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?
 A: 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.
Mathematica expression is here.
A: 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. 
