A generalized "Rare" integral involving $\operatorname{Li}_3$

Question

In my previous post, it can be shown that $$\int_{0}^{1} \frac{\operatorname{Li}_2(-x)- \operatorname{Li}_2(1-x)+\ln(x)\ln(1+x)+\pi x\ln(1+x) -\pi x\ln(x)}{1+x^2}\frac{\text{d}x}{\sqrt{1-x^2} } =\frac{\pi^3}{48\sqrt{2} }.$$ But how we verify this? $$\int_{0}^{1} \frac{\operatorname{Li}_3(1-z)+\operatorname{Li}_3 \left ( \frac{1}{1+z} \right ) +\frac{\pi^2}{3}\ln(1+z) -\frac{\pi z}{2}\ln(1+z)^2-\frac{1}{6}\ln(1+z)^3 - \frac{\pi z}{2}\ln(z)^2+\pi z\ln(z)\ln(1+z) }{1+z^2} \frac{\text{d}z}{\sqrt{1-z^2} } =\frac{35\pi\zeta(3)}{64\sqrt{2} }+\frac{\pi^3}{32\sqrt{2} }\ln(2).$$ Where $\operatorname{Li}_3$ is trilogarithm and $\zeta(3)=\operatorname{Li}_3(1)$ in the principal branch. The same method seems not quite powerful. Any suggestion will be appreciated.