On the integral $\int_{0}^{1/2}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz$

This is meant to be just a comment (too large to add it in the comments section)

$$I=\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{2}{3}\sum_{n=2}^{\infty}(-1)^{n-1}\frac{n(\overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (\overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$

A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.

By brutally evaluating such poly-logsine integrals, or those quadratic Euler sums @user 1591719 proposed, one have

• $\int_0^{\frac{1}{2}} \frac{\text{Li}_3(z)}{\sqrt{z (1-z)}} \, dz=-\frac{\pi ^2 C}{6}+C \log ^2(2)-56 \Im(\text{Li}_4(1+i))+4 \log (2) \Im(\text{Li}_3(1+i))+\pi \zeta (3)+\pi \log ^3(2)+\frac{17}{24} \pi ^3 \log (2)+\frac{7 \psi ^{(3)}\left(\frac{1}{4}\right)}{256}-\frac{7 \psi ^{(3)}\left(\frac{3}{4}\right)}{256}$
• $\int_0^{\frac{1}{2}} \frac{\text{Li}_3(1-z)}{\sqrt{z (1-z)}} \, dz=\frac{\pi ^2 C}{6}-C \log ^2(2)+56 \Im(\text{Li}_4(1+i))-4 \log (2) \Im(\text{Li}_3(1+i))+\pi \zeta (3)+\frac{1}{3} \pi \log ^3(2)-\frac{25}{24} \pi ^3 \log (2)-\frac{7 \psi ^{(3)}\left(\frac{1}{4}\right)}{256}+\frac{7 \psi ^{(3)}\left(\frac{3}{4}\right)}{256}$

See here for the methods of calculating quadratic Euler sums. This offers a far more systematic and advanced solution.