Find the Integral, $ I_{1,1I}=\int \frac{x\arcsin{x}}{c\arcsin{k}+\sqrt{1-x^2} + x\arcsin{x}}\mathrm{d}x $

Question

$$ I:=\int\arcsin\left(c+\frac{\sqrt{1-x^2} + x\arcsin{x}}{\arcsin{k}}\right)\mathrm{d}x $$ where $c$ and $k$ are constants with respect to $x$. Euler lets $$ I=\int \ln{(iu+\sqrt{1-u^2})}\mathrm{d}x $$ where $u:=c+\frac{\sqrt{1-x^2} + x\arcsin{x}}{\arcsin{k}}$. $I$ is separable: $$ I=\int\ln{(u)}\mathrm{d}x + \int\ln{(i+\sqrt{\frac{1}{u^2}-1})}\mathrm{d}x $$ The first of the sum of integrals, $I_{1}$, is found by naming $u$: $$ I_{1}=\int\ln{(c+\frac{\sqrt{1-x^2} + x\arcsin{x}}{\arcsin{k}})}\mathrm{d}x $$ $$ =\int\ln{(c\arcsin{k}+\sqrt{1-x^2} + x\arcsin{x})}\mathrm{d}x-\int\ln{(\arcsin{k})}\mathrm{d}x $$ Thus, $$ I_{1,2}=-x\ln\arcsin{k} $$ Per partes gives: $$ I_{1,1}=x\ln{(u\arcsin{k})}-\int \frac{x\arcsin{x}}{c\arcsin{k}+\sqrt{1-x^2} + x\arcsin{x}}\mathrm{d}x $$ Substitution provides $$ I_{1,1I}=\int \frac{e^{2y}\tan{e^y}}{c\arcsin{k}\cos{e^y}+\cos^2{e^y} + e^y\sin{e^y}\cos{e^y}}\mathrm{d}y $$ where $y:=\ln{\arcsin{x}}$. Substitution also provides $$ I_{1,1I}=\int \frac{y\sin{y}\cos{y}}{c\arcsin{k}+\cos{y}+y\sin{y}}\mathrm{d}y $$ where $y:=\arcsin{x}$. Where $$ J:= \int \frac{c\arcsin{k}}{c\arcsin{k}+\sqrt{1-x^2} + x\arcsin{x}}\mathrm{d}x $$ and $$ K:=\int \frac{\sqrt{1-x^2}}{c\arcsin{k}+\sqrt{1-x^2} + x\arcsin{x}}\mathrm{d}x, $$ $$ I_{1,1I}+J+K = x $$ I am trying to find $I_{1,1I}$, but my substitutions, per partes intgration, and additive integration techniques are not working.