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?
 A: Hint:
Another way:
Use Trigonometric substitution
As $2-x^2\ge0,$  WLOG $x=\sqrt2\sin t$
$\sqrt{2-x^2}=\sqrt2\cos t, dx=?$
A: Another way:
Integrate by parts
$$\int\dfrac{x^4}{(2-x^2)^{3/2}}dx=\int x^3\cdot\dfrac x{(2-x^2)^{3/2}}dx$$
$$=x^3\int\dfrac x{(2-x^2)^{3/2}}dx-\int\left[\dfrac{d(x^3)}{dx}\int\dfrac x{(2-x^2)^{3/2}}dx\right]dx$$
Finally, $$\dfrac{x^2}{\sqrt{2-x^2}}=\dfrac{x^2-2+2}{\sqrt{2-x^2}}=\dfrac2{\sqrt{2-x^2}}-\sqrt{2-x^2}$$
Use $\#1,\#8$ of INTEGRALS CONTAINING THE SQUARE ROOT OF $a^2-x^2$
A: $\int \frac{x^4}{(2-x^2)^{3/2}}dx$
Comparing with $\int x^m (a+bx^n)^p dx$
$m=4,a=2,b=-1,n=2,p=-\frac{3}{2}$
Here $\frac{m+1}{n}+p=\frac{5}{2}-\frac{3}{2}=1$, integer.
So we first transform the integral $ x^m (a+bx^n)^p $ by factoring $x^n$ i.e., by $x^2$ in this way
$\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int {x^4}{(2-x^2)^{-\frac{3}{2}}}dx=\int {x^4}{(x^2)^{-\frac{3}{2}}}{(2x^{-2}-1)^{-\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx$
Putting $2x^{-2}-1=z$ give $-4x dx =dz\implies x dx=-\frac{1}{4}dz$
Now $\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx=-\frac{1}{4}\int {z^{-\frac{3}{2}}}dz=-\frac{1}{4}\frac{z^{-\frac{1}{2}}}{-\frac{1}{2}}+c=\frac{1}{2} {(2x^{-2}-1)}^{-\frac{1}{2}}+c$
