Both Wolfram|Alpha and Maple can evaluate this integral for given values of $m$. For instance, for $m=0$, Maple gives
$$ I(m=0)=\frac{e^{-ic\cos(2x)}}{8c^2} \left( ic\cos(2x)-ic+1 \right) $$
Expressions for higher values of $m$ seem to suggest that something related to binomials is in play, but I'm not sure how to get there. Any ideas on evaluating this integral for $m\geq0$?
To progress, use the substitution $t=\cos x$, such that $dt=-\sin xdx$, which yields $$ \begin{align} I(m) &= -e^{ic}\int\sin^{2m+2}(\arccos t)\cos^{2m+1}(\arccos t)\exp(-2ict^2)\\ &=-e^{ic}\int (1-t^2)^{m+1}t^{2m+1}\exp(-2ict^2)dt \end{align} $$ where we used $\cos(2x)=2\cos^2(x)-1$ and $\sin(\arccos t)=\sqrt{1-t^2}$. Applying the binomial theorem on the first power yields $$ \begin{align} I(m) &= -e^{ic}\sum_{k=0}^{m+1}(-1)^k\binom{m+1}{k}\int t^{2m+1+2k}\exp(-2ict^2)dt \\ &= \frac{e^{ic}}{2}\sum_{k=0}^{m+1}(-1)^{k}\binom{m+1}{k}\frac{\Gamma(m+k+1,ict^2)}{(2ic)^{m+k+1}} \\ &= \frac{e^{ic}}{2}\sum_{k=0}^{m+1}(-1)^k\binom{m+1}{k}\frac{(m+k)!}{(2ic)^{m+k+1}}e^{-2ict^2}\sum_{l=0}^{m+k}\frac{(2ict^2)^l}{l!} \\ &= \frac{e^{-ic\cos(2x)}}{2}\sum_{k=0}^{m+1}\sum_{l=0}^{m+k}(-1)^{k}\binom{m+1}{k}\frac{(m+k)!}{(2ic)^{m+k+1-l}}\frac{\cos^{2l}(x)}{l!} \end{align} $$ where we recognized the definition of the incomplete gamma function on the first line. On the third ine, where we used the fact that the incomplete gamma function can be written as a sum for integer order, as in our case $m+k+1\in\mathbb{N}$.
This expression indeed reproduces the Maple result for $m=0$. Using the power reduction formula for the cosine, it is possible to express this directly in terms of multiple angles $$ \begin{multline} I(m) = \frac{e^{ic\cos(2x)}}{2}\sum_{k=0}^{m+1}(-1)^k\binom{m+1}{k}\frac{(m+k)!}{(2ic)^{m+k+1}}\\\times\sum_{l=0}^{m+k}\frac{(2ic)^l}{l!}\left[\frac{1}{4^l}\binom{2l}{l}+\frac{2}{4^l}\sum_{n=0}^{l-1}\binom{l}{n}\cos\left[2(l-n)x\right]\right] \end{multline} $$