The following question has a 500 points bounty on MSE that soon comes to an end, and no answer
as expected was given yet. How would a professional solve the problem? Wish you succcess.

11$\begingroup$ It's already expressed as a hypergeometric function. Is there any reason to expect anything more "closedform" than that? $\endgroup$ – Robert Israel May 11 '15 at 7:54

1$\begingroup$ @RobertIsrael my intuition strongly tells me that, yes, I should expect a closedform, hence the question on MSE, the bounty granted and then the post here. Perhaps after attending thousands of such questions is hard not to listen the inner voice of the intuition. Besides that, I think your question is far harder than my question, and to answer it as you expected I'd probably need to post another question and grant another bounty. I'll continue to rely on my intuition, it led me to amazing (to me) results so far. $\endgroup$ – user 1357113 May 11 '15 at 11:00

$\begingroup$ @RobertIsrael To give you a simple example, it's natural to ask yourself, after seeing the nice closed form of $$\sum_{n=1}^{\infty} B(n,n)=\frac{2\pi}{3\sqrt{3}}$$, if the quadratic version in terms of beta function has such a nice form too (avoiding the hypergeometric function), that is $$\sum_{n=1}^{\infty} (B(n,n))^2.$$ $\endgroup$ – user 1357113 May 11 '15 at 12:41

3$\begingroup$ @Chris'ssis, this may not be directly relevant and you may have already seen it, but the nice article What is a closedform number? (arXiv:math/9805045) might be of interest to you. $\endgroup$ – Igor Khavkine May 11 '15 at 16:28

2$\begingroup$ it's pretty close to $\frac{1}{3}\Gamma(2/3)I_0(2)$ $\endgroup$ – Carlo Beenakker May 11 '15 at 19:41
Not what you're really looking for, but here's a "word answer" in terms of a probabilistic interpretation.
Suppose Alice and Bob are playing a game where they are presented with $2n+1$ indistinguishable bees and have to separate them into
 1 queen bee,
 $n$ worker bees, and
 $n$ drones.
They both have to try to guess what groups the other has chosen. They play this game for increasing $n=0,1,2,3,\dots$
The probability that Alice is right is $$ \frac1{\binom{2n+1}{n}\binom{n}{1}} = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$
Thus the number you're interested in is the expected number of times Alice and Bob both guess correctly.
For the record here is the usual derivation relating the integral to the infinite series in terms of the beta function $B$.
Let $a(x)=x(1x)$ and $b(x,y)=a(x)a(y)$, and $$ c(x,y)=\frac1{1b(x,y)} = \sum_{n=0}^\infty b(x,y)^n. $$ In terms of the beta function $B$ $$ \int_0^1 a(x)^n\,dx = \frac{(n!)^2}{(2n+1)!} = B(n+1,n+1) $$ so $$ \int_{[0,1]^2} c(x,y)\,dx\,dy = \sum_{n=0}^\infty B(n+1,n+1)^2 = \sum_{n=1}^\infty B(n,n)^2 = \sum_{n=1}^\infty \left(\frac{\Gamma(n)^2}{\Gamma(2n)}\right)^2 $$ $$ =\sum_{n=1}^\infty \frac{\Gamma(n)^4}{\Gamma(2n)^2} = {_3}F_2\left(1,1,1; \frac32,\frac32; \frac1{16}\right). $$ (Wolfram Alpha link for the last equation)

1$\begingroup$ And along these lines, for $b(x_{1},.., x_{k})=\prod_{i=0}^{k} a(x_{i})$, $\int_{[0,1]^{k}} c(x_{1},..,x_{k}) dx_{1}..dx_{k} = \,_{k+1}F_{k}\left(1,..,1; \frac{3}{2},..,\frac{3}{2}; 4^{k}\right)$ with $k+1$ '1's and $k$ '$\frac{3}{2}$'s. $\endgroup$ – Johannes Trost May 19 '15 at 16:56