Integral $\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16\sqrt 2}$

Following Mhenni's suggestion, I will calculate $$I(\mu) = \int_0^{\infty} x^{\mu}\frac{1+x^2}{1+x^4} \,dx $$ and then take $I''(0)$. By the ubiquitous formula $$\int_0^{\infty}\frac{x^a}{1+x^b} \,dx =\frac{\pi}{b \sin(\pi(a+1)/b)}$$ we obtain $$I(\mu)=\frac{\pi}{4} \left[ \frac{1}{\sin(\pi(\mu+1)/4)} + \frac{1}{\sin(\pi(\mu+3)/4)} \right]$$ This gives (after some simple algebra)

$$I''(0) = \int_0^{\infty} \frac{1+x^2}{1+x^4} \log^2 x \,dx = \frac{\pi}{4}\left[\frac{3 \pi ^2}{8 \sqrt{2}}+\frac{3 \pi ^2}{8 \sqrt{2}} \right] = \frac{3 \pi ^3}{16 \sqrt{2}}$$ as was to be shown.

A related problem. Recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x)dx \implies F''(s)=\int_{0}^{\infty} \ln(x)^2x^{s-1}f(x)dx .$$

Now the whole problem boils down to finding the Mellin transform of $\frac{1+x^2}{1+x^4}$, differentiating twice, and then taking the limit as $s \to 1$.

Can you finish it?

Here is another way to calculate. It is much simpler. Clearly \begin{eqnarray*} I&=&2\int_0^1 \log^2 x\frac{1+x^2}{1+x^4}dx\\ &=&2\int_0^1\sum_{n=0}^\infty(1+x^2)(-1)^nx^{4n}\log^2xdx\\ &=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&4\sum_{n=0}^\infty(-1)^n(\frac{1}{(4n+1)^3}+\frac{1}{(4n+3)^3})\\ &=&4\sum_{n=-\infty}^\infty(-1)^n\frac{1}{(4n+1)^3}\\ &=&\frac{3\pi^3}{16\sqrt2} \end{eqnarray*} Here we use the following theorem $$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$ where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(4z+1)^3}$, $z=-\frac{1}{4}$ is the only pole and $$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{4})=-\frac{3\pi^2}{64\sqrt2}. $$ Thus we have the result.