Compute in a closed form the following sum : $\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$ Today Im going to  find the closed form of : 
$\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}$
My attempt : 
We know that : $\Gamma(z)=\int_0^{+\infty}t^{n-1}e^{-t}dt$ 
So :  $\Gamma^{4}(n+\frac{3}{4})=\int_{[0,+\infty[}(xyzt)^{n-\frac{1}{4}}e^{-x-y-z-t}dxdydzdt$ 
But the problems in : $\sum_{n=1}^{+\infty}\frac{x^{n}}{(4n+3)^{2}(n!)^{4}}$ I think related with hypergeomtric function 
if there some trick to compute this original sum drop here , thanks!
 A: As one could expect, the result must involve hypergeometric function.
A CAS gave for the  infinite summation
$$\frac{\Gamma \left(\frac{3}{4}\right)^4 \left(3136 \left(\,
   _5F_4\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4};1,1,1,\frac{7}{4};1\right)-1\right)-243 \,
   _6F_5\left(\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4};2,2,2,\frac{11}{4},\frac{11}{4};1\right)\right)}{28224}$$ and its numerical representation is $0.0211403036686719835443455214070$.
Inverse symbolic calculators do not identify this number but it seems to be very close to the positive root of 
$$ 43 x^2+141 x-3=0 \implies x=\frac{1}{86} \left(\sqrt{20397}-141\right)\approx  0.02114030330$$
Edit
In order to keep it, I shall mention that $\color{red}{\text{David H}}$ greatly simplified the expression to
$$\sum_{n=1}^{+\infty}\frac{\Gamma^{4}(n+\frac{3}{4})}{(4n+3)^{2}\Gamma^{4}(n+1)}=\frac{4 \pi ^4}{9 \Gamma \left(\frac{1}{4}\right)^4}\left(\,
   _6F_5\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{
   4};1,1,1,\frac{7}{4},\frac{7}{4};1\right)-1  \right)$$
