How to compute $\frac{1}{2\pi}\int_{-\pi}^\pi (\cos y + x)^{2k}dy$ and similar integrals For an integer $ k \ge 0$ and a scalar $x \in \mathbb R$, define
$$
\begin{split}
A_k(x) &:= \frac{1}{2\pi}\int_{-\pi}^\pi (\cos y + x)^{2k}dy,\\
B_k(x) &:= \frac{1}{2\pi}\int_{-\pi}^\pi (\cos y + x)^k(\sin y + x)^kdy,\\
C_k(x) &:= \frac{1}{2\pi}\int_{-\pi}^\pi\cos y (\cos y + x)^k(\sin y + x)^{k-1}dy
\end{split}
$$

Question. How to obtain analytic formulae for $A_k(x)$, $B_k(x)$, and $C_k(x)$ ?

Attempt
Let's attempt to compute $A_k(x)$ (since it looks the least intimidating). By Newton's binomial theorem, one can write $A_k(x) = \sum_{j=0}^k {2k\choose 2j} I_j(x) x^{2k-2j}$, where
$$
I_j(x) := \frac{1}{2\pi}\int_{-\pi}^\pi (\cos y)^{2j} dy=\frac{1}{2\pi}\int_0^{2\pi}(\cos y)^{2j}dy = \frac{1}{2^{2j}}{2j \choose j},
$$
Putting things together gives
$$
A_k(x) = \sum_{j=0}^k{2k\choose 2j}{2j\choose j}2^{-2j}x^{2k-2j} = (2k)!\sum_{j=0}^k \frac{1}{(j!2^j)^2}\frac{x^{2k-2j}}{(2k-2j)!} = \ldots
$$
and I don't know how to go from here (though it looks like the coefficient of a certain order term in a series expansion).
 A: The first of the three families of integrals admits an explicit representation in terms of Legendre polynomials. To wit, define a generating function for
$$A'_k(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}(x+\cos y)^kdy$$
as
$$G(t;x)=\sum_{k=0}^{\infty}A'_k(x)t^k=\frac{1}{2\pi}\int_{-\pi}^\pi\frac{dy}{1-t(x+\cos y)} $$
This integral can be computed elementarily in it's allowed convergence region:
$$G(t;x)=\frac{\text{sgn}(xt-1)}{\sqrt{(1-tx)^2-t^2}}~~,~~ \left|x-1/t\right|\geq 1$$
Of course $t=0$ is included in this region, and it's easy to see that for any finite $x$, we can perform the expansion around that point and when $t\to 0, \text{sgn}(xt-1)\to -1$. We can match this generating function to the one of trivially rescaled Legendre polynomials $a(x)^nP_n(b(x))$ if and only if $a(x)=i\sqrt{1-x^2}, b(x)=-ix/\sqrt{1-x^2}$  and hence we conclude by keeping only the even terms (the odd ones which should be trivially zero come out incorrect in this approach) that
$$A_k(x)=(-1)^k(1-x^2)^kP_{2k}\left(\frac{-ix}{\sqrt{1-x^2}}\right)$$
One can check easily that this formula results in a real expression and that it's range of validity goes past the original limitation of $|x|\leq 1$.
For the other two families of integrals I haven't found a good representation yet. However one can find strikingly simple expressions to the exponential generating function of all three families which I will leave here for future use. Defining
$$B_{kl}(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}dy (\cos y+x)^k(\sin y+x)^l~~,~~ C_{kl}(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}dy \cos y (\cos y+x)^k(\sin y+x)^l$$
$$A(t;x)=\sum_{k=0}^{\infty}A_k(x)\frac{t^k}{k!}~~,~~ B(a,b;x)=\sum_{k,l=0}^{\infty}B_{kl}(x)\frac{a^k b^l}{k!l!}~~,~~
C(a,b;x)=\sum_{k,l=0}^{\infty}C_{kl}(x)\frac{a^k b^l}{k!l!}$$
We can readily compute expressions for the 3 generating functions using the integral representations of Bessel functions
$$A(t;x)=e^{tx}I_0(t)$$
$$B(a,b;x)=e^{(a+b)x}I_0(\sqrt{a^2+b^2})$$
$$C(a,b;x)=e^{(a+b)x}\frac{a}{\sqrt{a^2+b^2}}I_1(\sqrt{a^2+b^2})$$
A: $$ A_k(x) = x^{2k} \sum_{j=0}^{k} \begin{pmatrix} 2k \\ 2j \end{pmatrix} \begin{pmatrix} 2j \\ j \end{pmatrix} \left( \frac{1}{2x} \right)^{2j}  $$
Consider the sum
$$ \Omega(x,k)= \sum_{j=0}^{2k} \begin{pmatrix} 2k \\ j \end{pmatrix} \begin{pmatrix} 2j \\ j \end{pmatrix} \left( \frac{1}{2x} \right)^{j} $$
It's interesting to note that
$$ A_k(x) = \frac{1}{2} x^{2k} \left( \Omega(x,k) + \Omega(-x,k) \right) $$
So I'm not going to focus on trying to sum this omega:
$$ \Omega(x,k)= \sum_{j=0}^{2k} \begin{pmatrix} 2k \\ j \end{pmatrix} \begin{pmatrix} 2j \\ j \end{pmatrix} \left( \frac{1}{2x} \right)^{j} $$
If you let $s = 2k$ ,$ u = \frac{1}{2x}$ then this is the same as
$$ \sum_{j=0}^{s} \begin{pmatrix} s \\ j \end{pmatrix}  \begin{pmatrix} 2j \\ j \end{pmatrix} u^j  $$
I'm like 99% sure this has a closed form. ill let you know if i find something.
So there's a closed form in terms of hypergeometric functions. I'm ab it surprised its not more elementary than this:
https://www.wolframalpha.com/input/?i=Sum%5B++%28k+choose+n%29*%282*n+choose+n%29*x%5En%2C+n+%3D+0+to+k%5D
