How to evaluate $\int_0^{\infty}\frac{\arctan{x}}{x^{1/4}(1+x)}\mathrm{d}x$? I want to evaluate
$ \displaystyle  \int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } \mathrm{d}x \tag*{} $
I know that it equals
$\displaystyle \frac{\pi^2}{2 \sqrt{2} } + \frac{\pi}{\sqrt{2}} \ln \left( \frac{2+\sqrt{2-\sqrt{2}} }{2-\sqrt{2-\sqrt{2}}}\right) \tag*{} $
But I can't seem to find a way to prove it.
After substituting $x=t^4$ , the integral is transformed to
$ \displaystyle I= 4\int_{0}^{\infty} \frac{t^2 \arctan(t^4) }{1+t^4} \mathrm{d}t \tag*{} $
After this I tried differentiating under the integral, but the resulting integrand is very complicated, and I couldn't make much progress futher on.
I'm not too familiar with contour integration, maybe that can help here?

Let $I(a)=\int_{0}^{\infty} \frac{\arctan{(ax)}}{x^{1/4}(1+x) } dx $ and evaluate
\begin{align}
I’(a)=& \int_{0}^{\infty} \frac{x^{3/4}}{(1+x)(1+a^2 x^2) } dx \\
=& \>\frac1{1+a^2}\int_0^\infty \left(\frac{a^2 x^{3/4}}{1+a^2 x^2} + \frac{x^{-1/4}}{1+a^2 x^2} - \frac{x^{-1/4}}{1+x}\right)dx\\
 =& \>\frac1{1+a^2}\left(\frac\pi2 \csc\frac\pi8\>a^{1/4}+ \frac\pi2 \sec\frac\pi8\>a^{-3/4} - \sqrt2\pi \right)dx\\
\end{align}
Then
\begin{align}
\int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } dx 
=& \>I(1) =\int_0^1\overset{a=y^4}{I’(a)\>da}\\
= &\>2\pi\int_0^1 \frac{\csc\frac\pi8 \>y^4 +\sec\frac\pi8 }{1+y^8}\>dy- \frac{\pi^2}{2\sqrt2}\\
 = &\>2\pi\cdot \left(\frac\pi{2\sqrt2}-\frac1{\sqrt2}\ln \tan\frac{3\pi}{16}\right)- \frac{\pi^2}{2\sqrt2}\\
= &\> \frac{\pi^2}{2\sqrt2} -\sqrt2\pi\ln \tan\frac{3\pi}{16} \
\end{align}
