Integral $\int \operatorname{sech}^4 x \, dx$ How we can solve this?$\newcommand{\sech}{\operatorname{sech}}$
$$
\int \sech^4 x \, dx.
$$
I know we can solve the simple case 
$$
\int \sech \, dx=\int\frac{dx}{\cosh x}=\int\frac{dx\cosh x}{\cosh ^2x}=\int\frac{d(\sinh x)}{1+\sinh^2 x}=\int \frac{du}{1+u^2}=\tan^{-1}\sinh x+C.
$$
I am stuck with the $\sech^4$ though. Thank you

Since
\begin{align}
\partial_{x} \left[ \tanh^{m}(ax) \right] = am\ sech^{2}(ax) \ \tanh^{m-1}(ax)
\end{align}
then
\begin{align}
sech^{4}(ax) &= sech^{2}(ax) (1 - \tanh^{2}(ax) ) \\ 
&= \partial_{x} \left[ \frac{1}{a} \tanh(ax) \right] -  \partial_{x} \left[ \frac{1}{3a} \tanh^{3}(ax) \right] \\
&= \partial_{x} \left[ \frac{1}{a} \tanh(ax) - \frac{1}{3a} \tanh^{3}(ax) \right].
\end{align}
Upon integration of both sides the result is
\begin{align}
\int sech^{4}(ax) \ dx = \frac{1}{a} \tanh(ax) - \frac{1}{3a} \tanh^{3}(ax).
\end{align}
