I stumbled onto this integral: $$I=\int_{-1}^1 (1-x)^{\alpha-1} (1+x)^{\beta-1} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x)\,dx$$ where $\alpha,\beta>1$.
The Jacobi polynomials satisfy the orthogonality condition: $$\int_{-1}^1 (1-x)^{\alpha} (1+x)^{\beta} P_m^{(\alpha,\beta)} (x)P_n^{(\alpha,\beta)} (x)\,dx =\frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1} \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1)}{\Gamma(n+\alpha+\beta+1)n!} \delta_{nm}, \qquad \alpha,\ \beta > -1$$ so my integral is almost this but not quite.
I thought of using: $$\frac{d^k}{dz^k} P_n^{(\alpha,\beta)} (z) = \frac{\Gamma (\alpha+\beta+n+1+k)}{2^k \Gamma (\alpha+\beta+n+1)} P_{n-k}^{(\alpha+k, \beta+k)} (z)$$ as: $$\frac{d}{dz} P_n^{(\alpha-1,\beta-1)} (z) = \frac{\Gamma (\alpha+\beta+n+1)}{2 \Gamma (\alpha+\beta+n)} P_{n-1}^{(\alpha, \beta)} (z)$$ but then I am stuck with: $$I=\frac{4 \Gamma (\alpha+\beta+n)\Gamma (\alpha+\beta+m)}{\Gamma (\alpha+\beta+n+1)\Gamma (\alpha+\beta+m+1)}\int_{-1}^1 (1-x)^{\alpha-1} (1+x)^{\beta-1}\frac{d}{du} P_{m-1}^{(\alpha-1,\beta-1)} (u) \frac{d}{du} P_{n-1}^{(\alpha-1,\beta-1)} (u)\,dx$$