Legendre Polynomials Triple Product

I think the best way to approach this is as follows, note that \begin{align} (x^2 -1 ) = \frac{P_2 - 2}{3} \end{align} You can then use the following definition \begin{align} P_kP_l = \sum_{m=|k-l|}^{k+l} \begin{pmatrix} k & l & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1)P_m \end{align} This allows the integral to be written as follows \begin{align} \int_{-1}^{1} (x^2-1)^3P_iP_jP_k \; dx &= \int_{-1}^{1} \frac{1}{9}\left(P_2^3 + . . .-8 \right) P_i P_j P_k \; dx \end{align} The most difficult term to deal with is the $ P_2^3 P_i P_j P_k$ \begin{align} P_2^3 P_i P_j P_k &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1)P_m P_2 P_i P_j P_k \\ &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) \sum_{n=|m-2|}^{m+2} \begin{pmatrix} 2 & m & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1)P_n P_i P_j P_k \\ &= \sum_{m=0}^{4} \begin{pmatrix} 2 & 2 & m \\ 0 & 0 & 0 \end{pmatrix}^2 (2m+1) \sum_{n=|m-2|}^{m+2} \begin{pmatrix} 2 & m & n \\ 0 & 0 & 0 \end{pmatrix}^2 (2n+1) \sum_{l=|n-i|}^{n+i} \begin{pmatrix} n & i & l \\ 0 & 0 & 0 \end{pmatrix}^2 (2l+1) P_l P_j P_k \end{align} Which can then make use of the usual triple integral. All other terms can be solved for in a similar manner.

You could integrate one of the $P_k(x)$ and take the derivative of the rest. The power of $(1-x^2)$ gets reduce by the fact that $$\partial_x P_l(x) = \frac{(1+l) [ P_{l+1}(x)-x P_l(x) ]}{x^2-1}.$$ You have to apply partial integration a few times and you will generate a bunch of Legendre triple product.