Deriving an explicit form for the nested sine integral $\int_{-\infty}^t \sin\left(A\sin(\omega t)-A\sin(\omega s)\right)e^{s-t}ds$ [closed]

For $A,\omega,t\in\mathbb{R}$ consider

\begin{equation} \begin{split} \mathcal{I}&=\int_{-\infty}^t \sin\left(A\sin(\omega t)-A\sin(\omega s)\right)e^{s-t}ds\\ &=\frac{e^{-t}}{\omega}\int_{-\infty}^{\omega t}\sin(v-A\sin u)e^{u/\omega}du,\\ &=\frac{e^{-t}}{\omega} \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}\int_{-\infty}^{\omega t} e^{u/\omega} (v-A\sin u)^{2n+1}du. \end{split} \end{equation}

such that $u=\omega s$ and $v=A\sin(\omega t)$. Does an explicit solution exist for the integral $\mathcal{I}$? Integrating by parts doesn't seem to simplify the expression. Perhaps a Taylor series form could work as an approximation? Any help would be much appreciated.

If you use the Binomial theorem and a reduction formulae for the powers of sine, you can obtain explicitly as many terms of your series as you want. Now, for a nice closed form, I think that is little hope for find any.