A friend of mine shared this problem with me. As he was told, this integral can be evaluated in a closed form (the result may involve polylogarithms). Despite all our efforts, so far we have not achieved anything, so I decided to ask for your advice.
$$\int_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$$
I found some similar questions here on MSE:
(1),
(2),
(3), (4),
(5),
(6),
(7),
(8), (9), (10),
(11), (12),
(13), (14).
Best Answer
\begin{align} & \int_0^1\ln(2+x)\,\ln(1+x)\,\ln\left(1+x^{-1}\right)\ln x\,dx\\ & \quad=\frac{71}{36}\,\ln^42+2\ln^32\cdot\ln3+4\ln2\cdot\ln^33-7\ln^22\cdot\ln^23-\frac23\,\ln^32-\frac23\,\ln^33-\ln^22\cdot\ln3\\ & \quad \quad +6\ln^22+3\ln^23-12\ln2-\frac{\pi^4}{216}+\pi^2\!\left(\frac{49}{36}\,\ln^22-2\ln2\cdot\ln3-\frac{\ln2}3+\frac{\ln3}3-\frac16\right)\\ & \quad \quad+\left(6-2\ln2-2\ln^22\right)\operatorname{Li}_2\!\left(\tfrac13\right)+(2-12\ln2)\left[\operatorname{Li}_3\!\left(\tfrac13\right)+\operatorname{Li}_3\!\left(\tfrac23\right)\right]-\frac23\,\operatorname{Li}_4\!\left(\tfrac12\right)+3\operatorname{Li}_4\!\left(\tfrac14\right)\\ & \quad \quad +\left(\frac54+\frac{221}{12}\ln2\right)\zeta(3). \end{align}