TITLE: How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
QUESTION [6 upvotes]: 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

REPLY [3 votes]: 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.