Yes, it has a closed form.
Edited according to your request. $$2\cdot n! (n-1)! \sum_{r=0}^{n-3} \frac{(r+2)(r+1)}{(n+r+2)!(n-r-2)!} =\frac2{\binom{2n}n}\left(2^{2n-2}+1-\binom{2n-1}{n-1}-n\right).$$
T. Amdeberhan
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