TITLE: Prove closed expression for $\int_0^1 \log(x) \log(1+x) \log(2+x)\,dx$
QUESTION [9 upvotes]: In Closed form of $\int_{0}^{1} \frac{\log(1+x)\log(2+x) \log(3+x)}{1+x}\,dx$ I have proposed an integral which I could not solve, and though there were some upvotes on the question no solution was provided. Hence I looked for simplifications which still are not trivial. 
Here's an example where I found a closed expression with the help of Mathematica which can be verified numerically but I'm lacking a proof. 
Hence my question is
Prove that
$$\int_0^1 \log(x)\log(x+1)\log(x+2)\,dx \\
= -6+3 \log ^3(2)-\frac{\log ^3(3)}{3}+\frac{\log ^2(2)}{2}-3 \log (3) \log (2)+6 \log (3)\\+\zeta(2) (1-2 \log (2))-\frac{13 \zeta (3)}{8}\\-\operatorname{Li}_2\left(-\frac{1}{2}\right)-6 \operatorname{Li}_2\left(-\frac{1}{2}\right) \log (2)+4 \operatorname{Li}_2\left(\frac{1}{4}\right) \log (2)\\-2 \operatorname{Li}_2\left(\frac{1}{3}\right) \log (3)+\operatorname{Li}_2\left(-\frac{1}{3}\right) \log (3)\\
-4 \operatorname{Li}_3\left(-\frac{1}{2}\right)-2 \operatorname{Li}_3\left(\frac{1}{3}\right)+\operatorname{Li}_3\left(-\frac{1}{3}\right)+2 \operatorname{Li}_3\left(\frac{1}{4}\right)\\\simeq -0.18403235664237885896 $$
Notice that the expression is composed of $\pi$, $\log(s)$, $\zeta(s)$, and $\operatorname{Li}_{s}(t)$.
Remark 1: Mathematica was able to find the antiderivative but it turned out to contain complex valued summands. These cancelled out numerically but I could not prove mathematically that their contribution vanishes. 
Remark 2: I have not found the present class of integrals (product of logs with successively shifted arguments) in the 60 problems the of the book "(Almost) Impossible Integrals, Sums, and Series" by
Cornel Ioan Valean (https://it.b-ok2.org/book/4996918/0df734) which is famous and frequently quoted in this forum. So this type of problem seems to be new.

REPLY [2 votes]: By the variable change $1/(1+x)=t$ and then integration by parts, it's easy to observe the main integral reduces easily to 
  $$\int_0^1\log(x)\log(1+x)\log(2+x)\textrm{d}x=\int_{1/2}^1\frac{\displaystyle \log\left(\frac{t}{1-t}\right)\log(t)\log\left(\frac{1+t}{t}\right)}{t^2}\textrm{d}t$$
$$=7 \log ^2(2)-\log ^3(2)-3 \log (2)\log (3)+\log ^2(2)\log (3)$$
$$+2\underbrace{\int_{1/2}^1 \frac{\log \left(1-t^2\right)}{t^2}\textrm{d}t}_{\text{Trivial}}-3\underbrace{\int_{1/2}^1\frac{\log ^2(t)}{t^2}\textrm{d}t}_{\text{Trivial}}-4\underbrace{\int_{1/2}^1 \frac{\log (t)}{1-t^2}\textrm{d}t}_{\text{Trivial}}-\underbrace{2\int_{1/2}^1 \frac{ \log ^2(t)}{1-t^2}\textrm{d}t}_{\text{Trivial}}$$
$$+\underbrace{\int_{1/2}^1\frac{\displaystyle \log \left(\frac{1-t}{2}\right)}{1+t}\textrm{d}t}_{\text{Trivial}}+\underbrace{\int_{1/2}^1\frac{\displaystyle \log \left(\frac{1+t}{2}\right)}{1-t}\textrm{d}t}_{\text{Trivial}}+\underbrace{\int_{1/2}^1\frac{\log (1+t)}{t}\textrm{d}t}_{\text{Trivial}}-\underbrace{\int_{1/2}^1\frac{\log (1-t)}{t}\textrm{d}t}_{\text{Trivial}} $$
$$+\underbrace{\int_{1/2}^1\frac{\log (1+t) \log (t)}{t}\textrm{d}t}_{\text{Trivial}}-\underbrace{\int_{1/2}^1\frac{\log (1-t) \log (t)}{t}\textrm{d}t}_{\text{Trivial}}$$
$$+\underbrace{\int_{1/2}^1\frac{\log (1-t) \log (1+t)}{t}\textrm{d}t}_{\displaystyle \mathcal{I}}+2\underbrace{\int_{1/2}^1\frac{ \log (1-t) \log (t)}{1+t}\textrm{d}t}_{\displaystyle \mathcal{J}}.$$
  The integral $\mathcal{I}$ is straightforward if we use that $$\log(1-t)\log(1+t)=\frac{1}{2} \left(\log ^2(1-t)+\log ^2(1+t)-\log ^2\left(\frac{1-t}{1+t}\right)\right),$$
  and when expanding the right-hand side and then taking the integral, it's clear that for the last integral we might like to make the change of variable $\displaystyle t\mapsto\frac{1-t}{1+t}$. The rest is known and easy.
With respect to the integral $\mathcal{J}$ we may use the same idea and use the algebraic identity above, but first we might want to make some rearrangements. Well, we see that 
  $$\int_{1/2}^1 \frac{\log (1-t) \log (t)}{1+t} \textrm{d}t=\underbrace{\int_0^1 \frac{\log (1-t) \log (t)}{1+t} \textrm{d}t}_{\displaystyle 13/8\zeta (3)- \pi ^2 \log (2)/4}-\underbrace{\int_0^{1/2} \frac{\log (1-t) \log (t)}{1+t} \textrm{d}t}_{\displaystyle \mathcal{K}},$$
  and for the last integral, which is $\mathcal{K}$, we make the change of variable $t\mapsto 1/2-t$ (that is in a way the magical part that soon allows us to connect things to the algebraic identity above), and then we have 
  $$\mathcal{K}=\int_0^{1/2} \frac{\log (1-t) \log (t)}{1+t} \textrm{d}t=\int_0^{1/2} \frac{\log (1/2-t) \log (1/2+t)}{3/2-t} \textrm{d}t$$
  and then let $t\mapsto t/2$ to arrive at 
  $$\mathcal{K}=\frac{1}{3} \int_0^1 \frac{(\log (1-t)-\log (2)) (\log (1+t)-\log (2))}{1-t/3} \textrm{d}t.$$
  What now? When expanding the integral, all the resulting integrals are easy to calculate except the integral 
  $$\int_0^1 \frac{\log (1-t)\log (1+t)}{1-t/3} \textrm{d}t.$$
  At this point we use again that  $$\log(1-t)\log(1+t)=\frac{1}{2} \left(\log ^2(1-t)+\log ^2(1+t)-\log ^2\left(\frac{1-t}{1+t}\right)\right),$$
  and the first two integrals are extracted with simple integration by parts or by using geometric series after arranging properly the argument of log in the numerator with the proper variable change, and for the last integral we use again the variable change $\displaystyle t\mapsto\frac{1-t}{1+t}$, and we are finally done.

End of story.
A first note: Here is the generalization of one of the integrals I used above (the case $n=1$), 

Let $n\ge1$ be a positive integer. Then
  $$\int_0^1 \frac{\log ^{2n-1}(x) \log(1-x)}{1+x} \textrm{d}x$$
$$=\frac{1}{2}(2n)!\zeta (2n+1)-2\log(2)(1 -2^{-2n})(2n-1)!\zeta (2n)$$
$$-2^{-1-2n} (2n+1-2^{1+2n})(2n-1)!\zeta(2n+1)$$
$$-(2n-1)!\sum_{k=1}^{n-1}\zeta (2k)\zeta (2n-2k+1)+2^{-2n}(2n-1)!\sum_{k=1}^{n-1}2^{2k}\zeta (2k)\zeta (2n-2k+1),$$
  where $\zeta$ represents the Riemann zeta function.

The generalization is presented and proved in the paper, A note presenting the generalization of a special logarithmic integral by C.I. Valean.
A second note: The calculation of the integral $\mathcal{K}$ is even simpler if we consider writing that 

$$\int_0^{1/2}\frac{\log(1-t)\log(t)}{1+t} \textrm{d}t$$
$$=\frac{1}{2}\int_0^{1/2}\frac{\log^2(1-t)}{1+t} \textrm{d}t+\frac{1}{2}\int_0^{1/2}\frac{\log^2(t)}{1+t} \textrm{d}t-\frac{1}{2}\int_0^{1/2}\frac{\displaystyle \log^2\left(\frac{t}{1-t}\right)}{1+t} \textrm{d}t,$$
  where in the last integral let $\displaystyle t\mapsto \frac{t}{1-t}$. This integral over the unit interval is also met in the book, (Almost) Impossible Integrals, Sums, and Series (and, of course, in the paper above which uses a very simple strategy for this particular case).