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.
Mathematica expression is here.
