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?