<s>[INST] Evaluation of Irrational Integral 
Evaluation of $$\int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx$$

$\bf{My\; Try::}$ Let $$I = \int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx = -\frac{1}{4}\int x\cdot \frac{-4x^3}{(1-x^{4})^{\frac{3}{2}}}dx$$
Using Integration by parts, We get
$$I =\frac{x}{2(1-x^4)^{\frac{1}{2}}}-\int\frac{1}{(1-x^4)^{\frac{1}{2}}}dx$$
Now How can I solve after that, Help required, Thanks
 [/INST] $$
\int\frac{1}{\sqrt{1-x^4}}\,dx=\int\frac{1}{\sqrt{1+x^2}}\frac{1}{\sqrt{1-x^2}}\,dx
$$
Let $u=\arcsin x$. Then $du=1/\sqrt{1-x^2}\,dx$, and we get
$$
\int\frac{1}{\sqrt{1+\sin^2 u}}\,du=F(-1\,|\, u)+C=F(-1\,|\,\arcsin x)+C
$$
where $F$ denotes an elliptic integral of first kind. Also, see wiki for notational variations.
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