circuit analysis of DC-source, inductance and modulated resistance
You won't (easily) get a closed form for \$\int e^{p(t)}dt\$.

The solution to the differential equation that you wrote is:

$$i(t)=\left(\frac{V}{L}\right)e^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}\left(\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau\right)+ Ce^{\frac{-R_{mod}\sin(\omega t)}{L\omega}-\frac{R_{offset}t}{L}}$$
where \$C\$ is an unknown determined by the initial condition.  Obviously the term involving integration is the troublesome bit so lets focus on that piece.

The power series for \$e^{x}\$ converges uniformly on bounded sets so we can exchange the power series with the integral giving
$$\int_0^t e^{\frac{R_{mod}\sin(\omega \tau)}{L\omega}+\frac{R_{offset}\tau}{L}} d\tau = \sum_{n=0}^{\infty}\frac{1}{n!}\int_0^t\left(\frac{R_{mod}\sin(\omega \tau)}{L \omega}+\frac{R_{offset} \tau}{L} \right)^n d\tau.$$

Integrating the inner terms involves integrating terms of the form \$\tau^k\sin(\omega \tau)^{n-k}\$ which have closed form solutions.  

The real upshot is that if we only want approximate solutions we can bound the tail of this series by bounding the integrands and using that $$\sum_{n=N+1}^{\infty}\frac{1}{n!}x^n = e^x-\frac{e^x \Gamma(N+1,x)}{N!}.$$

Then we will have a closed formed approximate solution \$\tilde{i}_N(t)\$ which approximates \$i(t)\$ and we will know a bound for the error \$|\tilde{i}_N(t) - i(t)|\$.