Integral ${\large\int}_0^1\ln^3\!\left(1+x+x^2\right)dx$ I'm interested in this integral:
$$I=\int_0^1\ln^3\!\left(1+x+x^2\right)dx.\tag1$$
Can we prove that
$$\begin{align}I&\stackrel{\color{gray}?}=\frac32\ln^33-9\ln^23+36\ln3+2\pi^2\ln3-\frac{4\pi^2}3+\left(8-\ln^23-4\ln3\right)\cdot\frac{\pi\sqrt3}2\\&-48-\frac{7\pi^3}{6\sqrt3}+(2-3\ln3)\cdot\psi^{(1)}\!\left(\tfrac13\right)+36\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\end{align}\tag2$$
of find a simpler closed form?
Also, can we prove that
$${_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\stackrel{\color{gray}?}=\frac{71\pi^3}{1296\sqrt3}+\frac{5\pi}{48\sqrt3}\ln^23-\frac1{\sqrt3}\,\Im\operatorname{Li}_3\!\left[\frac{(-1)^{\small1/6}}{\sqrt3}\right]\tag3$$
or find a simpler expression for it?
 A: I found the antiderivative:
$$\begin{align}\int\ln^3\!\left(1+x+x^2\right)dx&=\xi\,\sqrt3\,\Big[\alpha^3-6\alpha^2+24\alpha-48\Big]\\&-\beta\,\sqrt3\,\Big[4\beta^2-3\alpha\ln3+6\alpha-24+6\ln3\Big]\phantom{\Huge|}\\&+6\,\sqrt3\,\Im\Big[(\alpha-2-2\beta\,i)\,\operatorname{Li}_2(\gamma)-2\,\operatorname{Li}_3(\gamma)\Big]\phantom{\Huge|}\end{align}$$
where
$$\begin{align}&\color{maroon}{\alpha=\ln\!\left(1+x+x^2\right)}\\&\color{orange}{\beta=\arctan(2\,\xi)}\phantom{\Huge|}\\&\color{green}{\gamma=\frac12-i\,\xi}\phantom{\Huge|}\\&\color{blue}{\xi=\frac{1+2x}{2\sqrt3}}\phantom{\Huge|}\end{align}$$
It is valid and continuous at least for $x\ge0$, so it is good for our purposes.
Unfortunately, I cannot demonstrate any systematic approach that leads to this result, I got it with a series of lucky guesses and applications of PSLQ algorithm to determine rational coefficients, and finally proved its correctness using differentiation.
It yields the conjectured result modulo several polylogarithm identities that I'm still trying to prove.
