Is this closed-form of $\int_0^1 \operatorname{Li}_3^2(x)\,dx$ correct? According to Freitas' paper at page $11$.
$$\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)-\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$
I evaluated the LHS and it is $0.427714784290824$ to me, but the RHS is $-15.8071337213762487846272$. Where am I wrong? Is this closed-form correct? If not, what is the correct closed-form?
Edit. The correct closed-form is
$$\int_0^1 \operatorname{Li}_3^2(x)\,dx = 20-8\zeta(2)-10\zeta(3)+\frac{15}{2}\zeta(4)-2\zeta(2)\zeta(3)+\zeta^2(3).$$
 A: 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.
A: There is a wrong sign, the correct closed form is
$$20-8\,\zeta \left( 2 \right) -10\,\zeta \left( 3 \right) +{\frac {15}{2}}\,\zeta
 \left( 4 \right) -2\,\zeta \left( 2 \right) \zeta \left( 3 \right) +
 \left( \zeta \left( 3 \right)  \right) ^{2}
$$
