TITLE: Evaluate $\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx$
QUESTION [7 upvotes]: We know that (G&R): $$\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \, dx=2 \pi  ((a+b) \log (a+b)-a \log (a)-b \log (b))  $$  Where $a, b>0$. It can be proved by using Feynman's trick (i.e. differentiate w.r.t parameters) twice. The problem is: What are the closed-forms of following generalized integral: $$I=\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx $$
I've weakened the original problem and would like you to give some suggestions on it. Also this is related. Thank you.

REPLY [4 votes]: Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(a,b,c\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}.\tag{1}$$
Our objective is to derive a closed-form expression for $\mathcal{I}$ in terms of polylogarithms and elementary functions.
This function $\mathcal{I}$ has two important properties that follow almost immediately from definition $(1)$. First, it is symmetric under any permutation of its three parameters. Second, it obeys the scaling relation
$$\mathcal{I}{\left(a,b,c\right)}=d\,\mathcal{I}{\left(d^{-1}a,d^{-1}b,d^{-1}c\right)};~~~\small{\left(a,b,c,d\right)\in\mathbb{R}_{>0}^{4}}.$$

Consider the following derivative, which holds for fixed but arbitrary $x\in\mathbb{R}_{>0}$:
$$\frac{\partial}{\partial y}\ln{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{x^{2}}\cdot\frac{1}{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{\left(x^{2}+y^{2}\right)}.$$
Integrating both sides, we obtain the following integral representation:
$$\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}=\int_{0}^{a}\mathrm{d}y\,\frac{2y}{\left(x^{2}+y^{2}\right)};~~~\small{a\in\mathbb{R}_{>0}\land x\in\mathbb{R}_{>0}}.\tag{2}$$

Suppose $\left(a,b,c\right)\in\mathbb{R}_{>0}^{3}$. Using the integral representation in $(4)$ above to rewrite $\mathcal{I}$ as a multiple integral and changing the order of integration, we find that $\mathcal{I}$ can be expressed as a triple integral as follows:
$$\begin{align}
\mathcal{I}{\left(a,b,c\right)}
&=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\,\frac{2p}{\left(x^{2}+p^{2}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{2q}{\left(x^{2}+q^{2}\right)}\int_{0}^{c}\mathrm{d}r\,\frac{2r}{\left(x^{2}+r^{2}\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\
&~~~~~+\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\
&~~~~~+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{p}\\
&~~~~~+\frac{\left(r^{2}-p^{2}\right)}{q}+\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\int_{0}^{\infty}\mathrm{d}y\,\frac{1}{\left(y^{2}+1\right)}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}-\frac{\left(q^{2}-r^{2}\right)}{p}\\
&~~~~~-\frac{\left(r^{2}-p^{2}\right)}{q}-\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\frac{\pi}{2}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}.\tag{3}\\
\end{align}$$

Suppose $\left(a,b\right)\in\mathbb{R}_{>0}^{2}$, and consider the case where the third parameter of $\mathcal{I}$ is set equal to one. The integrand of our triple integral is a rational function, so at least one of the three integrations can be done in elementary terms. We find
$$\begin{align}
\mathcal{I}{\left(a,b,1\right)}
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{1}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\frac{\left(p-q\right)\left(p+q+r\right)}{\left(q+r\right)\left(r+p\right)}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\left[\frac{p}{\left(q+r\right)}-\frac{q}{\left(r+p\right)}\right]\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(q+r\right)}-q\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(r+p\right)}\right]\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\
&=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{2}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\
&=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{1+q}{q}\right)}+\ln{\left(\frac{1+p}{p}\right)}}{\left(p+q\right)}+\frac{\ln{\left(\frac{1+q}{q}\right)}-\ln{\left(\frac{1+p}{p}\right)}}{\left(p-q\right)}\bigg{]}\\
&=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}.\tag{4}\\
\end{align}$$

Set $A:=\frac{a}{1+a}\in\left(0,1\right)\land B:=\frac{b}{1+b}\in\left(0,1\right)$. Continuing from the double integral in the last line of $(4)$ above,
$$\begin{align}
\frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi}
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}\\
&=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\
&=\int_{0}^{\frac{a}{1+a}}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{\frac{b}{1+b}}\mathrm{d}y\,\frac{1}{\left(1-y\right)^{2}}\cdot\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(\frac{x}{1-x}-\frac{y}{1-y}\right)};~~~\small{\left[p=\frac{x}{1-x}\land q=\frac{y}{1-y}\right]}\\
&~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\
&=\int_{0}^{A}\mathrm{d}x\int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-x\right)\left(x-y\right)\left(1-y\right)}\\
&~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\int_{0}^{a}\mathrm{d}p\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\
&=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\frac{\left(1-x\right)}{\left(x-y\right)\left(1-y\right)}\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\
&~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{1}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\int_{0}^{a}\mathrm{d}p\,\frac{1}{\left(p+q\right)}\\
&=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{1}{\left(x-y\right)}-\frac{1}{\left(1-y\right)}\right]\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\
&~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\
&~~~~~-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\
&~~~~~+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}.\\
\end{align}$$
Given $0<x<1\land0<B<1$,
$$\begin{align}
\int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}
&=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,x\frac{\ln{\left(x\right)}-\ln{\left(xt\right)}}{\left(x-xt\right)};~~~\small{\left[y=xt\right]}\\
&=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,\frac{(-1)\ln{\left(t\right)}}{\left(1-t\right)}\\
&=\int_{1-\frac{B}{x}}^{1}\mathrm{d}u\,\frac{(-1)\ln{\left(1-u\right)}}{u};~~~\small{\left[t=1-u\right]}\\
&=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)},\\
\end{align}$$
and
$$\begin{align}
\int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}
&=\int_{1-B}^{1}\mathrm{d}t\,\frac{\ln{\left(x\right)}-\ln{\left(1-t\right)}}{t};~~~\small{\left[y=1-t\right]}\\
&=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-B\right)}-\ln{\left(1-B\right)}\ln{\left(x\right)}.\\
\end{align}$$
Then,
$$\begin{align}
\frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi}
&=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\
&~~~~~-\lim_{x\to0}\left(\frac{x}{1-x}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\
&~~~~~-\int_{0}^{A}\mathrm{d}x\,\left(\frac{x}{1-x}\right)\frac{d}{dx}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right];~~~\small{I.B.P.s}\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\
&~~~~~+\int_{0}^{A}\mathrm{d}x\,\left[\frac{B\ln{\left(\frac{B}{x}\right)}}{\left(1-x\right)\left(B-x\right)}-\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\right]\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\
&~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\frac{\left(1-B\right)\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\
&~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\left[\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\frac{\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)}\right]\\
&~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\
&~~~~~+\frac{B}{1-B}\left[-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(B\right)}-\ln{\left(x\right)}}{\left(1-x\right)}\right]\\
&~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(A\right)}\ln{\left(1-B\right)}\right]\\
&~~~~~+\frac{B}{1-B}\left[\operatorname{Li}_{2}{\left(1-A\right)}-\operatorname{Li}_{2}{\left(1-\frac{A}{B}\right)}+\ln{\left(B\right)}\ln{\left(1-A\right)}\right]\\
&~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)},\\
\end{align}$$
and then,
$$\begin{align}
\frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi}
&=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\
&~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\
&~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\
&~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{b+p}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\
&=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\
&~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\
&~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\
&~~~~~+a\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+ax}{ax}\right)}\ln{\left(\frac{b+ax}{ax}\right)}\\
&~~~~~+b\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+bx}{bx}\right)}\ln{\left(\frac{a+bx}{bx}\right)}\\
&=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\
&~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\
&~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\
&~~~~~+a\,\mathcal{J}{\left(a,\frac{a}{b}\right)}\\
&~~~~~+b\,\mathcal{J}{\left(b,\frac{b}{a}\right)},\\
\end{align}$$
where in the last line above we've defined another function $\mathcal{J}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the integral
$$\mathcal{J}{\left(p,q\right)}:=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}.$$
For $p>0\land q>0$, we obtain
$$\begin{align}
\mathcal{J}{\left(p,q\right)}
&=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}\\
&=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+qx}{qx}\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+px}{px}\right)}}{\left(1+qx\right)};~~~\small{I.B.P.s}\\
&=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+qx\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+px\right)}}{\left(1+qx\right)}\\
&~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+qx\right)}\\
&~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(q\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(p\right)}}{\left(1+qx\right)}\\
&=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\
&~~~~~+\frac{\operatorname{Li}_{2}{\left(\frac{p-q}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p}{1+p}\right)}-\operatorname{Li}_{2}{\left(-q\right)}}{p}+\frac{\operatorname{Li}_{2}{\left(\frac{q-p}{q+1}\right)}-\operatorname{Li}_{2}{\left(\frac{q}{1+q}\right)}-\operatorname{Li}_{2}{\left(-p\right)}}{q}\\
&~~~~~-\frac{\operatorname{Li}_{2}{\left(-p\right)}}{p}-\frac{\operatorname{Li}_{2}{\left(-q\right)}}{q}\\
&~~~~~-\frac{\ln{\left(q\right)}\ln{\left(1+p\right)}}{p}-\frac{\ln{\left(p\right)}\ln{\left(1+q\right)}}{q}.\\
\end{align}$$
And with that, our general evaluation of $\mathcal{I}$ is in principle complete, though the final expression is too cumbersome to bother writing out explicitly.
Cheers!