can you explain how can you solve this Integration

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From here we have:

$L_p^l\left(a x^2\right)=\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a x^2\right)^k}{k!}$

Then we have:

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$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\\\sum _{l=0}^{\infty } \frac{(c x y)^l \left(\sum _{k=0}^p \frac{(-1)^k \binom{p+l}{p-k} \left(a x^2\right)^k}{k!}\right) \sum _{j=0}^p \frac{(-1)^j \binom{p+l}{p-j} \left(b y^2\right)^j}{j!}}{(l+p)!}=\\\sum _{k=0}^p \sum _{j=0}^p \left(\sum _{l=0}^{\infty } \frac{(-1)^{j+k} \left(a x^2\right)^k (c x y)^l \left(b y^2\right)^j \binom{l+p}{-j+p} \binom{l+p}{-k+p}}{j! k! (l+p)!}\right)=\\\sum _{k=0}^p \sum _{j=0}^p \frac{(-1)^{j+k} \left(a x^2\right)^k \left(b y^2\right)^j \binom{p}{-j+p} \binom{p}{-k+p} \, _2F_2(1,1+p;1+j,1+k;c x y)}{j! k! p!}$

How did you get this form?

It can be expressed by The Srivastava and Daoust multivariable hypergeometric function. I don't know if this triple sum can be simplified.

$\sum _{l=0}^{\infty } \frac{(c x y)^l L_p^l\left(a x^2\right) L_p^l\left(b y^2\right)}{(l+p)!}=\sum _{m=0}^{\infty } \sum _{n=0}^{\infty } \sum _{u=0}^{\infty } \frac{\left((1)_u (-p)_m (-p)_n (1+p)_u\right) \left(\left(a x^2\right)^m (c x y)^u \left(b y^2\right)^n\right)}{\Gamma (1+p) \left((1)_{m+u} (1)_{n+u}\right) (m! n! u!)}$