Evaluating $\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$

Question

How to evaluate the integral

$$\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$$$a<b.$ I posted a similar question here.

Thanks in advance.

Answer 1

Hint:

$\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)x^2~dx$

$=\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)d\left(\dfrac{x^3}{3}\right)$

$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{x^3}{3}d\left(\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)\right)$

$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{2(a^2+b^2)x^4}{3(x^2-b^2)\sqrt{(x^2-b^2)^2-(x^2+a^2)^2}}dx$ (according to http://www.wolframalpha.com/input/?i=d%2Fdx%28arccos%28%28x%5E2%2Ba%5E2%29%2F%28x%5E2-b%5E2%29%29%29)

$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{2(a^2+b^2)x^4}{3(x^2-b^2)\sqrt{b^4-a^4-2(a^2+b^2)x^2}}dx$