Is there a closed form of $\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$? For naturals $n\ge m$, define
$$I(n,m):=\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$$
with $\text{arcsinh}\ x=\ln(x+\sqrt{1+x^2} )$, so $\text{arcsinh} \frac12=\ln \frac{\sqrt{5}+1}2 $.
Is it possible to find closed form expressions of $I(n,m)$? I mean closed form in a broad sense, i.e. involving any other "known" constants.  
Motivation: it is known that
$$I(2,1)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}xdx=\dfrac{\zeta(3)}{10},$$
further $$I(1,0)= \int_0^\frac12 \text{arcsinh}\  x\ dx=\frac12\left(2-\sqrt{5}+\ln \frac{\sqrt{5}+1}2 \right)$$
$$I(1,1)= \int_0^\frac12\dfrac{\text{arcsinh}\ x}xdx=\frac{\pi^2}{20}$$ and $$I(2,2)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}{x^2}dx=\frac{\pi^2}6-5\ln^2\frac{\sqrt{5}+1}2 $$
so there might be some hope that at least some others of the $I(n,m)$ have closed forms involving values of $\zeta(k)$, ideally odd zeta values.
 A: The following results are quoted in http://www.hindawi.com/journals/ijmms/2007/019381/abs/ (Integer Powers of Arcsin, by J.M. Borwein and M. Chamberland):
$$\large I(4,1)=-\frac{3}{2}\mathrm{Li}_5(g^2)+3\mathrm{Li}_4(g^2)\ln{g}+\frac{3}{2}\zeta(5)-\frac{12}{5}\zeta(3)\ln^2{g}-\frac{4}{15}\pi^2\ln^3{g}+\frac{4}{5}\ln^5{g},$$ where $g=(\sqrt{5}-1)/2$ is the golden ratio, and
$$\large \frac{2^n}{n!}I(n,1)=\zeta(n+1)-\frac{n(-\ln{g^2})^{n+1}}{2(n+1)!}-\sum\limits_{j=2}^{n+1}\frac{(-\ln{g^2})^{n+1-j}}{(n+1-j)!}\mathrm{Li}_j(g^2).$$
A: Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have 
$$I(n,m)=-\frac{2^{m-1}a^n}{m-1}+\frac n{m-1}\,[J(n-1,m-1;a)-J(n-1,m-1;0)]
$$
if $n\ge m\ge2$, where $a:=\text{arcsinh}\,\frac12$ and 
$$J(p,q;y):=\int\frac{y^p}{\sinh^qy}\,dy.$$
Formula 1.4.24.1 in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) tells us that
$$J(p,q;y)=
-\frac{py^{p-1}}{(q-1)(q-2)\sinh^{q-2}y}
-\frac{y^p\cosh y}{(q-1)\sinh^{q-1}y}$$
$$
+\frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2;y)
-\frac{q-2}{q-1}\,J(p,q-2;y).  
$$
Also, formulas 1.4.24.2 and 1.4.24.4 in PBM tell us that
$$J(p,1;y)=\sum_{k=0}^\infty\frac{(2-2^{2k})B_{2k}}{(2k)!(p+2k)}\,y^{p+2k}\quad (|y|<\pi,\ p>0)$$
and 
$$J(p,2;y)=-y^p\coth y+p\sum_{k=0}^\infty\frac{2^{2k}B_{2k}}{(2k)!(p+2k-1)}\,y^{p+2k-1}\quad (|y|<\pi,\ p>1).$$
These formulas in PBM should be easy to obtain/check. 
Since $|a|=a=\text{arcsinh}\,\frac12<1/2<\pi$, the above formulas provide a recursion to compute the values of $I(n,m)$ in terms of the Bernoulli numbers $B_{2k}$ or, alternatively, in terms of $J(p,1;a)-J(p,1;0)$ and $J(p,2;a)-J(p,2;0)$, that is, in terms of $I(n,2)$ and $I(n,3)$.
