We can generalize the integral by manipulating the Laplace transform of $J_{n}(bx)$, namely $$ \int_{0}^{\infty} J_{n}(bx) e^{-sx} \, dx = \frac{(\sqrt{s^{2}+b^{2}}-s)^{n}}{b^{n}\sqrt{s^{2}+b^{2}}}\ , \quad \ (n \in \mathbb{Z}_{\ge 0} \, , \text{Re}(s) >0 , \, b >0 )\tag{1}. $$
(See this question for a derivation of $(1)$ using contour integration.)
First let $s=p+ia$, where $p,a >0$.
A slight modification of the answer here shows that $\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx $ converges uniformly for all $p \in [0, \infty$).
This allows us to conclude that $$\begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \lim_{p \downarrow 0}\int_{0}^{\infty} J_{n}(bx) e^{-(p+ia)x} \, dx \\ &= \lim_{p \downarrow 0} \frac{\left(\sqrt{(-p+ia)^2+b^{2}}-p-ia\right)^{n}}{b^{n}\sqrt{(p+ia)^2+b^{2}}} \\ &= \frac{\left(\sqrt{b^{2}-a^{2}}-ia\right)^{n}}{b^{n}\sqrt{b^{2}-a^{2}}}. \end{align}$$
So if $ a < b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(\sqrt{b^{2}-a^{2}+a^{2}} e^{-i \arcsin \left(\frac{a}{b}\right)}\right)^{n}}{b^{n} \sqrt{b^{2}-a^{2}}} \\ &= \frac{e^{-in \arcsin \left(\frac{a}{b}\right)}}{\sqrt{b^{2}-a^{2}}} .\end{align}$$
And if $a >b$, $$ \begin{align} \int_{0}^{\infty} J_{n}(bx) e^{-iax} \, dx &= \frac{\left(i\sqrt{a^{2}-b^{2}}-ia \right)^{n}}{b^{n}i \sqrt{a^{2}-b^{2}}} \\ &= \frac{-i e^{i \pi n /2} \left(\sqrt{a^{2}-b^{2}}-a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}}. \end{align}$$
Therefore,
$$\int_{0}^{\infty} J_{n}(bx) \sin(ax) \, dx = \begin{cases}
\frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\
\frac{\cos \left(\frac{\pi n}{2} \right) \left(\sqrt{a^{2}-b^{2}} -a \right)^{n}}{b^{n} \sqrt{a^{2}-b^{2}}} & \quad a > b >0
\end{cases} $$
Let the parameter $a\in\mathbb R$, and write the integral as
$$I(a)=\int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx$$
Note that $I(a)=I(-a)$, and we get
$$I(a)=\frac12\int_{-\pi}^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx=\frac12\Re\int_{-\pi}^{\pi}e^{ia\sin x}e^{\cos (x)}\ dx$$
Let $z=e^{ix}$, and choose the unit circle as the contour,
$$I(a)=\frac12\Re\oint e^{ia\frac1{2i}(z-\frac1z)}e^{\frac12(z+\frac1z)}\frac{1}{iz}\ dz=\frac12\Re\oint e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{iz}\ dz$$
where $z=0$ is the only sigularity inside the contour
$$I(a)=\frac12\cdot\Re\left[2\pi i~\text{Res}(z=0)\right]=\pi\cdot \Re\left[\text{Res}\left( e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{ z},~ z=0\right)\right]$$
We take the Laurent series,
$$e^{\frac{a+1}{2}z}e^{\frac{1-a}2\cdot \frac1z}\frac{1}{ z}=\frac{1}{ z}\sum_{k=0}\frac{1}{k!}\left( \frac{a+1}{2}\right)^kz^k\sum_{n=0}\frac{1}{n!}\left( \frac{1-a}{2}\right)^nz^n$$
the coefficient of the $z^{-1}$ term is
$$\sum_{k=0}\frac{1}{k!}\left( \frac{a+1}{2}\right)^k\frac{1}{k!}\left( \frac{1-a}{2}\right)^k=\sum_{k=0}\frac{1}{(k!)^2}\left( \frac{1-a^2}{4}\right)^k$$
therefore,
$$\boxed{\begin{align} \int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi I_0(\sqrt{1-a^2}),~~~a\in (-1,1)\\
\\
\int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi,~~~a=\pm1\\
\\
\int_0^{\pi}\cos(a\sin x)e^{\cos (x)}\ dx&=\pi J_0(\sqrt{a^2-1}),~~~a\in (-\infty, -1)\cup(1, \infty)
\end{align}}$$
Best Answer
So after some computations, I was able to obtain a satisfactory answer in the case where c is an integer. For simplicity, let us define: \begin{align*} \forall n\in\mathbb N, \qquad I_n = \int_{0}^{2\pi} \cos(x\cos(\theta) + y\sin(\theta) + n\theta + d)d\theta. \end{align*} I already established that: \begin{align*} I_0 &= 2\pi J_0(\sqrt{x^2 + y^2})\cos(d),\\ I_1 &= -\frac{2\pi}{\sqrt{x^2 + y^2}}J_1\left(\sqrt{x^2 + y^2}\right)\left[x\sin(d) + y\cos(d)\right]. \end{align*} Let us rewrite $I_n$ as: \begin{align*} I_n &= \int_{0}^{2\pi} \cos(x\cos(\theta) + y\sin(\theta) + d)\cos(n\theta)d\theta - \int_{0}^{2\pi} \sin(x\cos(\theta) + y\sin(\theta) + d)\sin(n\theta)d\theta\\ &= Re\left[\int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + d)}\cos(n\theta)d\theta\right] - Im\left[\int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + d)}\sin(n\theta)d\theta\right]\\ &= Re\left[C_n\right] - Im\left[S_n\right], \end{align*} where I have defined the two sequences $C_n$ and $S_n$ that I am going to compute. We already know that: \begin{align*} C_0 &= 2\pi e^{id} J_0(\sqrt{x^2 + y^2})\\ C_1 &= 2\pi \frac{e^{id}}{i} \frac{\partial}{\partial x}[J_0(\sqrt{x^2 + y^2})]. \end{align*} Moreover, in the same manner, it can be shown that: \begin{align*} S_0 &= 0\\ S_1 &= 2\pi \frac{e^{id}}{i} \frac{\partial}{\partial y}[J_0(\sqrt{x^2 + y^2})]. \end{align*} Using the fact that $\cos(2x) = 2\cos^2(x) - 1$ and $\sin(2x) = 2\sin(x)\cos(x)$, it can be demonstrated that: \begin{align*} C_2 &= \frac{2}{i}\frac{\partial C_1}{\partial x} - C_0\\ S_2 &= \frac{2}{i}\frac{\partial S_1}{\partial x} - S_0. \end{align*} Using Chebyshev polynomials, we can demonstrate that: \begin{align*} \cos[(n+1)x] = 2\cos(x)\cos(nx) - \cos[(n-1)x]\\ \sin[(n+1)x] = 2\cos(x)\sin(nx) - \sin[(n-1)x]. \end{align*} Using induction reasoning alongside with these two relations, it is straightforwards to prove that: \begin{align*} C_{n+1} &= \frac{2}{i}\frac{\partial C_n}{\partial x} - C_{n-1}\\ S_{n+1} &= \frac{2}{i}\frac{\partial S_n}{\partial x} - S_{n-1}. \end{align*} To finish and in order to find an inductive relation for $I_n$, let us define: \begin{align*} \forall n\in\mathbb N, \qquad J_n = \int_{0}^{2\pi} \sin(x\cos(\theta) + y\sin(\theta) + n\theta + d)d\theta = Im\left[C_n\right] + Re\left[S_n\right], \end{align*} we can write: \begin{align*} I_{n+1} &= Re\left[C_{n+1}\right] - Im\left[S_{n+1}\right]\\ &= Re\left[\frac{2}{i}\frac{\partial C_n}{\partial x} - C_{n-1}\right] - Im\left[\frac{2}{i}\frac{\partial S_n}{\partial x} - S_{n-1}\right]\\ &= 2\frac{\partial}{\partial x}\left[Re\left[\frac{C_n}{i}\right] - Im\left[\frac{S_n}{i}\right]\right] - I_{n-1}\\ &= 2\frac{\partial}{\partial x}\left[Im\left[C_n\right] + Re\left[S_n\right]\right] - I_{n-1}\\ &= 2\frac{\partial J_n}{\partial x} - I_{n-1}. \end{align*} Similarly, we have: \begin{align*} J_{n+1} = -2\frac{\partial I_n}{\partial x} - J_{n-1}. \end{align*}
By defining $Z_n = I_n + iJ_n$ we can use the last two recurrence relations to obtain: \begin{align*} Z_{n+1} = -2i \frac{\partial Z_n}{\partial x} - Z_{n-1}, \end{align*} where $Z_n$ can also be defined using integral equation as: \begin{align*} \forall n\in\mathbb N, \qquad Z_n = \int_{0}^{2\pi} e^{i(x\cos(\theta) + y\sin(\theta) + n\theta + d)}d\theta. \end{align*}
Using brute force computation (see this post for the full answer), the solution is: \begin{align*} \quad Z_{\pm n} = 2\pi e^{id} \left(\frac{x\pm iy}{x^2+y^2}\right)^{n}\left[r^2 J_0(r) \sum_{k=1}^{u_n}a_k^{n} \left(x^2 + y^2\right)^{k-1} + i r J_1(r) \sum_{k=1}^{u_{n+1}}b_k^{n} \left(x^2 + y^2\right)^{k-1}\right] \end{align*} where $n\geq 1$ and $u_n$ is defined as $u_n = \left(2n+(-1)^n -1\right)/4$. The two coefficients can be computed using: \begin{equation*} \forall k\geq 1, n\geq 2k,\quad a_k^{n} = (2i)^{n-2k}\frac{(n-k)!(n-(k+1))!}{(k-1)!k!(n-2k)!} \end{equation*}
\begin{equation*} \forall k\geq 1, n\geq 2k-1,\quad b_k^{n} = (2i)^{n-(2k-1)}\frac{(n-k)!(n-k)!}{(k-1)!(k-1)!(n-(2k-1))!)} \end{equation*}
Remarks
(1) For negative integers, it is just a matter of sign: \begin{align*} \forall n\in\mathbb N, \qquad I_{-n} &= Re\left[C_n\right] + Im\left[S_n\right] \\ \forall n\in\mathbb N, \qquad J_{-n} &= Im\left[C_n\right] - Re\left[S_n\right]. \end{align*}
(2) For non integers number, considering the developments and all the computations that I have done, I suspect that factorial derivatives are involved but I have no idea how to do it.