# Integral $\int_0^1\arctan(x)\arctan\left(x\sqrt3\right)\ln(x)dx$

I need to evaluate this integral: $$\int_0^1\arctan(x)\arctan\left(x\sqrt3\right)\ln(x)dx$$ Apparently, Maple and Mathematica cannot do anything with it, but I saw similar integrals to be evaluated in terms of polylogarithms (unfortunately, I have not yet mastered them enough to do it myself). Could anybody please help me with it?

Following the method outlined in another answer, and simplifying the resulting expression, we get the following closed form: $$\frac{5 G}{6 \sqrt{3}}-\frac{\Im\operatorname{Li}_3(1+i)}{\sqrt{3}}+\Im\operatorname{Li}_3\left(i \sqrt{3}\right)-\frac{\Im\operatorname{Li}_3\left(i \sqrt{3}\right)}{4 \sqrt{3}}-\frac{1}{2} \Im\operatorname{Li}_3\left(1+i \sqrt{3}\right)\\ -3 \Im\operatorname{Li}_3\left(\left(-\frac{1}{2}+\frac{i}{2}\right) \left(-1+\sqrt{3}\right)\right)+\sqrt{3} \Im\operatorname{Li}_3\left(\left(-\frac{1}{2}+\frac{i}{2}\right) \left(-1+\sqrt{3}\right)\right)\\ +\frac{1}{\sqrt{3}}\Im\operatorname{Li}_3\left(\tfrac{(1+i) \sqrt{3}}{1+\sqrt{3}}\right)-3 \Im\operatorname{Li}_3\left(\left(\frac{1}{2}+\frac{i}{2}\right) \left(1+\sqrt{3}\right)\right)+\frac{2}{\sqrt{3}}\Im\operatorname{Li}_3\left(\left(\frac{1}{2}+\frac{i}{2}\right) \left(1+\sqrt{3}\right)\right)\\ -\frac{1}{288} \pi \left[-2 \left\{\vphantom{\Large|}3 \left(4+\sqrt{3}\right) \cdot \operatorname{Li}_2\left(\tfrac{1}{3}\right)+6 \ln ^23-6 \left(7 \sqrt{3}-24\right) \cdot \ln ^2\left(1+\sqrt{3}\right)\\+24 \ln 3 -4 \left(9+4 \sqrt{3}\right) \cdot \ln \left(1+\sqrt{3}\right)\right\}+3 \left(5 \sqrt{3}-36\right) \cdot \ln ^22\\ -4 \left\{\vphantom{\Large|}9+7 \sqrt{3}-6 \ln 3+3 \left(7 \sqrt{3}-24\right) \cdot \ln \left(1+\sqrt{3}\right)\right\}\cdot\ln 2\right]\\ -\frac{1}{216} \left(18+5 \sqrt{3}\right) \pi ^2+\left(\frac{5}{36}-\frac{31}{384 \sqrt{3}}\right) \pi ^3+\frac{5 \psi ^{(1)}\left(\frac{1}{3}\right)}{48 \sqrt{3}},$$ that might be possible to simplify further.
After converting to exponential form and expanding, Mathematica can find a (not pretty) closed form anti-derivative in terms of logs and polylogs. We can take the limit at $0$ and $1$. For all practical purposes I'm just going to provide a link to the text of the solution. This matches numerical estimates.