I am looking for an analytic solution of the integral
\begin{align} \int\limits_{-1}^{1} x^k \cdot (1-x^2)^{\alpha-1/2} C_n^{\alpha}(x) dx \end{align} where $C_n^{\alpha}(x)$ is a given Gegenbauer polynomial and $k$ is an integer.
A similar integral with integration range from 0 to 1 is calculated here, https://link.springer.com/article/10.1007/BF01650571, but I do not see how to calculate it in the range -1 to 1. A related question for Legendre polynomials is also answered here (https://mathoverflow.net/questions/244204/legendre-polynomial-integral) , but it also refers to the above paper, which only includes half the integration range, and only the special cases of $k=1$ and $k=2$ are given explicitly. Or am I missing something?
Since $$C_{n}^{\alpha}\left(-x\right)=\left(-1\right)^{n}C_{n}^{\alpha}\left(x\right)$$ which can be easily deduced by the explict representations of $C_{n}^{\alpha}\left(x\right)$, we can simply observe that \begin{align} \int_{-1}^{1}x^{k}\left(1-x^{2}\right)^{\alpha-1/2}C_{n}^{\alpha}\left(x\right)dx= & \int_{0}^{1}x^{k}\left(1-x^{2}\right)^{\alpha-1/2}C_{n}^{\alpha}\left(x\right)dx \\ + &\int_{-1}^{0}x^{k}\left(1-x^{2}\right)^{\alpha-1/2}C_{n}^{\alpha}\left(x\right)dx \\ = & \left(1+\left(-1\right)^{k+n}\right)\int_{0}^{1}x^{k}\left(1-x^{2}\right)^{\alpha-1/2}C_{n}^{\alpha}\left(x\right)dx \end{align} and now you can proceed as in the paper you cited.