Integral ${\large\int}_0^{\pi/2}\arctan^2\!\left(\frac{\sin x}{\sqrt3+\cos x}\right)dx$

$$I=\frac\pi{20}\ln^23+\frac\pi4\operatorname{Li_2}\left(\tfrac13\right)-\frac15\operatorname{Ti}_3\left(\sqrt3\right),$$ where $$\operatorname{Ti}_3\left(\sqrt3\right)=\Im\Big[\operatorname{Li}_3\left(i\sqrt3\right)\Big]=\frac{\sqrt3}8\Phi\left(-3,3,\tfrac12\right)=\frac{5\sqrt3}4\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac34\right)-\frac{5\pi^3}{432}-\frac\pi{16}\ln^23.$$

With the same approach I took in this answer, we have:

$$\begin{eqnarray*}\arctan\left(\frac{\sin x}{\sqrt{3}+\cos x}\right) &=& \text{Im}\log\left(\sqrt{3}+e^{ix}\right)\\&=&\text{Im}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,e^{inx}\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n\sqrt{3}^n}\,\sin(nx)\tag{1}\end{eqnarray*} $$ and by computing $\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx$ it follows that: $$\begin{eqnarray*}I&=&\int_{0}^{\pi/2}\arctan^2\left(\frac{\sin x}{\sqrt{3}+\cos x}\right)\,dx\\&=&\frac{\pi}{4}\,\text{Li}_2\left(\frac{1}{3}\right)+\sum_{m\neq n}\frac{(-1)^{n+m}}{nm\sqrt{3}^{n+m}}\int_{0}^{\pi/2}\sin(nx)\sin(mx)\,dx\tag{2}\end{eqnarray*}$$ but the last integral is zero if $n$ and $m$ have the same parity.

It follows that the last series in $(2)$ can be written as:

$$\begin{eqnarray*}-\sqrt{3}\sum_{a\geq 1}\sum_{b\geq 1}\frac{1}{(2b-1)3^{a+b}}\cdot\frac{(-1)^{a+b}}{(2a)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ and by reindexing the last double series on $a+b=s$ we get:

$$\begin{eqnarray*}-\sqrt{3}\sum_{s\geq 2}\frac{(-1)^s}{3^s}\sum_{b=1}^{s-1}\frac{1}{2b-1}\cdot\frac{1}{(2s-2b)^2-(2b-1)^2}\tag{3}\end{eqnarray*}$$ than can be dealt through partial fraction decomposition and dilogarithm identities.