Evaluating $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$

The values of the two Euler Sums are

$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n^{3}} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7}{4}\log(2) \zeta(3)+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log^4(2)$$ $$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n^{(2)}}{n^{2}} =-\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

Therefore the integral evaluates to

$$\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \end{align*}$$

Refer to this page for the evaluation of Euler Sums.

using an identity devoloped by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series": $$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n} $$ we get: \begin{align} I&=\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln x\ dx\\ &=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac1{(2n)^2}\right)=2\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}-\frac14\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42 \end{align}

where we used the value of the first sum proved here and the common value of the second sum which is $\frac54\zeta(4)$.

A solution by Cornel Ioan Valean (the nice, interesting part about this solution is that we calculate the integral without using harmonic series, Beta function at all)

Proof. Multiplying both sides of $i)$ (see below) by $1/(1-a)$ and integrating from $a=0$ to $a=1/2$, we obtain $$\int_0^{1/2}\left(\int_0^1 \frac{\log (x) \log (1-x)}{(1-a x)(1-a)} \textrm{d}x\right)\textrm{d}a=\int_0^1\left(\int_0^{1/2} \frac{\log (x) \log (1-x)}{(1-a x)(1-a)} \textrm{d}a\right)\textrm{d}x$$ $$=\int_0^1 \frac{\log (x) \log (1-x) \log (2-x)}{1-x} \textrm{d}x=\int_0^1 \frac{\log (1-x) \log (x) \log (1+x)}{x} \textrm{d}x$$ $$=\frac{\pi^2}{6}\int_0^{1/2}\frac{ \log (1-a)}{a(1-a)}\textrm{d}a+\frac{1}{6}\int_0^{1/2}\frac{\log ^3(1-a)}{a(1-a)}\textrm{d}a+\int_0^{1/2}\frac{\operatorname{Li}_3(a)}{a(1-a)}\textrm{d}a$$ $$-\int_0^{1/2}\frac{1}{a(1-a)}\operatorname{Li}_3\left(\frac{a}{a-1}\right) \textrm{d}a$$ $$=\frac{1}{12}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{4}\log(2)\zeta(3)-\frac{27}{16}\zeta(4)+2\operatorname{Li}_4\left(\frac{1}{2}\right),$$ and the solution is complete.

In the calculations we needed the following results:

Let $a<1$ be a real number. The following equality holds: $$i) \ \int_0^1 \frac{\log (x) \log (1-x)}{1-a x} \textrm{d}x=\frac{\pi^2}{6}\frac{ \log (1-a)}{a}+\frac{1}{6}\frac{\log ^3(1-a)}{a}+\frac{1}{a}\operatorname{Li}_3(a)-\frac{1}{a}\operatorname{Li}_3\left(\frac{a}{a-1}\right),$$ where $\operatorname{Li}_3$ is the Trilogarithm function. The result is stated and proved in the paper A special way of extracting the real part of the Trilogarithm, $ \operatorname{Li}_3\left(\frac{1\pm i}{2}\right)$ by Cornel Ioan Valean.

$$ii) \ \int_0^a \frac{\log (1-x)}{x (1-x)}\textrm{d}x=-\frac{1}{2} \log ^2(1-a)-\operatorname{Li}_2(a).$$

$$iii) \ \int_0^a \frac{\log ^3(1-x)}{x (1-x)} \textrm{d}x$$ $$=6 \operatorname{Li}_4(1-a)-6 \operatorname{Li}_3(1-a) \log (1-a)+3 \operatorname{Li}_2(1-a) \log ^2(1-a)$$ $$-\frac{1}{4} \log ^4(1-a)+\log (a) \log ^3(1-a)-\frac{\pi^4}{15},$$

which is straightforward with integration by parts.

$$iv) \ \int_0^a \frac{\operatorname{Li}_3(x)}{1-x} \textrm{d}x=-\frac{1}{2}(\operatorname{Li}_2(a))^2-\operatorname{Li}_3(a) \log (1-a),$$

and it's straightforward with integration by parts.

$$v) \ \int_0^a \frac{1}{x(1-x)}\operatorname{Li}_3\left(\frac{x}{x-1}\right)\textrm{d}x =\operatorname{Li}_4\left(\frac{a}{a-1}\right).$$

$$vi) \ \operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}(\zeta(2)-\log^2(2)).$$

$$ vii) \ \operatorname{Li}_3\left(\frac{1}{2}\right)=\frac{7}{8}\zeta(3)-\frac{1}{2}\log(2)\zeta(2)+\frac{1}{6}\log^3(2).$$

A first note: A generalization of the present integral with $\log^{2n}(x)$ is given in the book (Almost) Impossible Integrals, Sums, and Series (see page 6),

$$\int_0^1\frac{\log(1-x)\log^{2n}(x)\log(1+x)}{x} \textrm{d}x =\frac{1}{2}(2n)!\left(1-\frac{1}{2^{2n+1}}\right)\sum_{k=1}^{2n} \zeta(k+1)\zeta(2n-k+2)$$ $$-(2n)!\sum_{k=1}^{n}\left(1-\frac{1}{2^{2k-1}}\right)\zeta(2k)\zeta(2n-2k+3) + \frac{1}{2^{2n+3}} (2n+3-2^{2n+3})(2n)!\zeta(2n+3),$$ and the result is obtained by exploiting the series representation of $\log(1-x)\log(1+x)$.

A second note: By the strategy presented above more interesting integrals may be calculated (without Beta function, harmonic series). Another nice example is

$$\int _0^1 \frac{\log(1-x) \log(x)\log(1+x)}{1-x}\textrm{d}x$$ $$=\frac{17 }{16}\zeta(4)-\frac{1}{4} \log ^2(2)\zeta(2)+\frac{7}{8} \log (2)\zeta (3) -\frac{1}{12} \log ^4(2)-2 \text{Li}_4\left(\frac{1}{2}\right).$$

A second solution in large steps (we circumvent the use of harmonic series)

If we use the algebraic identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ which we combine with clever rearrangements of the resulting integrals, we arrive at

$$ \int _0^1 \frac{\log(1-x) \log(x)\log(1+x)}{x}\textrm{d}x$$ $$=\frac{1}{2}\int_0^1 \frac{\log^2(x)\log(1+x)}{1+x}\textrm{d}x-\frac{3}{8}\underbrace{\int_0^1 \frac{\log(1-x)\log^2(x)}{1-x}\textrm{d}x}_{\text{Beta function}},$$ where the first integral is calculated in the book, (Almost) Impossible Integrals, Sums, and Series, on pages $503-505$, and the second integral is a form of Beta function.