Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$ I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$
My attempt: Letting
$$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$
$$
\begin{aligned}
I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta)
\end{aligned}
$$
$$
I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha
$$
But I can't go further.

Taking integration by parts,
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$
Now
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx
= \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$
and the first integral is easily computed by integration by parts:
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx
= -\log 2 + \frac{\pi}{2}. $$
The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion
\begin{align*}
\log \sec\theta
 = -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right|
&= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\
&= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta),
\end{align*}
we have
\begin{align*}
\int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx
&= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\
&= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,}
\end{align*}
where $G$ is Catalan's constant. Combining altogether, we obtain
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$

Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields

