Term independent of x and y in this expansion

Find the term independent of $x$ and $y$ in
$$\left(x+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}+2\right)^{20}$$

My attempt

factorised it as$$\frac{(x+y+2)(xy+1)-1)}{xy}$$ but still the method is too lengthy

Solutions Collecting From Web of "Term independent of x and y in this expansion"

What about the brutal way?
$$ \left(x+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}+2\right)^{20} = \sum\binom{20}{a,b,c,d,e,f}2^f x^{a-c-e}y^{b-d-e} $$
where the sum is performed over the tuples $(a,b,c,d,e,f)\in\mathbb{N}^6$ such that $a+b+c+d+e+f=20$. If we are just interested in the constant term, we are just interested in the tuples for which $a=c+e$ and $b=d+e$, i.e. the triples $(c,d,e)\in\mathbb{N}^3$ such that $2c+2d+3e\leq 20$. The constant term is so given by
$$ \sum_{\substack{(c,d,e)\in\mathbb{N}^3\\2c+2d+3e\leq 20}}\frac{20!}{(c+e)!(d+e)!c!d!e!f!}2^f\quad \text{where}\;f=20-2c-2d-3d$$
that is trivially greater than $2^{20}+760\cdot 2^{18}$ (by restricting the sum just to $f=20$ and $f=18$, since $f=19$ is impossible). There are exactly $192$ triples contributing to the last sum. Thanks to Mathematica, the constant term is actually $\color{red}{737880155312536}$, between $2^{49}$ and $2^{50}$ and with a huge prime factor ($5957949707$).