For $n\ge m\ge 2$, define $$I(n,m):= \int _0^\infty\dfrac{\text{arcsinh}^nx}{x^m}dx$$ Computer algebra systems say that the indefinite integral can be expressed in terms of polylog functions (of rapidly increasing complexity), but I see no way of doing the limit $\to\infty$. On the other hand, I have found numerically that $I(n,m)$ can (conjecturally always) be written as a rational combination of zeta values, moreover this happens in a rather beautiful way.
Here are the values for small $m$
and then what seems to be the general formula.
Defining $\bar\zeta(n):=(2^n-1) \zeta(n)$, it seems like
$ I(n,2)=\dfrac{n!}{2^{n-1}} \bar\zeta(n)$
$ I(n,4)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1 {6}[ 4\bar\zeta(n-2)-\bar\zeta(n)]$
$ I(n,6)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{120}[ 16\bar\zeta(n-4)-40\bar\zeta(n-2)+9\bar\zeta(n)]$
$ I(n,8)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{5040}[ 64\bar\zeta(n-6)-560\bar\zeta(n-4) +1036\bar\zeta(n-2)-225\bar\zeta(n)]$
(you may want to stop here for a moment before reading on and look if the coefficients tell you something already)
and
$ I(n,3)=\dfrac{n!}{2^{n-1}} \zeta(n-1)$
$ I(n,5)=\dfrac{n!}{2^{n-1}} \cdot\dfrac13[ \zeta(n-3)-\zeta(n-1)]$
$ I(n,7)=\dfrac{n!}{2^{n-1}} \cdot\dfrac2{45}[\zeta(n-5)- 5 \zeta(n-3)+4\zeta(n-1)]$
Well, in umbral notation introducing a formal variable $\bar Z$, for even $m=2k$ $$ I(n,2k)=\frac{n!}{2^{n-1}} \cdot\frac1{(2k-1)!}\bar Z^{n}\prod_{j=-k+1}^{k-2} \left(\frac2{\bar Z}-(2j+1)\right)$$ where each power $\bar Z^r$ is to be replaced by $\bar\zeta (r)$. E.g. for $m=6$, the "zeta polynomial" is related to $16z^4-40z^2+9=(2z-3)(2z-1)(2z+1)(2z+3)$.
Likewise for odd $m=2k+3$ (sic to make notation more elegant), $$ I(n,2k+3)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}Z^{n}\prod_{j=-k}^{k}\left (\frac1{Z}-j\right)$$ where each power $Z^r$ is to be replaced by $\zeta (r)$. E.g. for $m=7$, the "zeta polynomial" is related to $z^5-5z^3+4z=(z-2)(z-1)z(z+1)(z+2)$.
We can rewrite the expression for even $m$ as follows to look amazingly similar to the one for odd $m$:
$$ I(n,2k+2)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}\bar Z^{n}\prod_{j=-k+1/2}^{k-1/2} \left(\frac1{\bar Z}-j\right)$$ where in the product $j$ runs over the half-integers.
So far, this is all speculation, but once these patterns found, it seems obvious that there must be some deeper connection (which might even yield an elegant proof). Any insights? Has anybody encountered similar families of integrals?
EDIT for the record: A similar pattern occurs for the integral $$J(n,m):=\int _0^\infty\dfrac{x^n}{\cosh^{m}x}dx,$$ which features for even $m$ the Dirichlet eta function $\eta(s)= \sum\limits_{n=1}^{\infty}\dfrac {(-1)^{n-1} }{ n^s}= \left(1-2^{1-s}\right) \zeta(s)$, and for odd $m$ the Dirichlet beta function $\beta(s) = \sum\limits_{n=0}^\infty \dfrac{(-1)^n} {(2n+1)^s}$.
Using the same umbral notation as above for $\eta$ and $\beta$ respectively, we have for even $m=2k$
$$\bbox[10px,#cFF]{ \int _0^\infty\dfrac{x^n}{\cosh^{2k}x}dx=\frac{2^{2k-n}~n!}{(2k-1)!} \eta^{n} \prod_{j=1}^{k-1}\left ( j^2-\frac { 1}{ \eta^2}\right) } $$ and for odd $m=2k+1$
$$\bbox[10px,#cFF]{\int _0^\infty\dfrac{x^n}{\cosh^{2k+1}x}dx=2\frac{n!}{(2k)!}\beta^n \prod_{j=1}^{k} \left ( (2j-1)^2-\frac { 1}{ \beta^2}\right).} $$
For example, if $m=8$, we get from $(x^2-1)(4x^2-1)(9x^2-1)=36x^6-49x^4+14x^2-1$ (note that these polynomials for even $m$ are essentially the same as the ones in the case of odd $m$ for the original integrals $I(n,m)$ above) $$J(n,8)=\int _0^\infty\dfrac{x^n}{\cosh^{8}x}dx=\frac{n!}{7!~2^{n-8}}\Bigl[36~\eta(n)-49~\eta(n-2)+14~\eta(n-4)-\eta(n-6)\Bigr]$$
and from $(x^2-1)(9x^2-1)(25x^2-1)=225x^6-259x^4+35x^2-1$, we get for $m=7$ $$J(n,7)=\int _0^\infty\dfrac{x^n}{\cosh^{7}x}dx=2\frac{n!}{6!} \Bigl[225~\beta(n)-259~\beta(n-2)+35~\beta(n-4)-\beta(n-6)\Bigr]$$