<s>[INST] Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$ I am looking for a closed form for: 
$$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$
I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/ hint would be greatly appreciated. Thanks!
 [/INST] A conjectural evaluation for the remaining alternating sum is given in equation (19) of https://arxiv.org/abs/1908.04770, namely
$$
\sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5}
= \frac{1}{13} \left( \frac{1}{3}\operatorname{Li}_6\left(-\frac{1}{8}\right) - 162 \operatorname{Li}_6\left(-\frac{1}{2}\right) - 126 \operatorname{Li}_6\left(\frac{1}{2}\right) \right)
-\frac{1787}{624} \zeta(6) + \frac{3}{8} \zeta(3)^2
\\
+\frac{31}{16}\zeta(5) \log(2) - \frac{15}{26} \zeta(4) \log^2(2) + \frac{3}{104} \zeta(2) \log^4(2) - \frac{1}{208} \log^6(2)
$$
Subtract this from $-\frac{9}{16} \zeta(6)$ to get the answer for the original question. Note the relation
$$
\int_0^1 \frac{\operatorname{Li}_3(-x)\operatorname{Li}_2(x)}{x} \mathrm{d} x
= -\frac{9}{16} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5}
= -\frac{49}{32} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n}}{n^5}
$$
where the denominator in the coefficient of $\zeta(6)$ in the above answer is corrected, the $-\frac{49}{28} \zeta(6)$ there should read $-\frac{49}{32}\zeta(6)$.
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