TITLE: Is there a closed form function for the integral $\int_{0}^{z} \frac{1}{x}\log(1-x)\log(x)\log(1+x)\,dx$?
QUESTION [9 upvotes]: Recently, in connection with the problem of calculating generating functions of the antisymmetric harmonic number (https://math.stackexchange.com/a/3526006/198592, and  What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$)  I stumbled on the beautiful integral
$$f(z) = \int_{0}^z \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{1}$$
which seems to be hard.
I tried the common procedure of partial integrations, variable transformations and antiderivatives hunting with Mathematica which generated a multitude of different variants of the integral but finally I could not solve it.
Question Can you calculate the integral $(1)$?
Notice that we are looking here for the integral as a function of the upper limit $z$, or equvalently, for an antiderivative. The problem counts as solved once $f(z)$ is expressed through known functions, we also say that $f(z)$ has a "closed functional form".
On the other hand there are myriads of integrations problems in this forum which are similar but have fixed limits, i.e. they are definite integrals which define a constant, and the question is then if this constant is expressible by known constants - has a "closed form".
Our problem also has a compagnion in the constant species ( Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$) which provided the closed form for $f(1)$.

REPLY [5 votes]: Introduction
Heureka! I have eventually solved the indefinite integral
$$a(x) = \int \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{1}$$
in other words, I have found the antidervative of the integrand
$$i(x) = \frac{\log(1-x)\log(x)\log(1+x)}{x}\tag{2}$$
This means that we have also the integral
$$f(z) = \int_{0}^z \frac{\log(1-x)\log(x)\log(1+x)}{x}\,dx\tag{3}$$
since it is convergent in the vicinity of $x=0$.
I was surprised that the use of "standard" integrations methods turned out to be sufficient (i.e. no (Euler-)series, no parametric derivatives were used).
For evaulating and checking the various integrals I made extensive use of Mathematica. The advantage of having indefinite integrals is that we can verify the result over a given range of the integration variable by calculating the derivative and comparing it to the integrand. This is in contrast to the case of constants (closed forms) obtained by assigning fixed values to the integration limits where the final result can only be checked numerically (most of the problems here are of the latter type).
Although an antiderivative was found it is available first of all as a Mathematica expression, and because the number of terms in it exceeds 1.000 printing it out is not really useful.
Solution
Step 1:
We simplify the integrand by the well-known decomposion of the product of two logarithms with different arguments into a sum of independent logarithmic terms
$$\log (1-x) \log (x+1)=\frac{1}{2} \log ^2\left(1-x^2\right)-\frac{1}{2} \log ^2(1-x)-\frac{1}{2} \log ^2(x+1)\tag{4}$$
The integral then becomes
$$a(x) = A(x) + B(x)$$
with
$$A=-\frac{1}{2} \int \frac{\log (x) \left(\log ^2(1-x)+\log ^2(x+1)\right)}{x} \, dx\tag{5a}$$
$$B= \frac{1}{2} \int \frac{\log (x) \log ^2\left(1-x^2\right)}{x} \, dx\tag{5b}$$
Step 2:
Doing the integrals.
$A(x)$ is immediately done by Mathematica producing a huge number of terms (leafcount=709) which consist of $\log$s and polylogs $\operatorname{Li}_2$ through $\operatorname{Li}_4$.
We normalize the antiderivative by subtracting a constant (obatined by calculating the limit for $x\to+0$ from above, i.e. from inside the integration interval) so that $A(0)=0$.
$B(x)$ is refused by Mathematica, so we have to prepare it manually. First we change the variable $x\to\sqrt{t}$:
$$\frac{\log (x) \log ^2\left(1-x^2\right)}{2 x}dx =\frac{ \log ^2(1-t) \log \left(\sqrt{t}\right)}{4 t}dt= \frac{ \log ^2(1-t) \log \left({t}\right)}{8 t}dt\tag{6}$$
Notice that the logarithm simplifies nicely.
Now the antiderivative is found by Mathematica, and in the end we have to reverse the variable transform by letting $t\to x^2$. It produces an expression with leafcount of 404 and consists of the same functions as $A(x)$. In the end we renormalize the result as above to $B(0)=0$.
The antiderivatives encountered here produced by Mathematica frequently contain complex terms.  They occur typically through logarithms and polylogarithms, as e.g. in $\log(x)$ for $x<0$ and in $\operatorname{Li_n}(x)$ for $x\gt1$. When the integral is real by definition in some interval we should seek to transform all terms into real terms in this interval. For the polylog functions with natural index we have employed systematically the transformation formula from  https://en.wikipedia.org/wiki/Polylogarithm thus making all terms real in the interval $0\lt x lt 1$.
$$\operatorname{Li}_n(x)\to (-1)^{n+1} \operatorname{Li}_n\left(\frac{1}{x}\right)-\frac{(2 \pi  i)^n}{n!} B_n\left(\frac{\log (x)}{2 \pi  i}+1\right)\text{ 
   for x>1}\tag{7}$$
where $B_n(z)$ are the Bernoulli polynomials of order $n$ (https://en.wikipedia.org/wiki/Bernoulli_polynomials).
In total the antiderivative consists of the appreciable amount of about 1.100 terms, and it is not useful to print them all out here. But I can provide the Mathematica code of it on demand.
It is remarkable that all polylogs except $\operatorname{Li_1}(x) = - \log(1-x)$ appear in first power and linearly, i.e. there are no products of these functions appearing.
Evaluations
Here is the plot of the exact function $f(z)$

This picture as such could have been produced, of course, also by numerical integration.
The knowledge of $f(z)$ permits us to calculate any definite integral we wish in the interval $0\le z \le 1$.
Example 1: $f(1)$
We obtain
$$f(1) = 2 \text{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4} \zeta (3) \log (2)-\frac{3 \pi ^4}{160}+\frac{\log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)\simeq 0.290721\tag{8}$$
in agreement with the expression for $f(1)$ calculated some time ago by others (Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$).
Example 2: $f(\frac{1}{2})$
This will be given in full length.
$$f(\frac{1}{2})= -\operatorname{Li}_4\left(\frac{1}{2}\right)+\operatorname{Li}_4\left(\frac{1}{3}\right)+\operatorname{Li}_4\left(\frac{2}{3}\right)-\frac{\operatorname{Li}_4\left(-\frac{1}{3}\right)}{4}-\frac{\operatorname{Li}_4\left(\frac{1}{4}\right)}{8}+\frac{\operatorname{Li}_4\left(\frac{3}{4}\right)}{4}+\frac{1}{2} \operatorname{Li}_2\left(-\frac{1}{2}\right) \log ^2(2)+\frac{3}{4} \operatorname{Li}_2\left(\frac{3}{4}\right) \log ^2(2)+\frac{1}{2} \operatorname{Li}_2\left(-\frac{1}{2}\right) \log ^2(3)+\frac{1}{2} \operatorname{Li}_2\left(\frac{1}{3}\right) \log ^2(3)-\frac{1}{6} \pi ^2 \left(3 \operatorname{Li}_2\left(-\frac{1}{2}\right)+3 \operatorname{Li}_2\left(\frac{1}{3}\right)-4 \log ^2(2)+5 \log ^2(3)-\log (3) \log (2)+4 (\log (3)-\log (2))^2-(\log (3)-\log (2)) (2 \log (2)+7 \log (3))\right)-\frac{1}{8} \operatorname{Li}_2\left(-\frac{1}{3}\right) \log ^2(3)+\frac{1}{8} \operatorname{Li}_2\left(\frac{3}{4}\right) \log ^2(3)-\operatorname{Li}_2\left(-\frac{1}{2}\right) \log (3) \log (2)-\frac{1}{2} \operatorname{Li}_2\left(\frac{3}{4}\right) \log (3) \log (2)+\operatorname{Li}_3\left(-\frac{1}{2}\right) \log (2)+\operatorname{Li}_3\left(\frac{1}{3}\right) \log (2)-\frac{1}{4} \operatorname{Li}_3\left(\frac{1}{4}\right) \log (2)+\frac{1}{2} \operatorname{Li}_3\left(\frac{3}{4}\right) \log (2)-\operatorname{Li}_3\left(-\frac{1}{2}\right) \log (3)-\frac{1}{4} \operatorname{Li}_3\left(-\frac{1}{3}\right) \log (3)-\frac{1}{4} \operatorname{Li}_3\left(\frac{3}{4}\right) \log (3)-\frac{15}{8} \zeta (3) \log (2)+\zeta (3) \log (3)-\frac{\pi ^4}{80}+\frac{4 \log ^4(2)}{3}+\frac{3 \log ^4(3)}{32}-\log (3) \log ^3(2)+\frac{5}{24} (\log (3)-\log (2))^4\simeq 0.155958\tag{9}$$
Discussion
There are several topics to be discussed in the wake of this type of calculations.

Generalization: solution of a whole class of trilog integrals

See Evaluating the indefinite integral $\int x^k \log (1-x) \log (x) \log (x+1) \, dx$
I have found that integrals of the type
$$f_k(x) = \int x^{k}\log(1-x)\log(x)\log(1+x)\,dx\tag{10}$$
with integer $k$ can be solved explicitly. For $k=-1$ it was shown  here, for other $k$ integration by parts leads to integrable exppressions.

Relation to generating function problem

I have mentioned here What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$ that the problem of the generating function of third order for the alternating harmonic series boils down to two integrals, one of which is the integral solved here.

Calculations tend to resemble a proof of existence

Because of the large number of terms involved the philosophy approaches more that of a proof of existence of an antiderivative. The procedure typically is: if Mathematica can't solve a given integral then perform manual (mathematical) actions on it leading to a number of integrals which hopefully Mathematica can solve. We could then disregard the actual results and only state that an antiderivative exists.