For first, we have:
$$I=\Re\,\text{Li}_2(1\pm i\sqrt{3})=\Re\int_{1}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt=\frac{\pi^2}{6}+\Re\int_{0}^{i\sqrt{3}}\frac{\log t}{1-t}\,dt\tag{1}$$
hence:
$$ I = \frac{\pi^2}{6}-\int_{0}^{\sqrt{3}}\frac{\frac{\pi}{2}+t\log t}{1+t^2}\,dt\tag{2}=-\int_{0}^{\sqrt{3}}\frac{t\log t}{1+t^2}\,dt $$
but:
$$\begin{eqnarray*} \int\frac{t\log t}{1+t^2}\,dt &=& \frac{\log(t)\log(1+t^2)}{2}+\text{Li}_2(it)+\text{Li}_2(-it)\\&=&\frac{\log(t)\log(1+t^2)}{2}+\frac{1}{4}\,\text{Li}_2(-t^2)\tag{3}\end{eqnarray*}$$
follows from integration by parts and the dilogarithm reflection formula, hence I simply get:
$$ \Re\,\text{Li}_2(1\pm i\sqrt{3})=-\frac{\log(3)\log(4)+\text{Li}_2(-3)}{4}\tag{4}$$
and it should not be too difficult to convert $(4)$ into your expression through the dilogarithm functional identities. In a similar fashion:
$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\int_{0}^{\sqrt{3}}\frac{\frac{\pi t}{2}-\log t}{1+t^2}\,dt = \frac{\pi}{2}\log 2+\int_{\frac{1}{\sqrt{3}}}^{+\infty}\frac{\log t}{1+t^2}\,dt\tag{5}$$
hence:
$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\frac{\pi}{2}\log 2+\frac{\pi}{12}\log 3+\Im\,\text{Li}_2\left(\frac{i}{\sqrt{3}}\right)\tag{6}$$
or:
$$ \Im\,\text{Li}_2(1+ i\sqrt{3})=\frac{\pi}{12}\log(192)+\frac{1}{\sqrt{3}}\sum_{n\geq 0}\frac{(-1)^n}{3^n(2n+1)^2}.\tag{7}$$
giving a value of the inverse tangent integral. It depends on the value of the Clausen function $\text{Cl}_2\left(\frac{\pi}{3}\right)$, and that quantity depends on the values of the Barnes $G$-function at $\frac{1}{3}$ and $\frac{2}{3}$.
