Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$

While going through the recent questions concerning tagged polylogarithms I stumbled upon this post which asks for a concrete evaluating of a polylogarithmic integral. However the post also states the equality

$$\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\,\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)\tag1$$

Together with the comment "It is not difficult to show that". Since I know the author of this post $-$ Jack D'Aurizio $-$ is familiar with integrals of this type I guess, indeed, for himself it is easily done. However, I have problems proving $(1)$

As we are dealing with an integral involving a Polylogarithm I thought about applying IBP to get rid of the the Polylogarithm. But I am not sure about the right choice of $u$ and $\mathrm dv$, respectively. My first guess was simply $u=\operatorname{Li}_3(1-z)$ and therefore $\displaystyle\mathrm dv=\frac1{\sqrt{z(1-z)}}$. From hereon the first problem occurs: integrating $\mathrm dv$. There are at least the two possibilities $v=\sin^{-1}(2z-1)$ and $v=2\sin^{-1}(\sqrt{z})$ which both lead to the same $\mathrm dv$ but on the other hand imply different results for the first IBP step. To be precise

$$\begin{align} \tag{1}\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\,\mathrm dz&=\left[\operatorname{Li}_3(1-z)\sin^{-1}(2z-1)\right]_0^1-\int_0^1\sin^{-1}(2z-1)\frac{\operatorname{Li}_2(z)}{z}\,\mathrm dz\\ &=\color{red}{\frac38\pi\zeta(3)}-\int_0^1\sin^{-1}(2z-1)\frac{\operatorname{Li}_2(z)}{z}\,\mathrm dz\\ \tag{2}\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\,\mathrm dz&=\left[\operatorname{Li}_3(1-z)2\sin^{-1}(\sqrt{z})\right]_0^1-\int_0^12\sin^{-1}(\sqrt{z})\frac{\operatorname{Li}_2(z)}z\,\mathrm dz\\ &=\color{red}{0}-\int_0^12\sin^{-1}(\sqrt{z})\frac{\operatorname{Li}_2(z)}{z}\,\mathrm dz \end{align}$$

I am in favor of the first option since it contains the value $\pi\zeta(3)$ but with the wrong coefficient. However, I am not capable of evaluating the remaining integrals which involves a combination of the inverse sine function and the Dilogarithm. Again I thought about IBP but I am totally confused what to choose as $u$ and $\mathrm dv$. Therefore I think I am on the wrong tack.

I have dealed with polylogarithmic and logarithmic integrals before but the square roots are causing me trouble. I tried to absorb at least the $\sqrt{1-z}$ within the Trilogarithm and then doing IBP which results in

$$\begin{align} \int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz&=\int_0^1\sum_{n=1}^{\infty}\frac{(1-z)^{n-1/2}}{n^3}\frac{dz}{\sqrt{z}}\\ &=\left[\sum_{n=1}^{\infty}\frac{(1-z)^{n-1/2}}{n^3}2\sqrt{z}\right]_0^1-\int_0^1\frac{\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)}{(1-z)^{3/2}}\sqrt{z}\mathrm dz\\ &=\color{red}{0}-\int_0^1\frac{\operatorname{Li}_3(1-z)-2\operatorname{Li}_2(1-z)}{(1-z)^{3/2}}\sqrt{z}\mathrm dz \end{align}$$

I am not sure whether this is helpful at all or if it does not make the whole problem more complicated. Honestly speaking, I am lost right now and do not know how to proceed of how to approach to the given equality at all.

Could someone explain me how to proceed with the given integrals including integrands combined out of inverse sine and polylogarithmic functions? Are these integrals even solvable; when yes how (maybe without using the given the given integral)? Or was my whole approach nonsense and another attempt is needed here? You can also share a link or refer to another post here on MSE in case I have overlooked something.

Thanks in advance!

Using the well-known identity: $${\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}}) + {\rm{L}}{{\rm{i}}_3}(1 - x) + {\rm{L}}{{\rm{i}}_3}(x) = \zeta (3) + \frac{{{\pi ^2}}}{6}\ln (1 - x) - \frac{1}{2}\ln x{\ln ^2}(1 - x) + \frac{1}{6}{\ln ^3}(1 - x)$$ we obtain (the integral on RHS can be easily evaluated by differentiating beta function): $$2\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(1 - x)}}{{\sqrt {x(1 - x)} }}dx} + \int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = - 2\pi \zeta (3) + \frac{8}{3}\pi {\ln ^3}2 - \frac{2}{3}{\pi ^3}\ln 2$$ By transformation $u=x/(1-x)$, we have $$\int_0^1 {\frac{{{\rm{L}}{{\rm{i}}_3}(\frac{{ - x}}{{1 - x}})}}{{\sqrt {x(1 - x)} }}dx} = \int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - u)}}{{(1 + u)\sqrt u }}du}$$ I claim this integral is $-6\pi \zeta(3)$.

To establish this value, it suffices to show, with $\zeta(\cdot,\cdot)$ Hurwitz zeta function, $$\int_0^\infty {\frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}}{x^{s - 1}}dx} = \frac{\pi }{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right] \qquad 0<s<1$$ by Mellin inversion theorem, this in turn is equivalent to, (which applies as the function tends to $0$ uniformly in the vertical strip $0<\Re(s)<1$ thanks to the $\csc(s\pi)$ factor) for an instance of $c$ with $0<c<1$: $$\tag{1} \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}} = \frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} \qquad x>0$$ Note that both sides of $(1)$ is an analytic function for $\Re(x) > 0$, hence it suffices to consider the case when $0<x<1$. When this is the case, we can draw a vertical semicircle on the left half-plane, with vertices $c \pm i\infty$, then the integral on the semicircle tends to $0$, calculating residues at $-1,-2,\cdots$ gives $$\begin{aligned}\frac{1}{{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\frac{{\pi {x^{ - s}}}}{{\sin (\pi s)}}\left[ {\zeta (3) - \zeta (3,1 - s)} \right]ds} &= \sum\limits_{n = 1}^\infty {{{( - x)}^n}\left[ {\zeta (3) - \zeta (3,1 + n)} \right]} \\ &=\sum\limits_{n = 1}^\infty {{{( - x)}^n}\sum\limits_{k = 1}^n {\frac{1}{{{k^3}}}} } = \frac{{{\rm{L}}{{\rm{i}}_3}( - x)}}{{1 + x}} \end{aligned}$$ where we exchanged two order of summations, completing the proof.

$$\int_{0}^{1}\frac{\text{Li}_3(1-z)}{\sqrt{z(1-z)}}\,dz=\int_{0}^{1}\frac{\text{Li}_3(z)}{\sqrt{z(1-z)}}\,dz=2\int_{0}^{1}\frac{\text{Li}_3(u^2)}{\sqrt{1-u^2}}\,du=2\int_{0}^{\pi/2}\text{Li}_3(\sin^2\theta)\,d\theta $$ by the very definition of $\text{Li}_3$, plus the identity $\int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n} $, equals $$ \pi\sum_{n\geq 1}\frac{\binom{2n}{n}}{n^3\cdot 4^n}, $$ i.e. a rather innocent hypergeometric series, namely $2\pi\cdot\phantom{}_5 F_4\left(1,1,1,1,\frac{3}{2};2,2,2,2;1\right)$, which can be evaluated in many ways, for instance through Fourier-Legendre series expansions, or by writing the thing above in terms of $$ \int_{0}^{1}\frac{\log^2(z)\,dz}{\sqrt{1-z}},\qquad \int_{0}^{1}\frac{\log^3(z)\,dz}{\sqrt{1-z}} $$ which clearly are the second and third derivatives of a Beta function.
In "higher" terms, any chain of identities of the $$ \int f(x)\omega(x)\,dx = \langle f,\omega\rangle \stackrel{\begin{array}{c}\text{series}\\[-0.2cm]\text{rearrengement}\end{array}}{=} \langle \tilde f,\tilde\omega\rangle=\int \tilde f(x)\tilde \omega(x)\,dx $$ kind induces a transformation $f\mapsto\tilde{f}$ which generalizes the binomial transform.
In our case $\text{Li}_3$ is essentially mapped into $\log^3$.