$\newcommand{\al}[1]{\begin{align}#1\end{align}} \renewcommand{\Im}{\operatorname{Im}}$Result:
$$I = \frac{\pi \left[2 \pi +\log \left(-4 \sqrt{17+13 \sqrt{2}}+8 \sqrt{2}+13\right)-4 \tan ^{-1}\left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}+\frac{1}{\sqrt{2}}+1\right)\right]}{8 \sqrt{2}} \\\approx 0.299397.$$
The evaluation of this integral by hand is not as tedious as I thought at first. It turns out I had already considered this integral last week, in a different form.
First substitute $\tan x = t$, as suggested by FDP in the comments. This gives
$$
I = \int_0^\infty dt \frac{t^2 \arctan \left( {\frac{1}{1+t^2}}\right)}{1+t^4}.
$$
Now observe that
$$
\al{
\arctan \left( {\frac{1}{1+t^2}} \right) &= \Im \log(i + 1 + t^2) = \Im \left[\log(i+1) + \log\left(1 + \frac{1-i}{2} t^2\right) \right] \\&= \frac{\pi}{4} + \Im \log\left(1 + \frac{1-i}{2} t^2\right).
}
$$
The integral splits up into a trivial part and a less trivial part. We will now find the latter.
Consider the integral
$$
J(a) = \int_0^\infty dx \frac{x^2}{1+x^4} \log(1+a x^2).
$$
We have $J(0) = 0$, and
$$
\al{
J'(a) &= \int_0^\infty dx \frac{x^4}{(1+x^4)(1+a x^2)}
\\&= \frac{1}{1+a^2} \int_0^\infty dx \left[ \frac{1}{1+ a x^2} + \frac{a x^2 - 1}{1+x^4} \right]
\\&= \frac{\pi /2}{1+a^2}\left[\frac{1}{\sqrt a} + \frac{a-1}{\sqrt2}\right].
}
$$
This is straightforward to integrate with respect to $a$. (For the first term, just substitute $\sqrt a = u$ and use partial fractions.)
The result is
$$
J(a) = \frac{\pi}{2 \sqrt 2} \left[ \log \left(a+\sqrt{2} \sqrt{a}+1\right)+2 \arctan\left(\sqrt{2} \sqrt{a}+1\right) \right]
$$
After plugging in the limit and tedious simplification (see Appendix to this answer), we obtain
$$
\Im J\left(\frac{1-i}{2} \right) = -\frac{\pi}{2 \sqrt{2}}\left\{\operatorname{arccoth}\left[\sqrt{7+5 \sqrt{2}}+\sqrt{2}+1\right] + \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right] \right\}
$$
Putting everything together,
$$
\al{
I &= \frac \pi 4 \int_0^\infty dt \frac{t^2}{1+t^4} + \Im \int_0^\infty dt \frac{t^2 \log\left(1 + \frac{1-i}{2} t^2\right)}{1+t^4}
\\&= \frac{\pi^2}{8 \sqrt 2} + \Im J\left(\frac{1-i}{2} \right)
\\&= \frac{\pi}{2 \sqrt{2}}\left\{ \frac \pi 4-\operatorname{arccoth}\left[\sqrt{7+5 \sqrt{2}}+\sqrt{2}+1\right] - \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right] \right\}.
}
$$
This is numerically equal to the claimed result, which I obtained via a completely different route.
Throughout I have not worried explicitly about choosing the correct branch of $\log$. If this led to any mistakes, please point it out to me.
Appendix
Here I sketch how $\Im J\left(\frac{1-i}{2} \right)$ can be calculated. For example, consider the term
$$
\Im \log\left(1 + \frac{1-i}{2} + \sqrt 2 \sqrt{\frac{1-i}{2}} \right) = \Im \log\left(\frac 3 2 - \frac i 2 + 2^{1/4}\left(\cos \frac \pi 8 - i \sin \frac \pi 8 \right) \right).
$$
Using the half-angle formulas for sine and cosine, we can write it as
$$
\al{
-\arctan \left(\frac{1+ \sqrt[4]{2} \sqrt{2-\sqrt{2}}}{3 + \sqrt[4]{2} \sqrt{2+\sqrt{2}}}\right) &= - \arctan\left(\frac{2 \sqrt{5 \sqrt{2}-7}+1}{7-2 \sqrt{2}}\right)
\\&= - \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right].
}
$$
To obtain the first equality, multiply numerator and denominator by a certain factor to get rid of the fourth roots. For the second, multiply by another factor to get rid of all the roots. Of course this simplification is just for aesthetics.
Moral: introduce a parameter in such a way that the integral with respect to the parameter becomes simple, and use Mathematica to simplify the end result.
