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.

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

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

$\text{Explanations:}$

$(2)$ 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

$$\small\frac{2-u^2}{(1+u)^2(u^2+1)}=-\frac{3 u}{2 \left(u^2+1\right)}+\frac{3}{2 (u+1)}+\frac{1}{2 (u+1)^2}$$

Perhaps the most promising approach is to rewrite the term $(2-x^2)(1+x^2)$ as follows:

$$(2 - x^2)(1 + x^2) = 2 + x^2 - x^4 = \frac{9}{4} - \left(x^2 - \frac{1}{2}\right)^2 = \frac{9}{4}\left(1 - \frac{4}{9}\left(x^2 - \frac{1}{2}\right)^2 \right)$$

This observation leads one to consider the substitution $y = \frac{2}{3}\left(x^2-\frac{1}{2}\right)$.

Result of the substitution is that the denominator cleans up considerably, becoming $(1 - y^2) + \sqrt{1 - y^2}$. Later on in the evaluation setting $y = \sin t$ may well be a promising idea.