This note discusses a possible route to prove that $$I=\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arccoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\,dx=\frac{\pi\log\pi}4-\frac{3\log2}8$$ and is based on the approach of de Reyna (2014) who proved the similar identity $$\int_0^1\frac{\operatorname{arccot}\left(1+\frac\pi{\operatorname{arctanh}x-\arctan x}\right)}x\,dx=\frac{\pi\log\pi}4-\frac{3\log2}8.$$
We first substitute $u=1/x$ to get$$I=\int_0^1\frac{\arctan\frac{\operatorname{arctanh}u-\arcsin u}{\operatorname{arctanh}u-\arcsin u+2\pi}}{u\sqrt{1-u^2}}\,du=\Im\int_0^1\frac{\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}u-\arcsin u)\right)}{u\sqrt{1-u^2}}\,du.$$ As we will find later, the $\log$ term on the numerator explains the appearances of $\log\pi$ and $\log2$ in the conjectured result. Another substitution $u=\sin\theta$ gives us $$I=\Im\int_0^{\pi/2}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin\theta)-\theta)\right)\csc\theta\,d\theta$$ where we note that the term inside the logarithm has real part $>1$ since $\operatorname{arctanh}\sin\theta-\theta>0$ on $(0,\pi/2)$; Bagul and Chesneau (2018) have determined optimal bounds for this inequality. Therefore, the only singularities on the first quadrant are at integer and half-integer multiples of $\pi$.
In a similar approach to de Reyna, we take a quarter-circular contour on the first quadrant and by letting the radius $R$ tend to infinity. Define $\gamma_0$ to be the line from $0$ to $\pi/2-\epsilon$, and $\gamma_k$ to be the line from $k\pi/2+\epsilon$ to $(k+1)\pi/2-\epsilon$ for each $1\le k\le\lfloor2R/\pi\rfloor$ as $\epsilon\to0^+$. Consequently, define $E_k$ to be the semicircular arc parametrised by $$p_k=\frac{k\pi}2+\epsilon e^{i\varphi};\quad\varphi\in[0,\pi]$$ taken clockwise to avoid each singularity. Define $C$ to be quarter-circular arc of radius $R$ from $\phi=0$ to $\pi/2$, and finally, define $\gamma_{-1}$ to be the line from $iR$ to $0$ to close the contour. Then by Cauchy's theorem, $$\left(\int_{\gamma_0}+\int_{\bigcup_{k\ge1}\gamma_k}+\int_{\bigcup_{k\ge1}E_k}+\int_C-\int_{\gamma_{-1}}\right)G(z)=0$$ where $$G(z)=\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin z)-z)\right)\csc z.$$
Conjecture 1. The integrals on $\gamma_0$ and $\gamma_{-1}$ are conjugates. In particular, $$2iI=\int_0^{\pi/2}G(z)\,dz-i\lim_{R\to\infty}\int_0^RG(iz)\,dz.$$ I do not have a proof of this yet, but it is extremely similar to $(3.8)$ of de Reyna, where the author exploited the identity $$\operatorname{arctanh}iz-\arctan iz=-i(\operatorname{arctanh}z-\arctan z)$$ to immediately arrive at the conclusion. In our case we are not so fortunate since $\sin iz=\sinh z$ has no immediate relation to $\tanh z$.
Proposition 2. As $R\to\infty$, $$\int_CG(z)\,dz\sim-\left(\pi+i\log\left(\frac{1+i}{2\pi}R\right)\right)\left(\frac\pi2+i\Re\log\cot\frac R2\right).$$
Proof: Substituting $z=Re^{i\phi}$ gives \begin{align}\int_CG(z)\,dz&=i\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin Re^{i\phi})-Re^{i\phi})\right)\,d\phi\\&\sim i\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\log\left(\frac{3+i}4-\frac{1+i}{2\pi}Re^{i\phi}\right)\,d\phi\end{align} as $\operatorname{arctanh}(\sin Re^{i\phi})\to\pi i/2$. To obtain its asymptotic behaviour, we expand the logarithm as a Taylor series \begin{align}\log\left(\frac{3+i}4-\frac{1+i}{2\pi}Re^{i\phi}\right)&=\log\left(\frac{1+i}{2\pi}Re^{i\phi}\right)-i\pi+\log\left(1-\frac{3+i}4\frac{2\pi}{1+i}\frac1{Re^{i\phi}}\right)\\&=\log\left(\frac{1+i}{2\pi}R\right)-i\pi+i\phi+{\cal O}(R^{-1})\end{align} so that \begin{align}\int_CG(z)\,dz&\sim\left(\pi+i\log\left(\frac{1+i}{2\pi}R\right)\right)\int_0^{\pi/2}\frac{Re^{i\phi}}{\sin(Re^{i\phi})}\,d\phi-i\int_0^{\pi/2}\frac{R\phi e^{i\phi}}{\sin(Re^{i\phi})}\,d\phi.\end{align} The first integral can be computed directly as $i\log\cot(Re^{i\phi}/2)\vert_0^{\pi/2}$ whereas the second integral tends to zero after integrating by parts. $\quad\square$
Proposition 3. As $\epsilon\to0^+$, $\int_{\bigcup_{k\,\text{odd}}E_k}G(z)\,dz\to0$. This accounts for all the singularities from the $\operatorname{arctanh}$ term.
Proof: The steps are similar to the proof of Proposition 2 so only an outline will be sketched. When $k$ is odd, we have $\sin p_k\to\pm1$ (where $\pm$ depends on whether $k\equiv1,3\pmod4$) and $$\operatorname{arctanh}\sin p_k=\pm\operatorname{arctanh}\cos(\epsilon e^{i\varphi})\sim\pm\log\epsilon$$ as $\epsilon\to0^+$. Thus for each odd $k$, \begin{align}\int_{E_k,k\,\text{odd}}G(z)\,dz&\sim\pm\int_0^\pi i\epsilon e^{i\varphi}\log\left(1+\frac{1+i}{2\pi}\left(\pm\log\epsilon-\frac{k\pi}2-\epsilon e^{i\varphi}\right)\right)\,d\varphi\to0\end{align} since the linear term $\epsilon$ dominates the logarithmic term multiplicatively. $\quad\square$
Proposition 4. As $\epsilon\to0^+$, $$\int_{\bigcup_{k\,\text{even}}E_k}G(z)\,dz\to i\pi\sum_{n=1}^{\lfloor R/\pi\rfloor}(-1)^n\log\left(1-\frac{1+i}2n\right).$$ This accounts for all the singularities from the $\csc$ term.
Proof: When $k=2n$ is even, we have \begin{align}\int_{E_k}G(z)\,dz&=\small i\int_0^{\pi}\frac{\epsilon e^{i\varphi}}{\sin(n\pi+\epsilon e^{i\varphi})}\log\left(1+\frac{1+i}{2\pi}(\operatorname{arctanh}(\sin(n\pi+\epsilon e^{i\varphi}))-n\pi-\epsilon e^{i\varphi})\right)\,d\varphi\\&\to i\int_0^{\pi}(-1)^n\log\left(1-\frac{1+i}2n\right)\,d\varphi\end{align} as $\epsilon\to0^+$ since $\operatorname{arctanh}\sin\epsilon-\epsilon\to0$. We immediately obtain the desired result upon noting that there are $\lfloor R/\pi\rfloor$ singularities that are integer multiples of $\pi$ for a fixed contour radius $R$. $\quad\square$
