Product of d-dimensional Legendre polynomials

Let $$P_n:\mathbb{R}\rightarrow\mathbb{R}$$ be the $$d$$-dimensional Legendre polynomials, that is they are orthonormal polynomials w.r.t the probability measure $$\mu_d$$ on $$[-1,1]$$ given by $$\mu_d= \frac{\Gamma\left(\frac{d}{2}\right)}{\sqrt{\pi}\Gamma\left(\frac{d-1}{2}\right)}\cdot (1-x^2)^{\frac{d-3}{2}}$$. In other words, $$\int_{-1}^1P_i(x)P_j(x)\mu_d(x) = \delta_{i,j}$$.

Let $$v_1,\dots,v_N\in \mathbb{S}^{d-1}$$ be some vectors. I need to calculate the following product:

$$\int_{x\sim\mathcal{U}(\mathbb{S}^{d-1})} P_{i_1}(\langle v_1,x\rangle)\cdots P_{i_N}(\langle v_N,x\rangle) dx~,$$

where $$\mathcal{U}(\mathbb{S}^{d-1})$$ is the uniform distribution over the unit sphere.

I think that I can prove that for two Legendre polynomials, we have $$\int_{x\sim\mathcal{U}(\mathbb{S}^{d-1})} P_i(\langle v_i,x\rangle)P_j(\langle v_j,x\rangle)dx = \delta_{i,j}P_i(\langle v_i,v_j\rangle)a_{i,d}$$ where $$a_{i,d}$$ is some normalization term that depends on $$i$$ and $$d$$. However, even for three polynomials, I couldn't calculate this integral.

Also, is it possible to bound this integral by some constant independent of the degrees of the polynomials? For example, in the case of a product of two polynomials, I think this is bounded by $$1$$.