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Using $$ \operatorname{Li}_3(-x) =-\frac{x}{2}\int_{0}^{1}\frac{\ln^2t}{1+tx} \text{d}t $$ It might be $$ -\frac{1}{2}\int_{0}^{1}\ln^2t \int_{1}^{\infty}\frac{x\ln(x-1)}{(1+tx)(1+x^2)}\text{d}x\text{d}t $$ Any suggestion is appreciated.

An integral relation between three integrals $$ \int_{0}^{1} \frac{\ln(x)^2\ln(1+x)^2}{1+x^2} \text{d}x -2\int_{0}^{1} \frac{\ln^3(x)\ln(1+x)}{1+x^2} \text{d}x +4\int_{1}^{\infty} \frac{\operatorname{Li}_3(-x)\ln(x-1)}{1+x^2}\text{d}x = -\frac{3\pi}{64}\zeta(3)\ln2-\frac{21\pi^5}{256} +\frac{\pi^3}{64}\ln^22 $$

Setness Ramesory
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    Presumably, there is no nice answer in terms of ordinary polylogarithm alone. Try running MZIntegrate[PolyLog[3,-1/x]*Log[1/x-1]/(1+x^2), {x, 0, 1}] – pisco Aug 29 '21 at 14:22
  • Also try playing around with this approach0 search which may give many hints. – Тyma Gaidash Aug 29 '21 at 14:25
  • @pisco What is the value of $$\int_{0}^{1} \frac{\zeta(3)\operatorname{Li}_5(1-x) \operatorname{Li}_7(1-x)\operatorname{Li}_9(1-x) -\operatorname{Li}_3(1-x)\zeta(5)\zeta(7)\zeta(9) }{x} \text{d}x$$? – Setness Ramesory Sep 15 '21 at 12:06
  • @aaaaaaaaabbbbbbbbbcccccc The first integral has weight $22$, far too high for an explicit calculation. Presumably there is no nice answer. – pisco Sep 15 '21 at 12:28
  • @pisco How to express $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^7\binom{2k}{k}}$? – Setness Ramesory Oct 08 '21 at 12:46
  • @aaaaaaaaabbbbbbbbbcccccc This is a CMZV of level 5 and weight $7$, again no reason to expect a nice answer, there could be a slim chance for it to be expressible via polylogs though. – pisco Oct 08 '21 at 23:04
  • @pisco is this true?$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^3\binom{2n}{n}} \left(H_n^{(3)}+\frac{1}{5n^3} \right) =\frac{2}{5}\zeta(3)^2$ – Setness Ramesory Oct 16 '21 at 10:01
  • @aaaaaaaaabbbbbbbbbcccccc This is still a conjecture, but I do have an idea on proving it. – pisco Oct 16 '21 at 10:18
  • @pisco Where do I find the value of $\zeta(3,\bar{1},\bar{1})$(all weight 5 MZVs)? – Setness Ramesory Nov 27 '21 at 11:49
  • @aaaaaaaaabbbbbbbbbcccccc The command MultiZeta[{3,1,1},{1,-1,-1}] using my package. – pisco Nov 27 '21 at 23:16
  • @pisco Does your algorithm can convert iterated integrals into MZVs? – Setness Ramesory Nov 29 '21 at 11:59
  • @aaaaaaaaabbbbbbbbbcccccc The process of converting iterated integrals into MZVs is used extensively in internal code of MZIntegrate. – pisco Nov 29 '21 at 12:10
  • @pisco Any explicit codes? – Setness Ramesory Nov 29 '21 at 12:12
  • @aaaaaaaaabbbbbbbbbcccccc What are you working on (e.g. what level and weight of MZV)? – pisco Nov 29 '21 at 12:19
  • @pisco Like this one $$\int_{0}^{1} \text{d}x \int_{x\ge t_2\ge t_1\ge0,x\ge u_3\ge u_2 \ge u_1\ge0}\frac{\text{d}t_1}{t_1} \frac{\text{d}t_2}{1-t_2} \frac{\text{d}u_1}{u_1} \frac{\text{d}u_2}{u_2} \frac{\text{d}u_3}{1+u_3}.$$ – Setness Ramesory Nov 29 '21 at 12:36
  • @aaaaaaaaabbbbbbbbbcccccc This is not a (homogenous) MZV, replace $dx$ by $dx/x$ or $dx/(1+x)$ to make it homogenous. If you really want to do this one, you need integration by parts to reduce it into a sum of homogenous MZV. – pisco Nov 29 '21 at 12:46

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