Prove that $\int_0^1\frac{\operatorname{Li}_3(1-z)}{\sqrt{z(1-z)}}\mathrm dz=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3)$
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 $\log^3$.
Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4.
$$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=2$ we get
$$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2(u)}{1-xu}\ du$$
Then \begin{align} I&=\int_0^1\frac{\operatorname{Li}_{3}(1-x)}{\sqrt{x(1-x)}}\ dx\overset{1-x\ \mapsto\ x}{=}\int_0^1\frac{\operatorname{Li}_{3}(x)}{\sqrt{x(1-x)}}\ dx\\ &=\int_0^1\frac{1}{\sqrt{x(1-x)}}\left(\frac12\int_0^1\frac{x\ln^2(u)}{1-xu}\ du\right)\ dx\\ &=\frac12\int_0^1\ln^2(u)\left(\int_0^1\frac{x}{(1-xu)\sqrt{x(1-x)}}\ dx\right)\ du\\ &=\frac12\int_0^1\frac{\ln^2(u)}{u}\left(\frac{\pi}{\sqrt{1-u}}-\pi\right)\ du\quad \text{apply integration by parts}\\ &=-\frac{\pi}{12}\int_0^1\ln^3(u)\ (1-u)^{-3/2}\ du\\ &=-\frac{\pi}{12}\lim_{\large\alpha\ \mapsto 1}\frac{\partial^3}{\partial \alpha^3}\text{B}\left(\alpha,-\frac12\right)\\ &=-\frac{\pi}{12}\left(-2\pi^2\left(\gamma+\psi\left(\frac12\right)\right)+2\left(\gamma+\psi\left(\frac12\right)\right)^3-2\left(\psi^{(2)}(1)-\psi^{(2)}\left(\frac12\right)\right)\right)\\ &=-\frac{\pi}{12}\left(-2\pi^2\left(-2\ln2\right)+2\left(-2\ln2\right)^3-2\left(12\zeta(3)\right)\right)\\ &=-\frac{\pi^3}{3}\log 2+\frac{4\pi}3\log^3 2+2\pi\zeta(3) \end{align}