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 $\dfrac{3-2\sqrt{2}}{6}$.
 A: 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
$$\small\frac{2-u^2}{(1+u)^2(u^2+1)}=-\frac{3 u}{2 \left(u^2+1\right)}+\frac{3}{2 (u+1)}+\frac{1}{2 (u+1)^2}$$
A: 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.
A: Let $x=\sin y$, and note that $(2-x^2)(1+x^2)$ simplifies to $2+\sin^2y\cos^2y$. Then integral becomes
$$I=\int_0^{\pi/2}\frac{\sin^3y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$ 
Now set $y=z-\frac{\pi}{2}$, to observe that 
$$I=I_0=\int_0^{\pi/2}\frac{\cos^3y\sin y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy.$$
With $\cos^3y\sin y+\cos y\sin^3 y=\cos y\sin y$ you obtain (with $2I=I+I_0$)
$$\begin{align}I&=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y\cos y}{2(2+\sin^2y\cos^2y)+3\sqrt{2+\sin^2y\cos^2y}}dy\\
&=\frac{1}{2}\int_0^{\pi/2}\frac{\sin y}{8+\sin^2y+3\sqrt{8+\sin^2y}}dy\end{align}$$
The rest should be manageable. 
Addendum:
$$\begin{align}I&=\int_0^{\pi/2}\frac{2\sin y}{17-\cos 2y+3\sqrt{34-2\cos2y}}dy\\
&=\int_0^{\pi/2}\Big(\frac{3\sec y \tan y}{\sqrt{34-2\cos 2y}}-\frac{1}{2}\sec y \tan y \Big) dy\\
&=0-\lim_{y\rightarrow 0}(-\frac{1}{2}\sec y+\frac{\sqrt{17-\cos 2y}\sec y}{6\sqrt{2}})
\end{align}$$
which results in what we need.
The first part of the second integral is clear. For the the second part note that 
$$\begin{align}\sec y\frac{3 \tan y}{\sqrt{34-2\cos 2y}}&=\sec y \Big( \frac{\sin 2y}{6\sqrt{34-2\cos 2y}}+\frac{\sqrt{34-2\cos 2y}\tan y}{12}\Big)\\
&=\Big(\frac{d \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}}}{dy}\Big)\sec y + \Big(\frac{d \sec y}{dy}\Big) \frac{\sqrt{17-\cos 2y}}{6\sqrt{2}} 
\end{align}$$
A: Let us first substitute $y = x^{2} + 1$. Then 
\begin{align*}
I
&:= \int_{0}^{1} \frac{x^{3}}{2(2-x^{2})(1+x^{2}) + 3\sqrt{(2-x^{2})(1+x^{2})}} \, dx \\
&= \frac{1}{2} \int_{1}^{2} \frac{y - 1}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy \tag{1}
\end{align*}
Using the substitution $y \mapsto 3-y$, it follows that 
$$ I = \frac{1}{2} \int_{1}^{2} \frac{2-y}{2y(3-y) + 3\sqrt{y(3-y)}} \, dy. \tag{2} $$
Thus adding (1) and (2) dividing by 2, we obtain
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}
 = \frac{1}{12} \left[ \frac{3 - 2\sqrt{(3-y)y}}{2y-3} \right]_{1}^{2}
 = \frac{3-2\sqrt{2}}{6}.
\end{align*}
Of course this may not be a comprehensive answer. My original approach utilized a chain of substitutions:
\begin{align*} I
&= \frac{1}{4} \int_{1}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}} \\
&= \frac{1}{2} \int_{3/2}^{2} \frac{dy}{2y(3-y) + 3\sqrt{y(3-y)}}, \qquad (\because \text{ by symmetry}) \\
&= \frac{1}{2} \int_{0}^{1} \frac{dy}{(9-s^{2}) + 3\sqrt{9-s^{2}}}, \qquad (s = 2x-3) \\
&= \frac{1}{6} \int_{0}^{\arcsin(1/3)} \frac{d\theta}{1+\cos\theta}, \qquad (s = 3\sin\theta) \\
&= \frac{1}{6} \int_{0}^{3-2\sqrt{2}} dt, \qquad (t = \tan(\theta/2)) \\
&= \frac{3-2\sqrt{2}}{6}.
\end{align*}
A: Here is what I have so far, so hopefully someone can take it further..
The above integral has a form of
$$
-\frac{1}{8}\int \frac{df}{dx}\frac{dg}{dx}dx
$$
Where
$$
f(x) = \ln\left(1-2x^2\right)\\
g(x) = \ln\left(\frac{2}{3}\sqrt{\left(2-x^2\right)\left(1+x^2\right)} +1\right)
$$
But as always double check the manipulation.
A: If you consider these imaginary parts of elliptic integrals as closed-forms, then
$$\frac{1}{2}-\frac{\sqrt 2}{3}-\frac{\sqrt{2} \log
   \left(\sqrt{2}-1\right)}{8}+\frac{\sqrt{2} \log
   \left(\sqrt{2}+1\right)}{8}-\frac{\sqrt 2}{8}\Im\left(F\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)+\frac{\sqrt{2}}{6}\Im\left(E\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)-\frac{3\sqrt{2}}{8
   }\Im\left(\Pi
   \left(-2;i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right),$$
where $F\left(x\,|\,m\right)$ is the elliptic integral of the first kind, $E\left(x\,|\,m\right)$ is the elliptic integral of the second kind, and $\Pi\left(n; x \, | \, m\right)$ is the elliptic integral of the third kind with parameters $m=k^2$.
A: 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.
