Given a product of Laguerre polynomials, $L_n(x) L_m(x)$, a particular question to ask is the expansion of this product in terms of the Laguerre polynomials $\{L_i(x)\}$ themselves. That is, we would seek the expansion coefficients $C_{nm}^i$ such that \begin{align} L_n(x) L_m(x) = \sum_{i=0}^\infty C_{nm}^i L_i(x). \end{align} Given the orthogonality relation of the Laguerre polynomials, the expression for $C_{nm}^i$ is \begin{align} C_{nm}^i = \int_0^\infty L_n(x)L_m(x) L_i(x) e^{-x}\,\text{d}x. \end{align} This problem has been explored and solved several times in the past. For instance, one may look at the paper "Products of Laguerre Polynomials" by Joseph Gillis and George Weiss from Mathematics of Computation in 1960.
What I am after now is an expression that is slightly modified from the integral expression of $C_{nm}^i$ above. That is, I would like to know how to solve the integral \begin{align} I_{nm}^i= \int_0^\infty L_n(x)L_m(x) L_i(x) e^{-3x/2}\,\text{d}x. \end{align} The only difference between $I_{nm}^i$ and $C_{nm}^i$ is the coefficient of the $x$ in the exponent of the integrand. Of course, one may expand each of the Laguerre polynomials using their explicit forms of \begin{align} L_a (x)= \sum_{k=0}^a \binom{a}{k} \frac{(-1)^k}{k!} x^k, \end{align} and evaluate $I_{nm}^i$ term by term. But the result will be some large chunky expression with three summations. Perhaps there is some resumming trickery that one may do, but I do not know of those tricks.
Any help in evaluating $I_{nm}^i$?
For Reference The formula for $C_{nm}^i$, is given by \begin{align} C_{nm}^i = (-1/2)^{n+m-i}\sum_j 2^{2j} \frac{(n+m-j)!}{(n-j)!(m-j)!(2j-n-m+i)!(n+m-i-j)!} \end{align} and the limits of the summation are defined to be that none of the arguments within any of the factorials are negative. This stipulates that $i$, in the summation of the expansion in the first equation in this question, only runs from $i = |n-m|$ to $n+m$. That is \begin{align} L_n(x) L_m(x) = \sum_{i=|n-m|}^{n+m} C_{nm}^i L_i(x). \end{align}
