion of the easy way (with $X=1/\sqrt(1-x^2)$); without it I would have applied the Euler substitution mechanically. – 2017-01-22