This question is related to the equivalence of formulas (1) and (2) below where formula (1) is from a post on the Harmonic Series Facebook group and formula (2) is based on evaluation of the integral using Mathematica. The right sides of formulas (2) and (3) below both seem to be equivalent numerically.
$$\int_0^{\frac{1}{2}}\frac{\text{Li}_2(-x)}{1-x}\,dx=-\text{Li}_3\left(-\frac{1}{2}\right)-\frac{13}{24}\zeta(3)\tag{1}$$
$$\int_0^{\frac{1}{2}}\frac{\text{Li}_2(-x)}{1-x}\,dx=-\text{Li}_3\left(\frac{1}{3}\right)+\text{Li}_3\left(\frac{2}{3}\right)+\text{Li}_3\left(-\frac{1}{3}\right)+\frac{1}{6} \left(6 \text{Li}_2\left(-\frac{1}{2}\right) (\log (2)-i \pi )+6 \text{Li}_2\left(\frac{2}{3}\right) \left(\log \left(\frac{9}{2}\right)+i \pi \right)+6 \text{Li}_2\left(-\frac{1}{3}\right) \log (3)-i \pi ^3+2 \log ^3(2)-3 i \pi \log ^2(2)+3 i \pi \log ^2(3)+4 \log ^2(3) \log \left(\frac{9}{8}\right)-\pi ^2 \log (3)\right)+\frac{5 \zeta (3)}{8}\tag{2}$$
Equating the right sides of formulas (1) and (2), rearranging terms, and simplifying a bit leads the following formula for $\zeta(3)$.
$$\zeta(3)=\frac{1}{7} \left(-6 \text{Li}_3\left(-\frac{1}{2}\right)+6 \text{Li}_3\left(\frac{1}{3}\right)-6 \text{Li}_3\left(\frac{2}{3}\right)-6 \text{Li}_3\left(-\frac{1}{3}\right)+\text{Li}_2\left(-\frac{1}{2}\right) (-\log (64)+6 i \pi )+6 \left(\text{Li}_2\left(\frac{1}{3}\right)-\text{Li}_2\left(-\frac{1}{3}\right)\right) \log (3)-\text{Li}_2\left(\frac{2}{3}\right) \left(12 \coth ^{-1}(5)+6 i \pi \right)+i \pi ^3-2 \log ^3(2)+3 i \pi \left(\log ^2(2)-\log ^2(3)\right)+\log ^2(3) \log \left(\frac{64}{9}\right)\right)\tag{3}$$
Question: Can formula (3) above be proven or simplified further?
There are a number of identities for the Dilogarithm $\text{Li}_2(z)$ and Trilogarithm $\text{Li}_3(z)$ functions, and I noticed some of the Trilogarithm identities involve $\zeta(3)$ terms, but formula (3) is as far as I've gotten at this point. I was hoping to at least reduce the number of $\text{Li}_k(z)$ terms in formula (3) above.
The formula
$$\int_0^{\frac{1}{2}}\frac{\text{Li}_2(-x)}{1-x}\,dx=-\frac{1}{6} \log^3 (2) -\frac{3}{2} \log (2) \log^2 (3) + \frac{4}{3} \log^3 (3) -\frac{\pi^2}{6} \log (6) + \log (3) \operatorname{Li}_2 \left(-\frac{1}{3}\right)+\log (9) \operatorname{Li}_2 \left(\frac{2}{3}\right)+\frac{1}{4} \operatorname{Li}_3 \left(\frac{1}{9}\right)-2\operatorname{Li}_3 \left( \frac{1}{3}\right)+\operatorname{Li}_3 \left(\frac{2}{3}\right)+\frac{5}{8}\zeta (3)\tag{4}$$
provided by @KStarGamer in a comment below seems to be equivalent numerically and leads to the simpler formula
$$\zeta (3)=-\frac{6 \text{Li}_3\left(-\frac{1}{2}\right)}{7}+\frac{12 \text{Li}_3\left(\frac{1}{3}\right)}{7}-\frac{6 \text{Li}_3\left(\frac{2}{3}\right)}{7}-\frac{3 \text{Li}_3\left(\frac{1}{9}\right)}{14}-\frac{6}{7} \text{Li}_2\left(\frac{2}{3}\right) \log (9)-\frac{6}{7} \text{Li}_2\left(-\frac{1}{3}\right) \log (3)+\frac{1}{7} \left(\log ^3(2)-8 \log ^3(3)+9 \log ^2(3) \log (2)+\pi ^2 \log (6)\right)\tag{5}$$
which is perhaps easier to investigate.
Best Answer
Using $\operatorname{Li}_2\left(1-z\right)+\operatorname{Li}_2\left(1-1/z\right)=-(\log^2z)/2$, \begin{align}\operatorname{Li}_2\left(1/4\right)+\operatorname{Li}_2\left(-1/3\right)&=-\frac{\log^2(3/4)}2\\\operatorname{Li}_2\left(2/3\right)+\operatorname{Li}_2\left(-2\right)&=-\frac{\log^23}2\end{align} so that $\displaystyle2\operatorname{Li}_2\left(2/3\right)+\operatorname{Li}_2\left(-1/3\right)=-\log^23-\frac{\log^2(3/4)}2-2\operatorname{Li}_2\left(-2\right)-\operatorname{Li}_2\left(1/4\right)$.
Using $\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(-z\right)=\operatorname{Li}_2\left(z^2\right)/2$ and $\operatorname{Li}_2\left(z\right)+\operatorname{Li}_2\left(1/z\right)=-\pi^2/6-\log^2(-z)/2$, \begin{align}2\operatorname{Li}_2\left(-2\right)+\operatorname{Li}_2\left(1/4\right)&=\operatorname{Li}_2\left(4\right)-2\operatorname{Li}_2\left(2\right)+\operatorname{Li}_2\left(1/4\right)\\&=-\frac{\pi^2}6-\frac{\log^2(-4)}2-\frac{\pi^2}2+2i\pi\log2,\end{align} Since $\operatorname{Li}_2\left(2\right)=\pi^2/4-i\pi\log2$, combining the identities above gives $$2\operatorname{Li}_2\left(2/3\right)+\operatorname{Li}_2\left(-1/3\right)=\frac{\pi^2}6-\frac32\log^23+2\log2\log3.$$ Thus it suffices to show that \begin{align}\zeta(3)&=\small\frac3{14}A-\frac67(2\operatorname{Li}_2\left(2/3\right)+\operatorname{Li}_2\left(-1/3\right))\log3+\frac{\log^32-8\log^33+9\log^23\log2+\pi^2\log6}7\\&=\frac3{14}A+\frac{\log^33-3\log^23\log2+\pi^2\log2+\log^23}7\end{align} where $A=-4\operatorname{Li}_3\left(-1/2\right)+8\operatorname{Li}_3\left(1/3\right)-4\operatorname{Li}_3\left(2/3\right)-\operatorname{Li}_3\left(1/9\right)$.
Using $\operatorname{Li}_3\left(z\right)+\operatorname{Li}_3\left(-z\right)=\operatorname{Li}_3\left(z^2\right)/4$, \begin{align}8\operatorname{Li}_3\left(1/3\right)-\operatorname{Li}_3\left(1/9\right)&=4\operatorname{Li}_3\left(1/3\right)-4\operatorname{Li}_3\left(-1/3\right)\end{align} so $A=4B$ where $B=-\operatorname{Li}_3\left(-1/2\right)+\operatorname{Li}_3\left(1/3\right)-\operatorname{Li}_3\left(2/3\right)-\operatorname{Li}_3\left(-1/3\right)$.
Using $\small\operatorname{Li}_3\left(z\right)+\operatorname{Li}_3\left(1-z\right)+\operatorname{Li}_3\left(1-1/z\right)=\zeta(3)+(\log^3z)/6+\pi^2(\log^2z)/6-\log^2z\log(1-z)/2$, \begin{align}\operatorname{Li}_3\left(2/3\right)+\operatorname{Li}_3\left(1/3\right)+\operatorname{Li}_3\left(-1/2\right)=\zeta(3)+\frac{\log^3(2/3)}6+\frac{\pi^2\log(2/3)}6+\frac{\log^2(2/3)\log3}2\end{align} and after several logarithmic manipulations, we have \begin{align}\zeta(3)&=\frac67B+\frac{\log^33-3\log^23\log2+\pi^2\log2+\log^23}7\\&=\frac67\left(2\operatorname{Li}_3\left(1/3\right)-\operatorname{Li}_3\left(-1/3\right)-\zeta(3)\right)+\frac{\pi^2\log3-\log^33}7\end{align} or equivalently, $$13\zeta(3)=6(2\operatorname{Li}_3\left(1/3\right)-\operatorname{Li}_3\left(-1/3\right))+\pi^2\log3-\log^33$$ which has a simple derivation.