Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?

$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$

This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.

Edit :

The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
By clever substitutions your professor probably meant Euler Substitutions.

$$\begin{align}
I
&amp;=\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx\tag{1}\\
&amp;=\frac12\int_0^1 \frac{t}{2(2-t)(1+t) + 3\sqrt{(2-t)(1+t)}}\,\mathrm dt\tag{2}\\
&amp;=\frac13\int_{\sqrt2}^{1/\sqrt2} \frac{u^2-2}{(1+u)^2(u^2+1)}\,\mathrm du\tag{3}\\
&amp;=\frac13\left[\int_{\sqrt2}^{1/\sqrt2} \frac{3 u}{2 \left(u^2+1\right)}\,\mathrm du
-\int_{\sqrt2}^{1/\sqrt2} \frac{3}{2 (u+1)}\,\mathrm du
-\int_{\sqrt2}^{1/\sqrt2} \frac{1}{2 (u+1)^2}\,\mathrm du\right]\tag{4}\\
&amp;=\frac13\left[\frac{3}{4}  \log \left(u^2+1\right)-\frac{3}{2} \log (u+1)+\frac{1}{2 (u+1)}\right]_{\sqrt2}^{1/\sqrt2}\tag{5}\\
&amp;=\frac{1}{3}\Bigg[\frac{3-2 \sqrt{2}}{2}\Bigg]\tag{6}\\
\end{align}$$


  $$\large\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx=\frac{1}{6}\left(3-2 \sqrt{2}\right)$$




$\text{Explanations:}$

$(2)$ Substitute $x^2=t\iff2x\,\mathrm dx=\,\mathrm dt$

$(3)$ Using Type $\rm III$ Euler Substitution

$$\small
\sqrt{(2-t)(t+1)}=(t+1)u
\iff t+1=\frac{3}{u^2+1}
\iff t=\frac{2-u^2}{u^2+1}
\iff  \,\mathrm dt=-\frac{6 u}{\left(u^2+1\right)^2}\,\mathrm du$$

$(4)$ Using Partial Fraction Decomposition

$$\small\frac{2-u^2}{(1+u)^2(u^2+1)}=-\frac{3 u}{2 \left(u^2+1\right)}+\frac{3}{2 (u+1)}+\frac{1}{2 (u+1)^2}$$
It's easy. Differentiate 

$$\frac{1}{4 (-1 + 2 x^2)} - \frac{\sqrt{2 + x^2 - x^4}}{6 (-1 + 2 x^2)} - 
 {1\over 4} \log[3 + 2 \sqrt{2 + x^2 - x^4}]$$

To get your integral.