Let $\mathcal{S}$ denote the sum of the double infinite series
$$\mathcal{S}:=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{2^{-m}3^{m-n-1}\pi\csc{\left(n\pi\right)}\,\Gamma{\left(m\right)}}{\left(n+1\right)^{2}\,\Gamma{\left(m-n\right)}\,\Gamma{\left(n+1\right)}}\approx-0.0042081,$$
where the following limiting value is used to handle the undefined terms
$$\frac{\pi\csc{\left(n\pi\right)}}{\Gamma{\left(m-n\right)}}=(-1)^{m-1}(n-m)!;~~~\small{m\in\mathbb{Z}\land n\in\mathbb{Z}\land1\le m\le n}.$$
We begin by converting $\mathcal{S}$ into a double integral. Making use of the integral representation for the beta function and using the technique of summation under the integral sign, we find
$$\begin{align}
\mathcal{S}
&=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{2^{-m}3^{m-n-1}\pi\csc{\left(n\pi\right)}\,\Gamma{\left(m\right)}}{\left(n+1\right)^{2}\,\Gamma{\left(m-n\right)}\,\Gamma{\left(n+1\right)}}\\
&=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{2^{-m}3^{m-n-1}(-1)^{m-1}(n-m)!\,\Gamma{\left(m\right)}}{\left(n+1\right)^{2}\,\Gamma{\left(n+1\right)}}\\
&=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{(-1)^{m-1}2^{-m}3^{m-n-1}\,\Gamma{\left(n-m+1\right)}\,\Gamma{\left(m\right)}}{\left(n+1\right)^{2}\,\Gamma{\left(n+1\right)}}\\
&=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{(-1)^{m-1}\left(\frac12\right)^{m}\left(\frac13\right)^{n-m+1}\operatorname{B}{\left(n-m+1,m\right)}}{\left(n+1\right)^{2}}\\
&=\sum_{m=2}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{m-1}\left(\frac12\right)^{m}\left(\frac13\right)^{n+1}\operatorname{B}{\left(n+1,m\right)}}{\left(n+m+1\right)^{2}}\\
&=\sum_{m=2}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m-1}\left(\frac12\right)^{m}\left(\frac13\right)^{n}\operatorname{B}{\left(n,m\right)}}{\left(n+m\right)^{2}}\\
&=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}\left(\frac12\right)^{m+1}\left(\frac13\right)^{n}\operatorname{B}{\left(n,m+1\right)}}{\left(n+m+1\right)^{2}}\\
&=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}\left(\frac12\right)^{m+1}\left(\frac13\right)^{n}}{\left(n+m+1\right)\,n}\,\operatorname{B}{\left(n+1,m+1\right)}\\
&=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}a^{m+1}b^{n}}{\left(n+m+1\right)\,n}\,\operatorname{B}{\left(n+1,m+1\right)};~~~\small{\left[a:=\frac12\land b:=\frac13\right]}\\
&=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}a^{m+1}b^{n}}{\left(n+m+1\right)\,n}\int_{0}^{1}\mathrm{d}t\,t^{n}(1-t)^{m}\\
&=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}a^{m+1}b^{n}}{n}\int_{0}^{1}\mathrm{d}t\,t^{n}(1-t)^{m}\int_{0}^{1}\mathrm{d}x\,x^{m+n}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{m}a^{m+1}b^{n}t^{n}(1-t)^{m}x^{m}x^{n}}{n}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\sum_{m=1}^{\infty}(-1)^{m}a^{m+1}(1-t)^{m}x^{m}\sum_{n=1}^{\infty}\frac{(btx)^{n}}{n}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\sum_{m=1}^{\infty}(-1)^{m}a^{m+1}(1-t)^{m}x^{m}\left[-\ln{\left(1-btx\right)}\right]\\
&=-a\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\ln{\left(1-btx\right)}\sum_{m=1}^{\infty}[(-1)a(1-t)x]^{m}\\
&=-a\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{[(-1)a(1-t)x]}{1-[(-1)a(1-t)x]}\ln{\left(1-btx\right)}\\
&=a\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{a(1-t)x}{1+a(1-t)x}\ln{\left(1-btx\right)}.\\
\end{align}$$
Note that $\frac{a}{1+a}=b\iff a=\frac{b}{1-b}$.
We can further reduce the double integral representation of $\mathcal{S}$ to a single-variable integral as follows:
$$\begin{align}
\mathcal{S}
&=a\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\,\frac{a(1-t)x}{1+a(1-t)x}\ln{\left(1-btx\right)}\\
&=a\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}y\,\frac{a(1-t)y}{t[t+a(1-t)y]}\ln{\left(1-by\right)};~~~\small{\left[tx=y\right]}\\
&=a\int_{0}^{1}\mathrm{d}y\int_{y}^{1}\mathrm{d}t\,\frac{a(1-t)y}{t[t+a(1-t)y]}\ln{\left(1-by\right)}\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\int_{y}^{1}\mathrm{d}t\,\frac{a(1-t)y}{t[t+a(1-t)y]}\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\int_{0}^{1-y}\mathrm{d}u\,\frac{ayu}{(1-u)(1-u+ayu)};~~~\small{\left[1-t=u\right]}\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\int_{0}^{1-y}\mathrm{d}u\,\left[\frac{1}{1-u}-\frac{1}{1-u+ayu}\right]\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\left[\int_{0}^{1-y}\mathrm{d}u\,\frac{1}{1-u}-\int_{0}^{1-y}\mathrm{d}u\,\frac{1}{1-u+ayu}\right]\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\left[-\int_{1}^{y}\mathrm{d}t\,\frac{1}{t}-\frac{1}{\left(1-ay\right)}\int_{0}^{(1-y)(1-ay)}\mathrm{d}v\,\frac{1}{1-v}\right];~~~\small{\left[u=\frac{v}{1-ay}\right]}\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\left[-\ln{\left(y\right)}+\frac{1}{1-ay}\int_{1}^{(1+a-ay)y}\mathrm{d}t\,\frac{1}{t}\right]\\
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\left[-\ln{\left(y\right)}+\frac{\ln{\left((1+a-ay)y\right)}}{1-ay}\right].\\
\end{align}$$
Then,
$$\begin{align}
\mathcal{S}
&=a\int_{0}^{1}\mathrm{d}y\,\ln{\left(1-by\right)}\left[-\ln{\left(y\right)}+\frac{\ln{\left((1+a-ay)y\right)}}{1-ay}\right]\\
&=-a\int_{0}^{1}\mathrm{d}y\,\ln{\left(y\right)}\ln{\left(1-by\right)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1+a-ay\right)}\ln{\left(1-by\right)}}{1-ay}+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(y\right)}\ln{\left(1-by\right)}}{1-ay}\\
&=a\int_{0}^{1}\mathrm{d}y\,y\frac{d}{dy}\left[\ln{\left(y\right)}\ln{\left(1-by\right)}\right];~~~\small{I.B.P.}\\
&~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1+a\right)}\ln{\left(1-by\right)}}{1-ay}+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-\frac{a}{1+a}y\right)}\ln{\left(1-by\right)}}{1-ay}\\
&~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(y\right)}\ln{\left(1-by\right)}}{1-ay}\\
&=a\int_{0}^{1}\mathrm{d}y\,y\left[\frac{\ln{\left(1-by\right)}}{y}-\frac{b\ln{\left(y\right)}}{1-by}\right]\\
&~~~~~+\ln{\left(1+a\right)}\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-by\right)}}{1-ay}+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(1-by\right)}}{1-ay}\\
&~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(y\right)}\ln{\left(1-by\right)}}{1-ay}\\
&=a\int_{0}^{1}\mathrm{d}y\,\left[\ln{\left(1-by\right)}-\frac{by\ln{\left(y\right)}}{1-by}\right]\\
&~~~~~+\ln{\left(1+a\right)}\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-by\right)}}{1-ay}+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(1-by\right)}}{1-ay}\\
&~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(y\right)}+a\ln^{2}{\left(1-by\right)}-a\ln^{2}{\left(\frac{y}{1-by}\right)}}{2\left(1-ay\right)}\\
&=-2a-\frac{a\left(1-b\right)}{b}\ln{\left(1-b\right)}+\frac{a}{b}\operatorname{Li}_{2}{\left(b\right)}\\
&~~~~~+\ln{\left(1+a\right)}\left[\operatorname{Li}_{2}{\left(b\right)}+\operatorname{Li}_{2}{\left(\frac{a}{a-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a-1}\right)}\right]+\operatorname{Li}_{3}{\left(a\right)}\\
&~~~~~+3\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(1-by\right)}}{2\left(1-ay\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(\frac{y}{1-by}\right)}}{2\left(1-ay\right)}\\
&=-2a-\frac{a\left(1-b\right)}{b}\ln{\left(1-b\right)}+\frac{a}{b}\operatorname{Li}_{2}{\left(b\right)}\\
&~~~~~+\ln{\left(1+a\right)}\left[\operatorname{Li}_{2}{\left(b\right)}+\operatorname{Li}_{2}{\left(\frac{a}{a-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a-1}\right)}\right]+\operatorname{Li}_{3}{\left(a\right)}\\
&~~~~~+\frac12\ln^{3}{\left(\frac{1}{1-b}\right)}-\frac32\ln^{2}{\left(\frac{a-b}{a}\right)}\ln{\left(\frac{1-a}{1-b}\right)}\\
&~~~~~+3\ln{\left(\frac{a-b}{a}\right)}\left[\operatorname{Li}_{2}{\left(\frac{b}{a}\right)}-\operatorname{Li}_{2}{\left(\frac{b(1-a)}{a(1-b)}\right)}\right]\\
&~~~~~+3S_{1,2}{\left(\frac{b}{a}\right)}-3S_{1,2}{\left(\frac{b(1-a)}{a(1-b)}\right)}+\operatorname{Li}_{3}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{3}{\left(\frac{a-b}{1-b}\right)}\\
&~~~~~+\ln{\left(1-b\right)}\left[\operatorname{Li}_{2}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{1-b}\right)}\right]+\frac12\ln^{2}{\left(1-b\right)}\ln{\left(1-a\right)},\\
\end{align}$$
where we've made use of a few integration formulas derived in appendices below.
Plugging back in the values $a=\frac12\land b=\frac13$ and making use of the appropriate polylogarithmic functional relations, one eventually obtains the following result:
$$\begin{align}
\mathcal{S}
&=-3\zeta{\left(3\right)}+3\operatorname{Li}_{3}{\left(\frac23\right)}+3\zeta{\left(2\right)}\ln{\left(\frac32\right)}+2\ln{\left(\frac32\right)}\operatorname{Li}_{2}{\left(-\frac12\right)}\\
&~~~~~-\frac32\ln{\left(2\right)}\ln^{2}{\left(\frac32\right)}-\frac12\ln^{3}{\left(\frac32\right)}+\frac32\operatorname{Li}_{2}{\left(\frac13\right)}+\ln{\left(\frac32\right)}-1.\blacksquare\\
\end{align}$$
Appendix 1. For $p<1\land q\le1$,
$$\begin{align}
-\int_{0}^{1}\mathrm{d}y\,\frac{p\ln{\left(1-qy\right)}}{1-py}
&=-\int_{0}^{1}\mathrm{d}y\,\frac{p}{1-py}\int_{0}^{1}\mathrm{d}x\,\frac{(-1)qy}{1-qxy}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{pqy}{\left(1-py\right)\left(1-qxy\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{pqy}{\left(1-py\right)\left(1-qxy\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{pq}{\left(p-qx\right)}\left[\frac{1}{\left(1-py\right)}-\frac{1}{\left(1-qxy\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(p-qx\right)x}\left[qx\int_{0}^{1}\mathrm{d}y\,\frac{p}{\left(1-py\right)}-p\int_{0}^{1}\mathrm{d}y\,\frac{qx}{\left(1-qxy\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(p-qx\right)x}\left[-qx\ln{\left(1-p\right)}+p\ln{\left(1-qx\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(p-qx\right)x}\left[\left(p-qx\right)\ln{\left(1-qx\right)}+qx\ln{\left(\frac{1-qx}{1-p}\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-qx\right)}}{x}+\int_{0}^{1}\mathrm{d}x\,\frac{q\ln{\left(\frac{1-qx}{1-p}\right)}}{\left(p-qx\right)}\\
&=-\operatorname{Li}_{2}{\left(q\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{q\ln{\left(\frac{1-qx}{1-p}\right)}}{\left(p-qx\right)}\\
&=-\operatorname{Li}_{2}{\left(q\right)}+\int_{0}^{q}\mathrm{d}y\,\frac{\ln{\left(\frac{1-y}{1-p}\right)}}{\left(p-y\right)};~~~\small{\left[qx=y\right]}\\
&=-\operatorname{Li}_{2}{\left(q\right)}+\int_{1-q}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{t}{1-p}\right)}}{\left(p-1+t\right)};~~~\small{\left[1-y=t\right]}\\
&=-\operatorname{Li}_{2}{\left(q\right)}+\int_{\frac{1-q}{1-p}}^{\frac{1}{1-p}}\mathrm{d}u\,\frac{\ln{\left(u\right)}}{u-1};~~~\small{\left[t=(1-p)u\right]}\\
&=-\operatorname{Li}_{2}{\left(q\right)}+\int_{\frac{p-q}{p-1}}^{\frac{p}{p-1}}\mathrm{d}v\,\frac{\ln{\left(1-v\right)}}{v};~~~\small{\left[1-u=v\right]}\\
&=-\operatorname{Li}_{2}{\left(q\right)}-\operatorname{Li}_{2}{\left(\frac{p}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p-q}{p-1}\right)}.\\
\end{align}$$
Appendix 2. For $0<b<a<1$,
$$\begin{align}
\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(1-by\right)}}{2\left(1-ay\right)}
&=\frac12\int_{0}^{b}\mathrm{d}t\,\frac{a\ln^{2}{\left(1-t\right)}}{b-at};~~~\small{\left[by=t\right]}\\
&=\frac12\int_{1-b}^{1}\mathrm{d}u\,\frac{a\ln^{2}{\left(u\right)}}{b-a(1-u)};~~~\small{\left[1-t=u\right]}\\
&=-\frac12\int_{1-b}^{1}\mathrm{d}u\,\frac{a\ln^{2}{\left(u\right)}}{a-b-au}\\
&=\frac12\int_{\frac{a(1-b)}{a-b}}^{\frac{a}{a-b}}\mathrm{d}v\,\frac{\ln^{2}{\left(\frac{(a-b)v}{a}\right)}}{v-1};~~~\small{\left[u=\frac{(a-b)v}{a}\right]}\\
&=\frac12\int_{\frac{a-b}{a}}^{\frac{a-b}{a(1-b)}}\mathrm{d}w\,\frac{\ln^{2}{\left(\frac{a-b}{aw}\right)}}{w\left(1-w\right)};~~~\small{\left[v=\frac{1}{w}\right]}\\
&=\frac12\int_{\frac{a-b}{a}}^{\frac{a-b}{a(1-b)}}\mathrm{d}w\,\frac{\ln^{2}{\left(\frac{a-b}{aw}\right)}}{w}+\frac12\int_{\frac{a-b}{a}}^{\frac{a-b}{a(1-b)}}\mathrm{d}w\,\frac{\ln^{2}{\left(\frac{a-b}{aw}\right)}}{1-w}\\
&=\frac12\int_{1}^{\frac{1}{1-b}}\mathrm{d}x\,\frac{\ln^{2}{\left(x\right)}}{x};~~~\small{\left[\frac{aw}{a-b}=x\right]}\\
&~~~~~+\frac12\int_{\frac{a-b}{a}}^{\frac{a-b}{a(1-b)}}\mathrm{d}w\,\frac{\left[\ln{\left(\frac{a-b}{a}\right)}-\ln{\left(w\right)}\right]^{2}}{1-w}\\
&=\frac16\ln^{3}{\left(\frac{1}{1-b}\right)}\\
&~~~~~+\int_{\frac{b(1-a)}{a(1-b)}}^{\frac{b}{a}}\mathrm{d}x\,\frac{1}{2x}\bigg{[}\ln^{2}{\left(\frac{a-b}{a}\right)}-2\ln{\left(\frac{a-b}{a}\right)}\ln{\left(1-x\right)}\\
&~~~~~+\ln^{2}{\left(1-x\right)}\bigg{]};~~~\small{\left[w=1-x\right]}\\
&=\frac16\ln^{3}{\left(\frac{1}{1-b}\right)}+\frac12\ln^{2}{\left(\frac{a-b}{a}\right)}\int_{\frac{b(1-a)}{a(1-b)}}^{\frac{b}{a}}\mathrm{d}x\,\frac{1}{x}\\
&~~~~~-\ln{\left(\frac{a-b}{a}\right)}\int_{\frac{b(1-a)}{a(1-b)}}^{\frac{b}{a}}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x}+\int_{\frac{b(1-a)}{a(1-b)}}^{\frac{b}{a}}\mathrm{d}x\,\frac{\ln^{2}{\left(1-x\right)}}{2x}\\
&=\frac16\ln^{3}{\left(\frac{1}{1-b}\right)}-\frac12\ln^{2}{\left(\frac{a-b}{a}\right)}\ln{\left(\frac{1-a}{1-b}\right)}\\
&~~~~~+\ln{\left(\frac{a-b}{a}\right)}\left[\operatorname{Li}_{2}{\left(\frac{b}{a}\right)}-\operatorname{Li}_{2}{\left(\frac{b(1-a)}{a(1-b)}\right)}\right]\\
&~~~~~+S_{1,2}{\left(\frac{b}{a}\right)}-S_{1,2}{\left(\frac{b(1-a)}{a(1-b)}\right)}.\\
\end{align}$$
Appendix 3. Suppose $a<1\land b<1$, and set $p:=\frac{b}{b-1}\land q:=\frac{a-b}{1-b}$. Then,
$$\begin{align}
-\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(\frac{y}{1-by}\right)}}{2\left(1-ay\right)}
&=-\frac12\int_{0}^{1}\mathrm{d}y\,\frac{a\ln^{2}{\left(\frac{y}{1-by}\right)}}{1-ay}\\
&=-\frac12\int_{0}^{\frac{1}{1-b}}\mathrm{d}t\,\frac{1}{\left(1+bt\right)^{2}}\cdot\frac{a\ln^{2}{\left(t\right)}}{1-a\left(\frac{t}{1+bt}\right)};~~~\small{\left[\frac{y}{1-by}=t\right]}\\
&=-\frac12\int_{0}^{\frac{1}{1-b}}\mathrm{d}t\,\frac{1}{\left(1+bt\right)}\cdot\frac{a\ln^{2}{\left(t\right)}}{1-\left(a-b\right)t}\\
&=-\frac12\int_{0}^{\frac{1}{1-b}}\mathrm{d}t\,\left[\frac{b}{1+bt}+\frac{\left(a-b\right)}{1-\left(a-b\right)t}\right]\ln^{2}{\left(t\right)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{1}{1-b}\left[\frac{b}{1+\left(\frac{b}{1-b}\right)u}+\frac{\left(a-b\right)}{1-\left(\frac{a-b}{1-b}\right)u}\right]\\
&~~~~~\times\ln^{2}{\left(\frac{u}{1-b}\right)};~~~\small{\left[t=\frac{u}{1-b}\right]}\\
&=\frac12\int_{0}^{1}\mathrm{d}u\,\left[\frac{\left(\frac{b}{b-1}\right)}{1-\left(\frac{b}{b-1}\right)u}-\frac{\left(\frac{a-b}{1-b}\right)}{1-\left(\frac{a-b}{1-b}\right)u}\right]\ln^{2}{\left(\frac{u}{1-b}\right)}\\
&=\frac12\int_{0}^{1}\mathrm{d}u\,\left[\frac{\left(\frac{b}{b-1}\right)}{1-\left(\frac{b}{b-1}\right)u}-\frac{\left(\frac{a-b}{1-b}\right)}{1-\left(\frac{a-b}{1-b}\right)u}\right]\ln^{2}{\left(u\right)}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\left[\frac{\left(\frac{b}{b-1}\right)}{1-\left(\frac{b}{b-1}\right)u}-\frac{\left(\frac{a-b}{1-b}\right)}{1-\left(\frac{a-b}{1-b}\right)u}\right]\ln{\left(1-b\right)}\ln{\left(u\right)}\\
&~~~~~+\frac12\int_{0}^{1}\mathrm{d}u\,\left[\frac{\left(\frac{b}{b-1}\right)}{1-\left(\frac{b}{b-1}\right)u}-\frac{\left(\frac{a-b}{1-b}\right)}{1-\left(\frac{a-b}{1-b}\right)u}\right]\ln^{2}{\left(1-b\right)}\\
&=\frac12\int_{0}^{1}\mathrm{d}u\,\left[\frac{p}{1-pu}-\frac{q}{1-qu}\right]\ln^{2}{\left(u\right)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}u\,\left[\frac{p}{1-pu}-\frac{q}{1-qu}\right]\ln{\left(1-p\right)}\ln{\left(u\right)}\\
&~~~~~+\frac12\int_{0}^{1}\mathrm{d}u\,\left[\frac{p}{1-pu}-\frac{q}{1-qu}\right]\ln^{2}{\left(1-p\right)}\\
&=\frac12\int_{0}^{1}\mathrm{d}u\,\frac{p\ln^{2}{\left(u\right)}}{1-pu}-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{q\ln^{2}{\left(u\right)}}{1-qu}\\
&~~~~~+\ln{\left(1-p\right)}\left[\int_{0}^{1}\mathrm{d}u\,\frac{p\ln{\left(u\right)}}{1-pu}-\int_{0}^{1}\mathrm{d}u\,\frac{q\ln{\left(u\right)}}{1-qu}\right]\\
&~~~~~+\frac12\ln^{2}{\left(1-p\right)}\left[\int_{0}^{1}\mathrm{d}u\,\frac{p}{1-pu}-\int_{0}^{1}\mathrm{d}u\,\frac{q}{1-qu}\right]\\
&=\operatorname{Li}_{3}{\left(p\right)}-\operatorname{Li}_{3}{\left(q\right)}-\ln{\left(1-p\right)}\left[\operatorname{Li}_{2}{\left(p\right)}-\operatorname{Li}_{2}{\left(q\right)}\right]\\
&~~~~~-\frac12\ln^{2}{\left(1-p\right)}\left[\ln{\left(1-p\right)}-\ln{\left(1-q\right)}\right]\\
&=\operatorname{Li}_{3}{\left(p\right)}-\operatorname{Li}_{3}{\left(q\right)}-\ln{\left(1-p\right)}\left[\operatorname{Li}_{2}{\left(p\right)}-\operatorname{Li}_{2}{\left(q\right)}\right]\\
&~~~~~+\frac12\ln^{2}{\left(1-p\right)}\ln{\left(\frac{1-q}{1-p}\right)}\\
&=\operatorname{Li}_{3}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{3}{\left(\frac{a-b}{1-b}\right)}+\ln{\left(1-b\right)}\left[\operatorname{Li}_{2}{\left(\frac{b}{b-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{1-b}\right)}\right]\\
&~~~~~+\frac12\ln^{2}{\left(1-b\right)}\ln{\left(1-a\right)}.\\
\end{align}$$
