$$\begin {aligned} \int _{0}^{\infty }\arctan (\cosh x)\ln (\tanh x)dx &=\int _{0}^{\infty }\frac {\arctan \sqrt {1+s^{2}}\ln \frac {s}{\sqrt {1+s^{2}}}}{\sqrt {1+s^{2}}}ds\quad \quad (s=\sinh x)\\ &=-\frac {1}2\int _{0}^{\infty }\frac {\arctan\frac {\sqrt {1+t^{2}}}t\ln (1+t^{2})}{t\sqrt {1+t^{2}}}dt\quad \quad \left (t=\frac {1}s\right )\\ &=-\frac {1}2\int _{0}^{\infty }\int _{0}^{1}\frac {\ln (1+t^{2})}{t^{2}+(1+t^{2})u^{2}}dudt\\ &=-\frac {1}2\int _{0}^{1}\int _{0}^{\infty }\frac {\ln (1+t^{2})}{u^2+(1+u^2)t^2}dtdu\\ &=-\frac {1}2\int _{0}^{1}\frac {\pi \ln \left (1+\frac {u}{\sqrt {1+u^{2}}}\right )}{\frac {u}{\sqrt {1+u^{2}}}}\frac {du}{1+u^{2}}\\ &=-\frac {\pi }2\int _{0}^{\frac {\pi }4}\frac {\ln (1+\sin \theta )}{\sin \theta }d\theta \quad \quad \left (u=\tan \theta \right )\\ &=-\frac {\pi }2\int _{0}^{\tan \frac {\pi }8}\frac {\ln \left (1+\frac {2v}{1+v^{2}}\right )}{\frac {2v}{1+v^{2}}}\frac {2dv}{1+v^{2}}\quad \quad \left (v=\tan \frac {\theta }2\right )\\ &=-\frac {\pi }2\int _{0}^{\sqrt 2-1}\frac {2\ln (1+v)-\ln (1+v^{2})}vdv\\ &=\frac {\pi }2\left [2\operatorname{Li} _{2}(-v)-\frac {1}2\operatorname{Li} _{2}(-v^{2}) \right ]_0^{\sqrt 2-1}\\ &=\textcolor {blue}{\pi \left (\operatorname{Li} _{2}\left (1-\sqrt 2\right )-\frac {1}4\operatorname {Li} _2 \left (-3+2\sqrt 2\right ) \right )}. \end {aligned}$$