I'm interested in the following definite integral: $$I=\int_0^1\frac{\ln^2\!\left(1+x+x^2\right)}x\,dx.\tag1$$ The corresponding antiderivative can be evaluated with Mathematica, but even after simplification is quite clumsy. It matches results of numerical integration, and its correctness can potentially be verified by hand using differentiation. So, we are assured that a closed form exists for $I$, albeit complicated one.
My program for a numerical search for closed forms found a much simpler candidate:
$$I\stackrel{\color{gray}?}=\frac{2\pi}{9\sqrt3}\psi^{\small(1)}\!\left(\tfrac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3).\tag2$$
Note that the trigamma value here can be expressed in terms of the dilogarithm of complex argument (see formula $(5)$ here) or of the $2^{nd}$ order harmonic number of fractional argument: $$\begin{align}\psi^{\small(1)}\!\left(\tfrac13\right)&=\frac{2\pi^2}3+2\sqrt3\,\Im\,\operatorname{Li}_2\!\left[(-1)^{\small1/3}\right],\tag3\\\psi^{\small(1)}\!\left(\tfrac13\right)&=\frac{\pi^2}6+9-H^{\small(2)}_{\small1/3}.\tag4\end{align}$$ Can we prove $(2)$, preferably not going through the huge intermediate antiderivative?
One possible direction that I thought of is to factor the polynomial under the logarithm: $$I=\int_0^1\Big[\ln\!\left(x+(-1)^{\small1/3}\right)+\ln\!\left(x-(-1)^{\small2/3}\right)\Big]^2x^{-1}\,dx.\tag5$$ After expanding the square brackets, Mathematica can find a simpler antiderivative for it. Can we reach $(2)$ following this direction manually?