How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$ 
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$

I use computer have this  reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{2}\dfrac{\pi\ln^2{2}}{\sqrt{2}}-\dfrac{3}{2}\dfrac{\pi^2\ln{2}}{\sqrt{2}}$$
where $C$ is Catalan constant.
My idea: let $$\tan{x}=t,\sin{2x}=\dfrac{2t}{1+t^2},dx=\dfrac{1}{1+x^2}$$
then
$$I=\int_{0}^{\infty}\ln^2{(1+t^4)}\dfrac{\dfrac{2}{1+t^2}}{2-\left(\dfrac{2t}{1+t^2}\right)^2}\cdot\dfrac{dt}{1+t^2}=\int_{0}^{\infty}\dfrac{\ln^2{(1+t^4)}}{(t^2-1)^2}dt$$
Then I can't.Thank you 
 A: Here is an approach. I'll continue from the step you have already reached with the correction suggested in Fabien comment which is considering the integral

$$ I = \int_{0}^{\infty} \frac{\ln^2(1+x^4)}{1+x^4} dx. $$

To evaluate the above integral we consider the integral
$$ F = \int_{0}^{\infty} (1+x^4)^{\alpha} dx = \frac{\pi}{2\sqrt{2}\Gamma( 3/4 )}{\frac {\Gamma  ( -1/4-\alpha ) }{
\Gamma  \left( -\alpha \right)  }},$$
which can be evaluated using the beta function techniques (use the substitution $(1+x^4)=\frac{1}{t}$). Now $I$ follows from $F$ as (see related techniques )

$$ I = \lim_{\alpha \to -1}\frac{d^2F(\alpha)}{d\alpha^2}=\frac{\pi \,\sqrt {2}}{48}\left({\pi}^{2}-36\,\pi \,\ln  \left( 2 \right) +108 \ln^2( 2 )+12\,\psi'\left( 3/4 \right)  \right) .$$

A: Hint: $\quad~I(n)~=~\displaystyle\int_0^\infty\Big(1+x^k\Big)^n~dx\quad~=>\quad~I''(-1)~=~\displaystyle\int_0^\infty\frac{\ln^2\big(1+x^k\big)}{1+x^k}~dx.~$ But, at 
the same time, $I(n)$ can be shown to equal $~\dfrac1k\cdot B\bigg(\dfrac1k~,-\dfrac1k-n\bigg),~$ with the help of the simple 
substitution $t=\dfrac1{1+x^k}$ . Then, expressing it in terms of the $\Gamma$ function, differentiating, and 
using the relationship between the polygamma function $\psi_0$ and harmonic numbers, along with 
Euler's formula for their generalization to non-natural arguments, we are thus finally able to 
arrive at an expression whose only “mysterious” quantity is $\psi_{_1}\Big(\frac34\Big).~$ Since this function has 
been studied for several centuries, I'm sure you will be able to find its closed form expression 
somewhere, along with the proof behind it. Also, Euler's reflection formula for the $\Gamma$ function 
might be of some assistance.
A: After performing the initial arctangent substitution, the resultant integral may be rewritten as the second derivative of a beta function and evaluated accordingly:
$$\begin{align}
I
&=\int_{0}^{\frac{\pi}{2}}\frac{2\cos^2{\theta}}{2-\sin^2{(2\theta)}}\log^2{\left(1+\tan^4{\theta}\right)}\,\mathrm{d}\theta\\
&=\int_{0}^{\infty}\frac{\log^2{(1+x^4)}}{1+x^4}\mathrm{d}x\\
&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\mathrm{d}x}{(1+x^4)^{\alpha}}\\
&=\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\int_{0}^{\infty}\frac{\frac14u^{-\frac34}}{(1+u)^{\alpha}}\mathrm{d}u\\
&=\frac14\frac{d^2}{d\alpha^2}\bigg{|}_{\alpha=1}\operatorname{B}{\left(\frac14,\alpha-\frac14\right)}\\
&=\frac14\Gamma{\left(\frac14\right)}\Gamma{\left(\frac34\right)}\left[-\zeta{(2)}+\left(\gamma+\Psi{\left(\frac34\right)}\right)^2+\Psi^\prime{\left(\frac34\right)}\right]\\
&=\frac{\sqrt{2}\,\pi}{4}\left[-\frac{\pi^2}{6}+\left(\frac{\pi}{2}-\log{8}\right)^2+\pi^2-8C\right]\\
&=-2\sqrt{2}\,\pi\,C+\frac{13\pi^3}{24\sqrt{2}}+\frac{\pi\log^2{8}}{2\sqrt{2}}-\frac{\pi^2\log{8}}{2\sqrt{2}}.~~~\blacksquare
\end{align}$$
A: You have made a mistake with the denominator, which cannot be zero. See, $2-sin(2x)^2$ is always bigger or equal than $1$.
I think that the mistake was in expanding $2-(\frac{2t}{1+t^2})^2$: you have changed that initial $2$ for a $1$ and then completed the square.  Repeat that expansion slowly.
