I want to find the indefinite integral of $\operatorname{sech^{2n}(n-m)}$ where $m$ is a constant and the function's argument is $n$. I know that
$$\operatorname{sech(n-m)}=\frac{1}{\operatorname{cosh(n)}\operatorname{cosh(m)}-\operatorname{sinh(n)}\operatorname{sinh(m)}}$$
and therefore
$$\operatorname{sech^{2n}(n-m)}\mathrm dn=\frac{1}{\bigl(\operatorname{cosh(n)}\operatorname{cosh(m)}-\operatorname{sinh(n)}\operatorname{sinh(m)}\bigr)^{2n}}$$
Also that
$$\operatorname{\int {sech^{2}(n)}}\mathrm dn=\operatorname{tanh(n)}+C$$
but I don't know how to proceed.