# For $h(t) \equiv e^{ct}$, express $\int_{-\pi}^\pi h(\cos x)h(t\cos x + (1-t^2)^{1/2}\sin x)dx$ in terms of special polynomials / functions

For any continuous function $$h:[-1,1] \to \mathbb R$$, define $$I_h:[-1,1] \to \mathbb R$$ by $$I_h(t) := \int_{-\pi}^\pi h(\cos\theta)h(t\cos\theta + (1-t^2)^{1/2}\sin\theta)d\theta = \int_{-\pi}^\pi h(\cos\theta)h(\sin(\theta + \alpha))d\theta,$$ where $$\alpha=\alpha(t) := \arcsin(t)$$.

For example, in the case $$h(t) \equiv t^n$$, the integral $$I_h(t)$$ has been computed here https://math.stackexchange.com/a/4242413/168758. Furthermore, the result in this case can be written compactly as $$I_h(t) = 2^{1-n}\pi(-i\cos\alpha)^nP_n(i\tan\alpha)$$, where $$P_n$$ is the $$n$$th Legendre polynomial.

Question. For $$h(t) \equiv e^{ct}$$ with $$c>0$$, is there a closed form solution in terms of special polynomials / functions (Legendre, Hermite, Bessel, etc.) ?

I'm really only interested in the values of $$I_h(0)$$, $$I_h(1)$$, and $$I_h'(0)$$.

Example. If $$c=1$$, then $$I_h(0) = 2\pi J_2(0)$$, where $$J_2$$ is the modified Bessel function of the second kind.

The integral is easily computable in terms of a Bessel function representation. For $$h(t)=e^{ct}$$ the integral in question reads
$$I_h(t)=\int_{-\pi}^{\pi}dx e^{c(1+t)\cos x+c\sqrt{1-t^2}\sin x}=\int_{-\pi}^{\pi}dx e^{c\sqrt{2(1+t)}\cos(x-\gamma)}$$
where the angle $$\gamma$$ is defined by $$\cos\gamma=\sqrt{\frac{1+t}{2}}~,~ \sin\gamma=\sqrt{\frac{1-t}{2}}$$. Because of the periodicity of the integrand and the fact that we are integrating over the exact period of the integrand we are allowed to shift away the angle $$\gamma$$. The resulting integral can be computed in terms of a modified Bessel function as follows
$$I_h(t)=\int_{-\pi}^{\pi}e^{c\sqrt{2(1+t)}\cos y}dy= 2\pi I_0\left(c\sqrt{2(1+t)}\right)$$