Following the same approach as the given link, we have $\displaystyle\operatorname{arctanh}(x)=\frac12\ln\left(\frac{1+x}{1-x}\right)$, and with the same substitution $\displaystyle t=\frac{1+x}{1-x}$ we get
$$I(a,b)=-\frac1{2^{a-1}}\int_0^1\frac{(1+t)^{b-2}}{(1-t)^b}\ln^a(t)~\mathrm dt$$
Using generalized binomial expansion theorem, we can expand this as
$$I(a,b)=-\frac1{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\sum_{k=0}^\infty(k+1)\dots(k+b-1)\int_0^1t^{k+j}\ln^a(t)~\mathrm dt$$
The integral at the end can be viewed as the $a$th derivative of $\displaystyle\int_0^1t^u~\mathrm dt=\frac1{u+1}$ at $u=k+j$, which is given by $\displaystyle\frac{(-1)^aa!}{(k+j+1)^{a+1}}$, which leaves us with the series
$$I(a,b)=\frac{(-1)^{a+1}a!}{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\sum_{k=0}^\infty\frac{(k+1)\dots(k+b-1)}{(k+j+1)^{a+1}}$$
which boils the problem down to computing the last series. See that we have
$$\sum_{k=0}^\infty\frac{(k+1)\dots(k+b-1)}{(k+j+1)^{a+1}}=\sum_{k=j+1}^\infty\frac{P_{b,j}(k)}{k^{a+1}}=\sum_{m=0}^{b-1}\alpha_{b,j,m}\zeta(a-m+1)-\sum_{k=1}^j\frac{P_{b,j}(k)}{k^{a+1}}$$
for a polynomial $\displaystyle P_{b,j}(k)=\sum_{m=0}^{b-1}\alpha_{b,j,m}k^m$. This leaves us with the solution given by
$$I(a,b)=\frac{(-1)^{a+1}a!}{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\left[\sum_{m=0}^{b-1}\alpha_{b,j,m}\zeta(a-m+1)-\sum_{k=1}^j\frac{P_{b,j}(k)}{k^{a+1}}\right]$$
The rest is just multiplying out $(k-j)\dots(k+b-j-2)$ to extract the coefficients of $P_{b,j}$. These are Stirling numbers of the first kind.
Start by splitting the integral into two convergent parts:
$$I=\int_0^\infty \frac{\tanh x-x+x-x\operatorname{sech} x}{x^3}dx=\int_0^\infty \frac{\tanh x-x}{x^3}dx+\int_0^\infty \frac{1-\operatorname{sech} x}{x^2}dx$$
$$I_1=\int_0^\infty \frac{\tanh x-x}{x^3}dx\overset{IBP}=\frac12 \int_0^\infty \frac{\operatorname{sech}^2 x -1}{x^2}dx$$
$$\overset{IBP}=-\int_0^\infty \frac{\tanh x\operatorname{sech}^2 x}{x}dx=-\frac{7\zeta(3)}{\pi^2}$$
$$I_2=\int_0^\infty \frac{1-\operatorname{sech} x}{x^2}dx\overset{IBP}=\int_0^\infty \frac{\tanh x\operatorname{sech} x}{x}dx=\frac{4G}{\pi}$$
$$I_2=\int_0^\infty \frac{\tanh x \operatorname{sech} x}{x}dx\overset{x=\ln t}=2\int_1^\infty \frac{(t^2-1)}{(t^2+1)\ln t}dt\overset{t=\frac{1}{x}}=\int_0^\infty \frac{x^2-1}{(x^2+1)^2\ln x}dx$$
Above the two integrals were averaged after the reciprocal subtitution was done.
Now we will use Feynman's trick alongside beta function: $$I(a)=\int_0^\infty \frac{x^a-1}{(x^2+1)^2 \ln x}dx\Rightarrow I'(a)=\int_0^\infty \frac{x^a}{(x^2+1)^2}dx=\frac12 \left(\frac{1-a}{2}\right)\frac{\pi}{\sin\left(\frac{\pi(a+1)}{2}\right)}$$
$$I(0)=0\Rightarrow I_2=\frac{\pi}{4}\int_0^2 \frac{1-a}{\sin\frac{\pi(a+1)}{2}}da =\frac{2}{\pi} \int_{0}^\frac{\pi}{2}\frac{t}{\sin t}dt=\frac{4 G}{\pi}$$
$I_1$ can be found here.
Best Answer
Part 2.
Given $\left(a,b\right)\in\mathbb{R}^{2}$ such that $0<a\le b<1$ and using the same auxiliary parameters and substitution employed for $\mathcal{F}$, the integral $\mathcal{G}$ can be rewritten as
$$\begin{align} \mathcal{G}{\left(a,b\right)} &=\frac{b}{\sqrt{1-a^{2}}}\int_{a}^{1}\mathrm{d}u\,\frac{2\left(u-a\right)\ln{\left(\frac{u+a}{u-a}\right)}}{\left(u^{2}-b^{2}\right)\sqrt{1-u^{2}}}\\ &=\frac{b}{\sqrt{1-a^{2}}}\int_{a}^{1}\mathrm{d}u\,\frac{2\left(u-a\right)\ln{\left(\frac{u+a}{u-a}\right)}}{\left(u+b\right)\left(u-b\right)\left(1+u\right)\sqrt{\frac{1-u}{1+u}}}\\ &=\frac{b}{\sqrt{1-a^{2}}}\int_{1}^{0}\mathrm{d}x\,\frac{(-1)4A^{2}x}{\left(1+A^{2}x^{2}\right)^{2}}\cdot\frac{\left(1+B^{2}\right)^{2}\left(1+A^{2}x^{2}\right)^{2}}{2\left(1-A^{2}B^{2}x^{2}\right)\left(B^{2}-A^{2}x^{2}\right)}\\ &~~~~~\times\frac{\left(A^{2}-A^{2}x^{2}\right)}{\left(1+A^{2}\right)}\cdot\frac{\ln{\left(\frac{1-A^{4}x^{2}}{A^{2}-A^{2}x^{2}}\right)}}{\sqrt{A^{2}x^{2}}};~~~\small{\left[u=\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{A^{2}\left(1-B^{2}\right)\left(1+B^{2}\right)\left(1-x^{2}\right)}{\left(1-A^{2}B^{2}x^{2}\right)\left(B^{2}-A^{2}x^{2}\right)}\ln{\left(\frac{1-A^{4}x^{2}}{A^{2}-A^{2}x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(1-A^{2}B^{2}\right)}{\left(1-A^{2}B^{2}x^{2}\right)}-\frac{\left(B^{2}-A^{2}\right)}{\left(B^{2}-A^{2}x^{2}\right)}\right]\ln{\left(\frac{1-A^{4}x^{2}}{A^{2}-A^{2}x^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(1-p^{2}\right)}{\left(1-p^{2}x^{2}\right)}-\frac{\left(1-q^{2}\right)}{\left(1-q^{2}x^{2}\right)}\right]\ln{\left(\frac{1-p^{2}q^{2}x^{2}}{pq-pqx^{2}}\right)}.\\ \end{align}$$
Assuming $b<a$, we have $A<B\implies q<1$, and we obtain
$$\begin{align} \mathcal{G}{\left(a,b\right)} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(1-p^{2}\right)}{\left(1-p^{2}x^{2}\right)}-\frac{\left(1-q^{2}\right)}{\left(1-q^{2}x^{2}\right)}\right]\ln{\left(\frac{1-p^{2}q^{2}x^{2}}{pq-pqx^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\left(1-p^{2}\right)}{\left(1-p^{2}x^{2}\right)}-\frac{\left(1-q^{2}\right)}{\left(1-q^{2}x^{2}\right)}\right]\left[\ln{\left(1-p^{2}q^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(pq\right)}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\left(1-p^{2}\right)}{\left(1-p^{2}x^{2}\right)}\left[\ln{\left(1-p^{2}q^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(pq\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\left(1-q^{2}\right)}{\left(1-q^{2}x^{2}\right)}\left[\ln{\left(1-p^{2}q^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(pq\right)}\right]\\ &=\frac{1-p^{2}}{2p}\int_{0}^{1}\mathrm{d}x\,\frac{2p}{1-p^{2}x^{2}}\left[\ln{\left(1-p^{2}q^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(pq\right)}\right]\\ &~~~~~-\frac{1-q^{2}}{2q}\int_{0}^{1}\mathrm{d}x\,\frac{2q}{1-q^{2}x^{2}}\left[\ln{\left(1-p^{2}q^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(pq\right)}\right]\\ &=\frac{1-p^{2}}{2p}\bigg{[}\int_{0}^{1}\mathrm{d}x\,\frac{2p\ln{\left(1-p^{2}q^{2}x^{2}\right)}}{1-p^{2}x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2p\ln{\left(1-x^{2}\right)}}{1-p^{2}x^{2}}\\ &~~~~~-2\ln{\left(pq\right)}\int_{0}^{1}\mathrm{d}x\,\frac{p}{1-p^{2}x^{2}}\bigg{]}\\ &~~~~~-\frac{1-q^{2}}{2q}\bigg{[}\int_{0}^{1}\mathrm{d}x\,\frac{2q\ln{\left(1-p^{2}q^{2}x^{2}\right)}}{1-q^{2}x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2q\ln{\left(1-x^{2}\right)}}{1-q^{2}x^{2}}\\ &~~~~~-2\ln{\left(pq\right)}\int_{0}^{1}\mathrm{d}x\,\frac{q}{1-q^{2}x^{2}}\bigg{]}\\ &=\frac{1-p^{2}}{2p}\bigg{[}2\operatorname{Li}_{2}{\left(\frac{p}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-pq}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p+pq}{p-1}\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(\frac{p}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+pq}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{p-pq}{p+1}\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(\frac{p}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}+2\operatorname{Li}_{2}{\left(\frac{p}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}\\ &~~~~~-2\ln{\left(pq\right)}\operatorname{artanh}{\left(p\right)}\bigg{]}\\ &~~~~~-\frac{1-q^{2}}{2q}\bigg{[}2\operatorname{Li}_{2}{\left(\frac{q}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q-pq}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q+pq}{q-1}\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(\frac{q}{q+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+pq}{q+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q-pq}{q+1}\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(\frac{q}{q-1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+1}{q-1}\right)}+2\operatorname{Li}_{2}{\left(\frac{q}{q+1}\right)}-\operatorname{Li}_{2}{\left(\frac{q-1}{q+1}\right)}\\ &~~~~~-2\ln{\left(pq\right)}\operatorname{artanh}{\left(q\right)}\bigg{]}\\ &=\frac{1-p^{2}}{2p}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{p-pq}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p+pq}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+pq}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{p-pq}{p+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}-2\ln{\left(pq\right)}\operatorname{artanh}{\left(p\right)}\bigg{]}\\ &~~~~~-\frac{1-q^{2}}{2q}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{q-pq}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q+pq}{q-1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+pq}{q+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q-pq}{q+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{q+1}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q-1}{q+1}\right)}-2\ln{\left(pq\right)}\operatorname{artanh}{\left(q\right)}\bigg{]}.\\ \end{align}$$
In the special case $b=a$, we have $B=A\implies q=1\land p=\frac{1-a}{1+a}$, and so
$$\begin{align} \mathcal{G}{\left(a,a\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{\left(1-p^{2}\right)}{\left(1-p^{2}x^{2}\right)}\ln{\left(\frac{1-p^{2}x^{2}}{p-px^{2}}\right)}\\ &=\frac{1-p^{2}}{2p}\int_{0}^{1}\mathrm{d}x\,\frac{2p}{1-p^{2}x^{2}}\left[\ln{\left(1-p^{2}x^{2}\right)}-\ln{\left(1-x^{2}\right)}-\ln{\left(p\right)}\right]\\ &=\frac{1-p^{2}}{2p}\left[\int_{0}^{1}\mathrm{d}x\,\frac{2p\ln{\left(1-p^{2}x^{2}\right)}}{1-p^{2}x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2p\ln{\left(1-x^{2}\right)}}{1-p^{2}x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2p\ln{\left(p\right)}}{1-p^{2}x^{2}}\right]\\ &=\frac{1-p^{2}}{2p}\bigg{[}2\operatorname{Li}_{2}{\left(\frac{p}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{2p}{p-1}\right)}-2\operatorname{Li}_{2}{\left(\frac{p}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{2p}{p+1}\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(\frac{p}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}+2\operatorname{Li}_{2}{\left(\frac{p}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}\\ &~~~~~-2\ln{\left(p\right)}\operatorname{artanh}{\left(p\right)}\bigg{]}\\ &=\frac{1-p^{2}}{2p}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{2p}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{2p}{p+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}+\ln{\left(\frac{1-p}{1+p}\right)}\ln{\left(p\right)}\bigg{]}\\ &=\frac{2a}{1-a^{2}}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{a-1}{a}\right)}+\operatorname{Li}_{2}{\left(1-a\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(-\frac{1}{a}\right)}-\operatorname{Li}_{2}{\left(-a\right)}+\ln{\left(a\right)}\ln{\left(\frac{1-a}{1+a}\right)}\bigg{]}\\ &=\frac{2a}{1-a^{2}}\bigg{[}2\operatorname{Li}_{2}{\left(1-a\right)}+\frac12\ln^{2}{\left(a\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(-a\right)}-\frac12\ln^{2}{\left(a\right)}+2\operatorname{Li}_{2}{\left(-1\right)}+\ln{\left(a\right)}\ln{\left(\frac{1-a}{1+a}\right)}\bigg{]}\\ &=\frac{2a}{1-a^{2}}\bigg{[}2\operatorname{Li}_{2}{\left(1\right)}-2\ln{\left(a\right)}\ln{\left(1-a\right)}-2\operatorname{Li}_{2}{\left(a\right)}\\ &~~~~~-2\operatorname{Li}_{2}{\left(-a\right)}+2\operatorname{Li}_{2}{\left(-1\right)}+\ln{\left(a\right)}\ln{\left(\frac{1-a}{1+a}\right)}\bigg{]}\\ &=\frac{2a}{1-a^{2}}\bigg{[}2\operatorname{Li}_{2}{\left(1\right)}+2\operatorname{Li}_{2}{\left(-1\right)}-2\operatorname{Li}_{2}{\left(a\right)}-2\operatorname{Li}_{2}{\left(-a\right)}\\ &~~~~~-\ln{\left(a\right)}\left[2\ln{\left(1-a\right)}+\ln{\left(\frac{1+a}{1-a}\right)}\right]\bigg{]}\\ &=\frac{2a}{1-a^{2}}\left[\operatorname{Li}_{2}{\left(1\right)}-\ln{\left(a\right)}\ln{\left(1-a^{2}\right)}-\operatorname{Li}_{2}{\left(a^{2}\right)}\right].\\ \end{align}$$
Given $\left(a,b\right)\in\mathbb{R}^{2}$ such that $0<a\le b<1$ and again reusing the auxiliary parameters and initial substitution employed for $\mathcal{F}$ and $\mathcal{G}$, the integral $\mathcal{H}$ can be rewritten as
$$\begin{align} \mathcal{H}{\left(a,b\right)} &=\int_{a}^{1}\mathrm{d}u\,\frac{2\sqrt{1-b^{2}}\,u\ln{\left(\frac{u+a}{2a}\right)}}{\left(u^{2}-b^{2}\right)\sqrt{1-u^{2}}}\\ &=\int_{a}^{1}\mathrm{d}u\,\frac{2\left(1+b\right)\sqrt{\frac{1-b}{1+b}}\,u\left[\ln{\left(u+a\right)}-\ln{\left(2\right)}-\ln{\left(a\right)}\right]}{\left(u+b\right)\left(u-b\right)\left(1+u\right)\sqrt{\frac{1-a}{1+a}}\sqrt{\frac{(1+a)(1-u)}{(1-a)(1+u)}}}\\ &=\int_{\frac{1-A^{2}}{1+A^{2}}}^{1}\mathrm{d}u\,\frac{2\left(1+\frac{1-B^{2}}{1+B^{2}}\right)Bu\left[\ln{\left(u+\frac{1-A^{2}}{1+A^{2}}\right)}-\ln{\left(2\right)}-\ln{\left(\frac{1-A^{2}}{1+A^{2}}\right)}\right]}{\left(u+\frac{1-B^{2}}{1+B^{2}}\right)\left(u-\frac{1-B^{2}}{1+B^{2}}\right)\left(1+u\right)A\sqrt{\frac{1-u}{A^{2}(1+u)}}}\\ &=\int_{1}^{0}\mathrm{d}x\,\frac{(-1)4A^{2}x}{\left(1+A^{2}x^{2}\right)^{2}}\cdot\frac{2\left(1+\frac{1-B^{2}}{1+B^{2}}\right)B\left(\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}\right)}{\left(\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}+\frac{1-B^{2}}{1+B^{2}}\right)\left(\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}-\frac{1-B^{2}}{1+B^{2}}\right)\left(1+\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}\right)A\sqrt{x^{2}}}\\ &~~~~~\times\left[\ln{\left(\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}+\frac{1-A^{2}}{1+A^{2}}\right)}-\ln{\left(2\right)}-\ln{\left(\frac{1-A^{2}}{1+A^{2}}\right)}\right];~~~\small{\left[u=\frac{1-A^{2}x^{2}}{1+A^{2}x^{2}}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2AB\left(1+B^{2}\right)\left(1-A^{2}x^{2}\right)}{\left(1-A^{2}B^{2}x^{2}\right)\left(B^{2}-A^{2}x^{2}\right)}\ln{\left(\frac{1-A^{4}x^{2}}{\left(1-A^{2}\right)\left(1+A^{2}x^{2}\right)}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2\left(p+q\right)\left(1-pqx^{2}\right)}{\left(1-p^{2}x^{2}\right)\left(1-q^{2}x^{2}\right)}\ln{\left(\frac{1-p^{2}q^{2}x^{2}}{\left(1-pq\right)\left(1+pqx^{2}\right)}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2p}{\left(1-p^{2}x^{2}\right)}+\frac{2q}{\left(1-q^{2}x^{2}\right)}\right]\ln{\left(\frac{1-p^{2}q^{2}x^{2}}{\left(1-pq\right)\left(1+pqx^{2}\right)}\right)}.\\ \end{align}$$
Set $\beta:=2\arctan{\left(B\right)}$. Then, $0<\beta<\frac{\pi}{2}\land B=\tan{\left(\frac{\beta}{2}\right)}\land\frac{q-p}{p+q}=\frac{1-B^{2}}{1+B^{2}}=\cos{\left(\beta\right)}$.
Also, set $r:=\frac{1-p}{1+p}\land s:=\frac{1-q}{1+q}$. Then, $0\le s<r<1$ with $s=0\iff q=1$, and we have
$$\begin{align} \mathcal{H}{\left(a,b\right)} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2p}{1-p^{2}x^{2}}+\frac{2q}{1-q^{2}x^{2}}\right]\ln{\left(\frac{1}{1-pq}\cdot\frac{1-p^{2}q^{2}x^{2}}{1+pqx^{2}}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2p}{1-p^{2}x^{2}}\ln{\left(\frac{1}{1-pq}\cdot\frac{1-p^{2}q^{2}x^{2}}{1+pqx^{2}}\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2q}{1-q^{2}x^{2}}\ln{\left(\frac{1}{1-pq}\cdot\frac{1-p^{2}q^{2}x^{2}}{1+pqx^{2}}\right)}\\ &=\int_{0}^{p}\mathrm{d}y\,\frac{2}{1-y^{2}}\ln{\left(\frac{p}{1-pq}\cdot\frac{1-q^{2}y^{2}}{p+qy^{2}}\right)};~~~\small{\left[x=p^{-1}y\right]}\\ &~~~~~+\int_{0}^{q}\mathrm{d}y\,\frac{2}{1-y^{2}}\ln{\left(\frac{q}{1-pq}\cdot\frac{1-p^{2}y^{2}}{q+py^{2}}\right)};~~~\small{\left[x=q^{-1}y\right]}\\ &=\int_{\frac{1-p}{1+p}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{p}{1-pq}\cdot\frac{1-q^{2}\left(\frac{1-t}{1+t}\right)^{2}}{p+q\left(\frac{1-t}{1+t}\right)^{2}}\right)}\\ &~~~~~+\int_{\frac{1-q}{1+q}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{q}{1-pq}\cdot\frac{1-p^{2}\left(\frac{1-t}{1+t}\right)^{2}}{q+p\left(\frac{1-t}{1+t}\right)^{2}}\right)};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\ &=\int_{\frac{1-p}{1+p}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{p}{1-pq}\cdot\frac{\left(1+t\right)^{2}-q^{2}\left(1-t\right)^{2}}{p\left(1+t\right)^{2}+q\left(1-t\right)^{2}}\right)}\\ &~~~~~+\int_{\frac{1-q}{1+q}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{q}{1-pq}\cdot\frac{\left(1+t\right)^{2}-p^{2}\left(1-t\right)^{2}}{q\left(1+t\right)^{2}+p\left(1-t\right)^{2}}\right)}\\ &=\int_{\frac{1-p}{1+p}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{p}{1-pq}\cdot\frac{\left[\left(1+t\right)-q\left(1-t\right)\right]\left[\left(1+t\right)+q\left(1-t\right)\right]}{p\left(1+2t+t^{2}\right)+q\left(1-2t+t^{2}\right)}\right)}\\ &~~~~~+\int_{\frac{1-q}{1+q}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{q}{1-pq}\cdot\frac{\left[\left(1+t\right)-p\left(1-t\right)\right]\left[\left(1+t\right)+p\left(1-t\right)\right]}{q\left(1+2t+t^{2}\right)+p\left(1-2t+t^{2}\right)}\right)}\\ &=\int_{\frac{1-p}{1+p}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{p}{1-pq}\cdot\frac{\left[\left(1-q\right)+\left(1+q\right)t\right]\left[\left(1+q\right)+\left(1-q\right)t\right]}{\left(p+q\right)-2\left(q-p\right)t+\left(p+q\right)t^{2}}\right)}\\ &~~~~~+\int_{\frac{1-q}{1+q}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{q}{1-pq}\cdot\frac{\left[\left(1-p\right)+\left(1+p\right)t\right]\left[\left(1+p\right)+\left(1-p\right)t\right]}{\left(p+q\right)+2\left(q-p\right)t+\left(p+q\right)t^{2}}\right)}\\ &=\int_{\frac{1-p}{1+p}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{p\left(1+q\right)^{2}\left[\left(\frac{1-q}{1+q}\right)+t\right]\left[1+\left(\frac{1-q}{1+q}\right)t\right]}{\left(1-pq\right)\left(p+q\right)\left[1-2\left(\frac{q-p}{p+q}\right)t+t^{2}\right]}\right)}\\ &~~~~~+\int_{\frac{1-q}{1+q}}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{q\left(1+p\right)^{2}\left[\left(\frac{1-p}{1+p}\right)+t\right]\left[1+\left(\frac{1-p}{1+p}\right)t\right]}{\left(1-pq\right)\left(p+q\right)\left[1+2\left(\frac{q-p}{p+q}\right)t+t^{2}\right]}\right)}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)\left(1+st\right)}{\left(r+s\right)\left(1-rs\right)\left[1-2t\cos{\left(\beta\right)}+t^{2}\right]}\right)}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)\left(1+rt\right)}{\left(r+s\right)\left(1-rs\right)\left[1+2t\cos{\left(\beta\right)}+t^{2}\right]}\right)},\\ \end{align}$$
and then,
$$\begin{align} \mathcal{H}{\left(a,b\right)} &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)\left(1+st\right)}{\left(r+s\right)\left(1-rs\right)\left[1-2t\cos{\left(\beta\right)}+t^{2}\right]}\right)}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)\left(1+rt\right)}{\left(r+s\right)\left(1-rs\right)\left[1+2t\cos{\left(\beta\right)}+t^{2}\right]}\right)}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{r}^{1}\mathrm{d}t\,\frac{\ln{\left(1+st\right)}}{t}\\ &~~~~~-\int_{r}^{1}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\beta\right)}+t^{2}\right)}}{t}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{s}^{1}\mathrm{d}t\,\frac{\ln{\left(1+rt\right)}}{t}\\ &~~~~~-\int_{s}^{1}\mathrm{d}t\,\frac{\ln{\left(1+2t\cos{\left(\beta\right)}+t^{2}\right)}}{t}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+st\right)}}{t}-\int_{0}^{r}\mathrm{d}t\,\frac{\ln{\left(1+st\right)}}{t}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\beta\right)}+t^{2}\right)}}{2t}+2\int_{0}^{r}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\beta\right)}+t^{2}\right)}}{2t}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1+rt\right)}}{t}-\int_{0}^{s}\mathrm{d}t\,\frac{\ln{\left(1+rt\right)}}{t}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\pi-\beta\right)}+t^{2}\right)}}{2t}+2\int_{0}^{s}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\pi-\beta\right)}+t^{2}\right)}}{2t}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-s\right)}+\operatorname{Li}_{2}{\left(-rs\right)}\\ &~~~~~+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-r\right)}+\operatorname{Li}_{2}{\left(-rs\right)}\\ &~~~~~+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}.\\ \end{align}$$
For $b<a$, we have $A<B\implies q<1\implies 0<s$, and then
$$\begin{align} \mathcal{H}{\left(a,b\right)} &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)\left(s+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-s\right)}+\operatorname{Li}_{2}{\left(-rs\right)}\\ &~~~~~+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)\left(r+t\right)}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-r\right)}+\operatorname{Li}_{2}{\left(-rs\right)}\\ &~~~~~+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{r}^{1}\mathrm{d}t\,\frac{\ln{\left(1+s^{-1}t\right)}}{t}\\ &~~~~~-\operatorname{Li}_{2}{\left(-s\right)}+\operatorname{Li}_{2}{\left(-rs\right)}+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~+\int_{s}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}+\int_{s}^{1}\mathrm{d}t\,\frac{\ln{\left(1+r^{-1}t\right)}}{t}\\ &~~~~~-\operatorname{Li}_{2}{\left(-r\right)}+\operatorname{Li}_{2}{\left(-rs\right)}+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\\ &=-\ln{\left(r\right)}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-s^{-1}\right)}+\operatorname{Li}_{2}{\left(-rs^{-1}\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(-s\right)}+\operatorname{Li}_{2}{\left(-rs\right)}+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~-\ln{\left(s\right)}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}-\operatorname{Li}_{2}{\left(-r^{-1}\right)}+\operatorname{Li}_{2}{\left(-sr^{-1}\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(-r\right)}+\operatorname{Li}_{2}{\left(-rs\right)}+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\\ &=-2\operatorname{Li}_{2}{\left(-1\right)}+\frac12\ln^{2}{\left(r\right)}-2\operatorname{Li}_{2}{\left(-1\right)}+\frac12\ln^{2}{\left(s\right)}+2\operatorname{Li}_{2}{\left(-1\right)}-\frac12\ln^{2}{\left(\frac{r}{s}\right)}\\ &~~~~~-\ln{\left(r\right)}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}-\ln{\left(s\right)}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}+2\operatorname{Li}_{2}{\left(-rs\right)}\\ &~~~~~+\frac12\left(\pi-\beta\right)^{2}+2\operatorname{Li}_{2}{\left(-1\right)}+\frac12\beta^{2}+2\operatorname{Li}_{2}{\left(-1\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}+\ln{\left(r\right)}\ln{\left(s\right)}-\ln{\left(r\right)}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}-\ln{\left(s\right)}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}\\ &~~~~~+\frac{\pi^{2}}{2}-\pi\beta+\beta^{2}+2\operatorname{Li}_{2}{\left(-rs\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\\ &=\frac{\pi^{2}}{3}-\ln{\left(r\right)}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}-\ln{\left(s\right)}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}\\ &~~~~~+\ln{\left(r\right)}\ln{\left(s\right)}-\beta\left(\pi-\beta\right)+2\operatorname{Li}_{2}{\left(-rs\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}.\\ \end{align}$$
In the special case $b=a\implies B=A\implies q=1\implies s=0$, we have
$$\begin{align} \mathcal{H}{\left(a,a\right)} &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{\left(1-r^{2}\right)t}{r}\right)}-\operatorname{Li}_{2}{\left(0\right)}+\operatorname{Li}_{2}{\left(0\right)}+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{r+t}{r}\right)}-\operatorname{Li}_{2}{\left(-r\right)}+\operatorname{Li}_{2}{\left(0\right)}+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}-2\operatorname{Li}_{2}{\left(0,\pi-\beta\right)}\\ &=\int_{r}^{1}\mathrm{d}t\,\frac{1}{t}\ln{\left(\frac{1-r^{2}}{r}\right)}+\int_{r}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{t}+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(-r^{-1}\right)}-\operatorname{Li}_{2}{\left(-r\right)}+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}\\ &=-\ln{\left(r\right)}\ln{\left(\frac{1-r^{2}}{r}\right)}-\frac12\ln^{2}{\left(r\right)}+2\operatorname{Li}_{2}{\left(1,\beta\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(-r^{-1}\right)}-\operatorname{Li}_{2}{\left(-r\right)}+2\operatorname{Li}_{2}{\left(1,\pi-\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-\ln{\left(r\right)}\ln{\left(\frac{1-r^{2}}{r}\right)}+\frac12\beta^{2}+\frac12\left(\pi-\beta\right)^{2}-2\operatorname{Li}_{2}{\left(r,\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-\ln{\left(\frac{1-p}{1+p}\right)}\ln{\left(\frac{4p}{1-p^{2}}\right)}+\frac12\beta^{2}+\frac12\left(\pi-\beta\right)^{2}-2\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-\ln{\left(\frac{1-A^{2}}{1+A^{2}}\right)}\ln{\left(\frac{4A^{2}}{1-A^{4}}\right)}+\frac12\beta^{2}+\frac12\left(\pi-\beta\right)^{2}-2\operatorname{Li}_{2}{\left(\frac{1-B^{2}}{1+B^{2}},\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-\ln{\left(a\right)}\ln{\left(\frac{1-a^{2}}{a}\right)}+\frac12\beta^{2}+\frac12\left(\pi-\beta\right)^{2}-2\operatorname{Li}_{2}{\left(\cos{\left(\beta\right)},\beta\right)}\\ &=2\operatorname{Li}_{2}{\left(-1\right)}-\ln{\left(a\right)}\ln{\left(\frac{1-a^{2}}{a}\right)}+\frac12\beta^{2}+\frac12\left(\pi-\beta\right)^{2}-\left(\frac{\pi}{2}-\beta\right)^{2}-\frac12\operatorname{Li}_{2}{\left(\cos^{2}{\left(\beta\right)}\right)}\\ &=\frac{\pi^{2}}{12}-\ln{\left(a\right)}\ln{\left(\frac{1-a^{2}}{a}\right)}-\frac12\operatorname{Li}_{2}{\left(a^{2}\right)}.\\ \end{align}$$
Putting everything together, for $\frac{1}{\sqrt{2}}\le a<1$ with $b=\sqrt{1-a^{2}}$ we obtain the following nightmarish result:
$$\begin{align} 60a^{4}\mathcal{I}{\left(a\right)} &=\mathcal{E}{\left(a,b\right)}+\mathcal{F}{\left(a,b\right)}+4a^{5}\mathcal{G}{\left(a,b\right)}-8a^{4}\mathcal{H}{\left(a,b\right)}\\ &=2a\left(2-3a^{2}+3b^{2}\right)\sqrt{1-a^{2}}+\frac{2a\left(a^{2}-b^{2}\right)\left(a^{2}-3b^{2}\right)}{\sqrt{1-b^{2}}}\operatorname{artanh}{\left(\frac{\sqrt{1-a^{2}}}{\sqrt{1-b^{2}}}\right)}\\ &~~~~~+12\mathcal{J}_{4}{\left(a\right)}-4\left(5a^{2}-3b^{2}\right)\mathcal{J}_{2}{\left(a\right)}-4\left(5a^{2}-3b^{2}\right)b^{2}\mathcal{J}_{0}{\left(a\right)}\\ &~~~~~+\frac{2\left(a^{2}-b^{2}\right)\left(2a^{4}+2a^{2}b^{2}-3b^{4}\right)}{b\sqrt{1-b^{2}}}\bigg{[}-2\ln{\left(pq\right)}\operatorname{artanh}{\left(\frac{q-p}{1-pq}\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+1}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q-1}{q+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{p-pq}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+pq}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p+pq}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-pq}{p+1}\right)}\\ &~~~~~-\operatorname{Li}_{2}{\left(\frac{q-pq}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q+pq}{q-1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+pq}{q+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q-pq}{q+1}\right)}\bigg{]}\\ &~~~~~+\frac{4a^{5}(1-p^{2})}{2p}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{p-pq}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p+pq}{p-1}\right)}+\operatorname{Li}_{2}{\left(\frac{p+pq}{p+1}\right)}+\operatorname{Li}_{2}{\left(\frac{p-pq}{p+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{p+1}{p-1}\right)}-\operatorname{Li}_{2}{\left(\frac{p-1}{p+1}\right)}-2\ln{\left(pq\right)}\operatorname{artanh}{\left(p\right)}\bigg{]}\\ &~~~~~-\frac{4a^{5}(1-q^{2})}{2q}\bigg{[}-\operatorname{Li}_{2}{\left(\frac{q-pq}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q+pq}{q-1}\right)}+\operatorname{Li}_{2}{\left(\frac{q+pq}{q+1}\right)}+\operatorname{Li}_{2}{\left(\frac{q-pq}{q+1}\right)}\\ &~~~~~+\operatorname{Li}_{2}{\left(\frac{q+1}{q-1}\right)}-\operatorname{Li}_{2}{\left(\frac{q-1}{q+1}\right)}-2\ln{\left(pq\right)}\operatorname{artanh}{\left(q\right)}\bigg{]}\\ &~~~~~-8a^{4}\bigg{[}2\operatorname{Li}_{2}{\left(1\right)}-\ln{\left(r\right)}\ln{\left(\frac{\left(1-r^{2}\right)s}{\left(r+s\right)\left(1-rs\right)}\right)}-\ln{\left(s\right)}\ln{\left(\frac{\left(1-s^{2}\right)r}{\left(r+s\right)\left(1-rs\right)}\right)}\\ &~~~~~+\ln{\left(r\right)}\ln{\left(s\right)}-\beta\left(\pi-\beta\right)+2\operatorname{Li}_{2}{\left(-rs\right)}-2\operatorname{Li}_{2}{\left(r,\beta\right)}-2\operatorname{Li}_{2}{\left(s,\pi-\beta\right)}\bigg{]}.\\ \end{align}$$
Finally, we obtain an explicit value for the integral in the $a=\frac{1}{\sqrt{2}}$ special case. Considering how cumbersome the general case turned out to be, the final result is very pleasingly concise!
$$\begin{align} 15\mathcal{I}{\left(\frac{1}{\sqrt{2}}\right)} &=\mathcal{E}{\left(a,a\right)}+\mathcal{F}{\left(a,a\right)}+4a^{5}\mathcal{G}{\left(a,a\right)}-8a^{4}\mathcal{H}{\left(a,a\right)}\\ &=4a\sqrt{1-a^{2}}+12\mathcal{J}_{4}{\left(a\right)}-8a^{2}\mathcal{J}_{2}{\left(a\right)}-8a^{4}\mathcal{J}_{0}{\left(a\right)}\\ &~~~~~+4a^{5}\left(\frac{2a}{1-a^{2}}\right)\left[\operatorname{Li}_{2}{\left(1\right)}-\ln{\left(a\right)}\ln{\left(1-a^{2}\right)}-\operatorname{Li}_{2}{\left(a^{2}\right)}\right]\\ &~~~~~-8a^{4}\left[\frac12\operatorname{Li}_{2}{\left(1\right)}-\ln{\left(a\right)}\ln{\left(\frac{1-a^{2}}{a}\right)}-\frac12\operatorname{Li}_{2}{\left(a^{2}\right)}\right]\\ &=11a\sqrt{1-a^{2}}+\left(9-2a^{2}\right)a\sqrt{1-a^{2}}\ln{\left(\frac{a}{\sqrt{1-a^{2}}}\right)}\\ &~~~~~+\frac12\left(9-8a^{2}-16a^{4}\right)\left[\operatorname{Cl}_{2}{\left(2\arccos{\left(a\right)}\right)}+\operatorname{Cl}_{2}{\left(\pi-2\arccos{\left(a\right)}\right)}\right]\\ &~~~~~+\frac{8a^{6}}{1-a^{2}}\left[\operatorname{Li}_{2}{\left(1\right)}-\ln{\left(a\right)}\ln{\left(1-a^{2}\right)}-\operatorname{Li}_{2}{\left(a^{2}\right)}\right]\\ &~~~~~-4a^{4}\left[\operatorname{Li}_{2}{\left(1\right)}-2\ln{\left(a\right)}\ln{\left(\frac{1-a^{2}}{a}\right)}-\operatorname{Li}_{2}{\left(a^{2}\right)}\right]\\ &=\frac{11}{2}+\operatorname{Li}_{2}{\left(1\right)}-\frac12\ln^{2}{\left(2\right)}-\operatorname{Li}_{2}{\left(\frac12\right)}+\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}\\ &=\frac{11}{2}+\frac{\pi^{2}}{12}+C,\\ \end{align}$$
where here $C$ denotes the Catalan constant.
Whew! If you made it here to the bottom, do like me and go take a well deserved nap. Cheers :p