Is this closed-form of $\int_0^1 \operatorname{Li}_3^2(x)\,dx$ correct?

The trilogarithm has the antiderivative,

$$\int\operatorname{Li}_{3}{\left(x\right)}\,\mathrm{d}x=x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}+\color{grey}{constant}.$$

So, integrating by parts we find after integrating all the terms with simple antiderivatives:

$$\begin{align} \int_{0}^{1}\operatorname{Li}_{3}^2{\left(x\right)}\,\mathrm{d}x &=\left[\left(x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}\right)\operatorname{Li}_{3}{\left(x\right)}\right]_{0}^{1}\\ &~~~~~ -\int_{0}^{1}\left(x\operatorname{Li}_{3}{\left(x\right)}-x\operatorname{Li}_{2}{\left(x\right)}+x-x\ln{\left(1-x\right)}+\ln{\left(1-x\right)}\right)\frac{\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\zeta{(3)}-\zeta{(2)}\zeta{(3)}+\zeta{(3)}^2-\int_{0}^{1}\operatorname{Li}_{3}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x+\int_{0}^{1}\operatorname{Li}_{2}^2{\left(x\right)}\,\mathrm{d}x\\ &~~~~ -\int_{0}^{1}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x+\int_{0}^{1}\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=4-2\zeta{(2)}-\zeta{(3)}+\frac54\zeta{(4)}-\zeta{(2)}\zeta{(3)}+\zeta{(3)}^2\\ &~~~~ +\int_{0}^{1}\operatorname{Li}_{2}^2{\left(x\right)}\,\mathrm{d}x-\int_{0}^{1}\operatorname{Li}_{3}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}\,\mathrm{d}x.\\ \end{align}$$

For the remaining two integrals, you may substitute the values reported in the paper you cited.

There is a wrong sign, the correct closed form is

$$20-8\,\zeta \left( 2 \right) -10\,\zeta \left( 3 \right) +{\frac {15}{2}}\,\zeta \left( 4 \right) -2\,\zeta \left( 2 \right) \zeta \left( 3 \right) + \left( \zeta \left( 3 \right) \right) ^{2} $$