Integral $\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx$ 
Evaluate the integral $$P=\int_0^{\pi/2}x\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)dx.$$

Context:
I started trying to evaluate the integral $$J=\int_0^\infty \frac{\arctan(x)^2}{x^2+x+1}dx,$$
and the integral $P$ is part of the process. At first, I tried $x\mapsto 1/x$, but it just ended up showing that $$J=\frac{\pi^2}{4}\int_0^\infty \frac{dx}{x^2+x+1}-\pi\int_0^\infty \frac{\arctan x}{x^2+x+1}dx+J,$$ 
which is of no use. Next, I tried integration by parts, using 
$$\int\frac{dx}{x^2+x+1}=\frac{2}{\sqrt3}\arctan\frac{2x+1}{\sqrt3},$$
so that 
$$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}\int_0^\infty\arctan(x)\arctan\left(\tfrac1{\sqrt3}+\tfrac{2}{\sqrt3}x\right)\frac{dx}{1+x^2}.$$
Then with $x\mapsto \tan x$ we have $$J=\frac{\pi^3}{4\sqrt3}-\frac{4}{\sqrt3}P.$$
Theoretically, integration by parts if possible from this point, as Wolfram provides an awful closed form for the anti-derivative of $\arctan\left(\tfrac{1}{\sqrt3}+\tfrac{2}{\sqrt3}\tan x\right)$, but I do not think this is really that realistic of an approach. Is there a better way to evaluate the integral $P$? 
 A: $$J=\int_0^\infty \frac{\arctan^2 x}{1+x+x^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2}\frac{t^2}{1+\sin t\cos t}dt\overset{2t=\frac{\pi}{2}-x}=\frac14\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{\left(\frac{\pi}{2}-x\right)^2}{2+\cos x}dx$$
$$=\frac12\int_0^\frac{\pi}{2}\frac{\frac{\pi^2}{4}+x^2}{2+\cos x}dx=\frac{\pi^3}{24\sqrt 3}+\frac12\int_0^\frac{\pi}{2}\frac{x^2}{2+\cos x}dx$$
Now we will use the following Fourier series (refer to this thread):
$$\frac{1}{2+\cos x}=\frac{1}{\sqrt 3}+\frac{2}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\cos(nx)$$
$$\Rightarrow J=\frac{\pi^3}{16\sqrt 3}+\frac{1}{\sqrt 3}\sum_{n=1}^\infty (-1)^n(2-\sqrt 3)^n\int_0^\frac{\pi}{2}x^2\cos(nx)dx$$
$$=\frac{\pi^3}{16\sqrt 3}+\frac{\pi^2}{4\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n}$$
$$+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\cos\left(\frac{n\pi}{2}\right)}{n^2}-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^n\sin\left(\frac{n\pi}{2}\right)}{n^3}$$
$$\small =\frac{\pi^3}{16\sqrt 3}-\frac{\pi^2}{4\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{2n+1}+\frac{\pi}{\sqrt 3}\sum_{n=1}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n}}{(2n)^2}+\frac{2}{\sqrt 3}\sum_{n=0}^\infty \frac{(-1)^n(2-\sqrt 3)^{2n+1}}{(2n+1)^3}$$
$$=\boxed{\frac{\pi^3}{24\sqrt 3}+\frac{\pi}{4\sqrt 3}\operatorname{Li}_2\left(-\left(2-\sqrt 3\right)^2\right)+\frac{2}{\sqrt 3}\operatorname{Ti}_3\left(2-\sqrt 3\right)}$$
$$\text{where }\operatorname{Li}_k(x)=\sum\limits_{n=1}^\infty \frac{x^n}{n^k},\  \operatorname{Ti}_k(x)=\Im\operatorname{Li}_k(ix)=\sum\limits_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^k}.$$
