<s>[INST] Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$ 
Evaluate the following integral:
  $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$

I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$. 
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\frac{2}{t^2+1}$
It go me there: $\int\frac{2}{t^2+1}^2(2-\frac{2}{t^2+1})^{-3/2}-\sqrt{2}t(\frac{1}{t^2+1})^{3/2}dt$ which is just more complicated expression. Where I went wrong?
 [/INST] Hint:
Another way:
Use Trigonometric substitution
As $2-x^2\ge0,$  WLOG $x=\sqrt2\sin t$
$\sqrt{2-x^2}=\sqrt2\cos t, dx=?$
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