Accidentally while trying to evaluate a similar integral, I think originally found here, I have taken the denominator instead of $x^4+4x^2+1$ as $x^4+x^2+1$ and stumbled into the following integral: $$J=\int_0^\infty \frac{(x^2-1)\arctan(x^2)}{x^4+x^2+1}dx$$ I think this have a closed form because the linked one has a simple closed form: $\displaystyle{\frac{\pi^2}{12\sqrt{2}}}$, also if there is $\arctan x $ instead of $\arctan(x^2) $ then we have: $$\int_0^\infty \frac{\arctan x}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt{3}}-\frac{2}{3}G+\frac{\pi}{12}\ln(2+\sqrt{3})$$
Some proofs are found here: Using $\int_0^{\infty} \frac{\tan^{-1}(x^2)}{x^2+a^2} \ dx$ or using residues. Anyway I started by splitting into two integrals and substituting $x=\frac{1}{t}$: $$\int_0^\infty \frac{x^2\arctan(x^2)}{x^4+x^2+1}dx=\int_0^\infty \frac{\frac{\pi}{2}-\arctan\left(x^2\right)}{x^4+x^2+1}dx\Rightarrow J=\frac{\pi^2}{4\sqrt{3}}-2\int_0^\infty \frac{\arctan(x^2)}{x^4+x^2+1}dx$$ Well, now the main issue is to evaluate: $\displaystyle{I=\int_0^\infty \frac{\arctan(x^2)}{x^4+x^2+1}dx}$
Using the same method as in the second link I arrived at: $$I=\left(\frac{1-i\sqrt 3}{2}\right)f\left(\sqrt{\frac{1+i\sqrt 3}{2}}\right)+\left(\frac{1+i\sqrt 3}{2}\right)f\left(\sqrt{\frac{1-i\sqrt 3}{2}}\right)$$ Where $\displaystyle{f(a)=\int_0^{\infty} \frac{\tan^{-1}(x^2)}{x^2+a^2}=\frac{\pi}{2a}\left(\tan^{-1}(\sqrt{2}a+1)+\tan^{-1}(\sqrt{2}a-1)-\tan^{-1}(a^2)\right)},\,$ but I don't see how to simplify further.
I also tried the "straight forward" way, by employing Feynman's trick to the following integral: $$I(b)=\int_0^\infty \frac{\arctan(bx)^2}{x^4+x^2+1}dx\rightarrow \frac{d}{db}I(b)=\int_0^\infty \frac{2bx^2}{(x^4+x^2+1)(1+b^4x^4)}dx$$ $$=\frac{2b}{b^8-b^4+1}\int_0^\infty \frac{x^2}{x^4+x^2+1}dx-\frac{b^5}{b^8-b^4+1}\int_0^\infty \frac{dx}{x^2+x+1}+$$$$+\frac{2b^5}{b^8-b^4+1}\int_0^\infty\frac{dx}{1+b^4x^4}+\frac{b^9-b^5}{b^8-b^4+1}\int_0^\infty \frac{x^2}{1+b^4x^4}dx$$ $$=\frac{\pi}{\sqrt 3}\frac{b}{b^8-b^4+1}-\frac{2\pi}{3\sqrt 3}\frac{b^5}{b^8-b^4+1}+\frac{\pi}{\sqrt 2}\frac{b^4}{b^8-b^4+1}+\frac{\pi}{2\sqrt 2}\frac{b^6-b^2}{b^8-b^4+1}$$ Now since $I(0)=0$ we have: $ \displaystyle{I=I(1)-I(0)=\int_0^1 \left(\frac{d}{db}I(b)\right)db}$
Integrating the first two parts is okay-ish, but for the last two I have no idea on how to proceed, also it seems that only elementary constants appear thus I believe the integral can be approached in a nicer way. I would love to get some help, if it's possible without using residues since I am not great there.