<s>[INST] 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 
 [/INST] 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 )

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

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