Mathworld's discussion of the Gamma function has the pleasant formula:

$$ \frac{\Gamma(\frac{1}{24})\Gamma(\frac{11}{24})}{\Gamma(\frac{5}{24})\Gamma(\frac{7}{24})} = \sqrt{3}\cdot \sqrt{2 + \sqrt{3}} $$

This may have been computed algorithmically, according to the page. So I ask how one might derive this?

My immediate thought was to look at $(\mathbb{Z}/24\mathbb{Z})^\times = \big( \{ 1,5,7,11 \big| 13, 17, 19 , 23 \}, \times \big)$ where $1,5,7,11$ are relatively prime to 24. And the other half?

We could try to use the mirror formula $$ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$ or the Euler beta integral but nothing has come up yet: $$ \int_0^1 x^a (1-x)^b \, dx = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} $$

I am lucky the period integral of some Riemann surface will be the ratio of Gamma functions: $$ \int_0^1 (x - a)^{1/12} (x - 0)^{11/12} (x - 1)^{-5/12} (x - d)^{-7/12} \, dx $$ these integrals appear in the theory of hypergeometric function

In light of comments, I found a paper of Benedict Gross and the paper of Selberg and Chowla

$$ F( \tfrac{1}{4},\tfrac{1}{4};1;\tfrac{1}{64}) = \sqrt{\frac{2}{7\pi}} \times \left[\frac{ \Gamma(\frac{1}{7})\Gamma(\frac{2}{7})\Gamma(\frac{4}{7}) }{ \Gamma(\frac{3}{7})\Gamma(\frac{5}{7})\Gamma(\frac{6}{7}) }\right]^{1/2} $$

so in our case we are looking at quadratic residues mod 12. However, however it does not tell us that LHS evaluates to RHS.