What is this integral
$\displaystyle \int\left(\frac{\mathrm{arcsinh}(ax)}{ax}\right)^{ b}dx$
where a and b are constants.
Let $\displaystyle Sin(X)=ax$ then $\displaystyle dx=\frac{Cos(X)}{a}dX$
the integral becomes:
$\displaystyle \int\left(\frac{X}{Sin(X)}\right)^{b}\frac{Cos(X)} {a}dX$
now
$\displaystyle \int\left({Sin(X)}\right)^{-b}{Cos(X)}dX= \frac {Sin(X)^{1-b}} {1-b}=u_1$
and the integral becomes
$\displaystyle \int \frac {X^bu_1'}{a}dX=\frac{u_1X}{a}-\int \frac{u_1bX^{b-1}}{a}dX=\frac{u_1X}{a}-\int \left( \frac{X}{Sin(X)}\right)^{b-1}} \frac {b}{a(1-b)}dX$
This looks like a dead end . But I include it because it might give you an idea?
Perhaps there is no closed form solution? In my last line I am trying to integrate sinc(X) raised to the power of 1-b. But if you google 'integral of sinc' it seems like this is far from trivial. See here Sine Integral -- from Wolfram MathWorld for the integral of sinc.