Evaluating $\int_0^\infty \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$ So, my friend has challenged me to solve the following integral
$\displaystyle \tag*{} I=\int_0^\infty  \frac{\arctan (x^2) \ln(1+x^4)}{x(x^8+x^4+1)} \mathrm dx$
I started by doing $x^2=t$, so
$\displaystyle \tag{1} I=\frac 12 \int_0^\infty  \frac{\arctan (x) \ln(1+x^2)}{x(x^4+x^2+1)} \mathrm dx$
Now, I first tried to solve this generalized integral:
$\displaystyle \tag*{} \mathcal L(m,n,a) = \int_0^\infty \frac{\arctan(mx)\ln(1+n^2x^2)}{x(a^2+x^2)} \ \mathrm dx$
We have $\mathcal L(0,1,a)=0$ and $\mathcal L(1,0,a)=0$. We now differentiate $\mathcal L(m,n,a)$ w.r.t $m$ first and then w.r.t $n$. We get:
$\displaystyle \tag{2} \begin{align} \partial m \ \partial n  \ \mathcal L(m,n,a) &= \int_0^\infty \frac{2nx^2}{(a^2+x^2)(1+m^2x^2)(1+n^2x^2)} \ \mathrm dx \\\\
 &= \frac{\pi n}{(m+n)(am+1)(an+1)} \\\\
 &= \frac{\pi}{2(am+1)(an+1)} , \ \ \text{via symmetry} \end{align}$
And now taking the double integeral gives
$\displaystyle \tag{3} \mathcal L(1,1,a) = \int_0^1\int_0^1 \frac{\pi}{2(am+1)(an+1)} \ \mathrm dm \ \mathrm dn$
Using the elementary result:
$\displaystyle \tag*{} \int_0^1 \frac{1}{ay+1} \ \mathrm dy = \frac{\log(a+1)}{a}$
We find the value of $(3)$ as
$\displaystyle \tag*{} \mathcal L(1,1,a) = \frac{\pi \log^2(a+1)}{2a^2}$
Finally, we conclude that
$\displaystyle \tag*{}\int_0^\infty \frac{\arctan(x)\ln(1+x^2)}{x(a^2+x^2)} \ \mathrm dx={\color{Red} { \frac{\pi \log^2(a+1)}{2a^2}}}:= {\color{Blue}{ f(a)}}$
Now by completing square and performing, we have:
$\displaystyle \tag*{} \begin{align}\frac{1}{x^4+x^2+1} &=\frac{1}{\left(x^2+\frac 12\right)^2+\frac 34 }\\ &= \frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)} \\ &= \frac{2}{i \sqrt 3} \left(\frac{1}{\left(x^2+\frac 12 - \frac{i \sqrt 3}{2}\right)}-\frac{1}{\left(x^2+\frac 12 + \frac{i \sqrt 3}{2}\right)}\right)\end{align} $
Adding all the pieces together, we get
$\displaystyle \tag{4} I=\frac{1}{i \sqrt 3} \left(f\left(\sqrt{\frac{1-i\sqrt3}{2}}\right) - f\left(\sqrt{\frac{1+i\sqrt3}{2}}\right)\right)$
My question is to simplify $(4)$ (if all my steps are correct) and also is there any short ways (methods others than contour integration) to destroy this integral?
Thanks :).

Note that $a_\pm =\sqrt{\frac{1\pm i\sqrt3}{2}}=\frac{\sqrt3\pm i}2$
and
\begin{align}
 I=&\ \frac 12 \int_0^\infty  \frac{\arctan x\ln(1+x^2)}{x(x^4+x^2+1)} dx\\
=& \ \frac{1}{2\sqrt 3 i} \int_0^\infty \frac{\arctan x\ln(1+x^2)}{x(x^2+a_-^2)}-\frac{\arctan x\ln(1+x^2)}{(x^2+a_+^2)}\ dx\\
= & \ \frac\pi{4\sqrt3 i}\bigg(\frac{\ln^2(a_- +1)}{a_-^2}- \frac{\ln^2(a_+ +1)}{a_+^2} \bigg)\\
=& \ \frac\pi{4\sqrt3 i}\bigg(\frac{\ln(a_- +1)}{a_-}- \frac{\ln(a_++1)}{a_+} \bigg)\bigg(\frac{\ln(a_- +1)}{a_-}+\frac{\ln(a_+ +1)}{a_+} \bigg)\\
=& \ \frac\pi{4\sqrt3 i}\cdot\frac{\sqrt3 \pi -6\ln(2+\sqrt3)}{12i} \cdot \frac{6\sqrt3\ln(2+\sqrt3)+\pi}{12}\\
= &\ \frac\pi{16}\ln^2(2+\sqrt3)-\frac{\pi^2}{48\sqrt3}\ln(2+\sqrt3)-\frac{\pi^3}{576}
\end{align}
