Solve analytically integral $\int\frac{y^4}{\sqrt{(1-y^2)(y^2-r^2)}}dy$

Question

I tried to solving $\int\frac{y^4}{\sqrt{(1-y^2)(y^2-r^2)}}dy$ analytically, someone has any ideas?

Answer 1

If you assume $0 < r < 1$ $$I(y)=\int\frac{y^4}{\sqrt{(1-y^2)(y^2-r^2)}}dy$$ $$I(y)=-\frac{1}{3} y \sqrt{\left(1-y^2\right) (y^2-r^2)}+\frac{1}{3} i \left(\left(r^2+2\right) F\left(\sin ^{-1}\left(\frac{y}{r}\right)|r^2\right)-2 \left(r^2+1\right) E\left(\sin ^{-1}\left(\frac{y}{r}\right)|r^2\right)\right)$$ This makes $$J(r)=\int_r^1 \frac{y^4}{\sqrt{(1-y^2)(y^2-r^2)}}dy=$$ $$\frac{i}{3} \left(\frac{\left(r^2+2\right)}{r} \left(K\left(\frac{1}{r^2}\right)-r K\left(r^2\right)\right)+2 \left(r^2+1\right)\left( E\left(r^2\right)- E\left(\csc ^{-1}(r)|r^2\right)\right)\right)$$ which, for sure,is a real $(\frac 23 \leq J(r) \leq \frac \pi 2)$.