About Integration $ \int \frac{\tanh(\sqrt{1+z^2})}{\sqrt{1+z^2}}dz $ How to calculate the following integral
$$
\int \frac{\tanh(\sqrt{1+z^2})}{\sqrt{1+z^2}}dz
$$
Is there any ways to calculate those integral in analytic? (Is $[0,\infty]$, case the integral is possible?)
How about using numerical method? Is there are good numerical scheme to complete this integral?
From the answer by @Lucian, 
The integral of
$\displaystyle\int_0^\infty\bigg[1-\tanh(\cosh x)\bigg]~dx$ is converges. 
How one can evaluate this integral? 

For $\int_0^\infty(1-\tanh\cosh x)~dx$ ,
Similar to How to evaluate $\int_{0}^\infty \frac{{x^2}}{e^{\beta {\big(\sqrt{x^2 + m^2}}- \nu\big)} + 1} dx$,
$\int_0^\infty(1-\tanh\cosh x)~dx$
$=\int_0^\infty\left(1-\dfrac{1-e^{-2\cosh x}}{1+e^{-2\cosh x}}\right)~dx$
$=\int_0^\infty\dfrac{2e^{-2\cosh x}}{1+e^{-2\cosh x}}~dx$
$=\int_0^\infty\sum\limits_{n=0}^\infty2(-1)^ne^{-2(n+1)\cosh x}~dx$
$=\sum\limits_{n=0}^\infty2(-1)^nK_0(2(n+1))$
For $\int\dfrac{\tanh\sqrt{1+z^2}}{\sqrt{1+z^2}}~dz$ ,
$\int\dfrac{\tanh\sqrt{1+z^2}}{\sqrt{1+z^2}}~dz$
$=\int\dfrac{1-e^{-2\sqrt{1+z^2}}}{(1+e^{-2\sqrt{1+z^2}})\sqrt{1+z^2}}~dz$
$=\int\dfrac{1}{\sqrt{1+z^2}}~dz-\int\dfrac{2e^{-2\sqrt{1+z^2}}}{(1+e^{-2\sqrt{1+z^2}})\sqrt{1+z^2}}~dz$
$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n+1}e^{-2(n+1)\sqrt{1+z^2}}}{\sqrt{1+z^2}}~dz$
$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n+1}e^{-2(n+1)\sqrt{1+\sinh^2u}}}{\sqrt{1+\sinh^2u}}~d(\sinh u)$ $(\text{Let}~z=\sinh u)$
$=\sinh^{-1}z+\int\sum\limits_{n=0}^\infty2(-1)^{n+1}e^{-2(n+1)\cosh u}~du$
$=\sinh^{-1}z+\sum\limits_{n=0}^\infty2(-1)^{n+1}J(2(n+1),0,u)+C$ (according to https://www.cambridge.org/core/services/aop-cambridge-core/content/view/9C572E5CE44E9E0DE8630755DF99ABAC/S0013091505000490a.pdf/incomplete-bessel-functions-i.pdf)
$=\sinh^{-1}z+\sum\limits_{n=0}^\infty2(-1)^{n+1}J(2(n+1),0,\sinh^{-1}z)+C$
