Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wonder if a generalization to weight $5$ can be made.

Beta+IBP $3$ times+log factorization yields

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*$S=\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=\int_0^1 \frac2x \text{Li}_4\left(\frac{x^2(1-x)}2\right) dx\\ =\int_0^1 -\frac{2\left((3 x-2) \text{Li}_3\left(\frac{1}{2} (1-x) x^2\right)\right) \log (x)}{(x-1) x}  dx\\ =\int_0^1 \frac{2\left((3 x-2) \text{Li}_2\left(\frac{1}{2} (1-x) x^2\right)\right) \left(\log ^2(x)-\text{Li}_2(1-x)\right)}{(x-1) x}  dx\\ =\int_0^1 \left(\frac{2}{x}-\frac{1}{1-x}\right) f(x) \left(\log \left((1-x)^2+1\right)+\log (x+1)-\log (2)\right) dx$
Where

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*$\small f(x)=2\left(-\text{Li}_3(1-x)+2 \text{Li}_3(x)-2 \text{Li}_2(1-x) \log (x)-2 \text{Li}_2(x) \log (x)+\frac{2 \log ^3(x)}{3}-\log (1-x) \log ^2(x)\right)$
Apply reflection one obtain

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*$S=\int_0^1 \left(\frac{2 f(1-x) \left(\log \left(x^2+1\right)-\log (2)\right)}{1-x}-\frac{f(1-x) \log \left(x^2+1\right)}{x}+\frac{2 f(x) \log (x+1)}{x}-\frac{f(x) (\log (x+1)-\log (2))}{1-x}\right) \, dx$
Which has $4$-admmisible integrand thus solvable via MZVs of level $4$ (see arXiv $2007.03957$)

$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}=3 \pi  \beta (4)+4 \pi  \operatorname{Im} \left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{51 \text{Li}_5\left(\frac{1}{2}\right)}{2}-15 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{\pi ^2 \zeta (3)}{4}+\frac{9 \zeta (5)}{2}-3 \zeta (3) \log ^2(2)-\frac{97 \log ^5(2)}{240}+\frac{41}{144} \pi ^2 \log ^3(2)-\frac{61}{960} \pi ^4 \log (2)$$

