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

An elliptic integral (of the first kind) is one of the form
$$
F(t\mid k^2)=\int\frac{1}{\sqrt{1-k^2\sin^2t}}\,dt
$$
or by substituting $x=\sin t$
$$
G(x\mid k^2)=\int\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\,dx
$$
In our case, the integral 
$$
\int\frac{1}{\sqrt{1-x^4}}\,dx
$$
is $G(x\mid-1)$.
