Wolfram Alpha tells me for instance $$ \frac{\Gamma(6-1/4)}{\Gamma(12+5/4)}-\frac{\Gamma(12-1/4)}{\Gamma(9+5/4)}=\frac{133259008 \sqrt{2} \pi}{1020857565\Gamma(1/4)^2}. $$ I am now looking for a general formula for the constant $C_{ij}$ such that $$ \frac{\Gamma(|i-j|-1/4)}{\Gamma(|i-j|+5/4)}-\frac{\Gamma(i+j-1/4)}{\Gamma(i+j+5/4)}=C_{ij}\frac{\sqrt{2} \pi}{\Gamma(1/4)^2} $$ for $i,j\in\mathbb{N}$, i.e. $C_{9,3}=C_{3,9}= 133259008/1020857565$.
Thanks for your help.
The rising factorial is defined by
$$x^{(n)}=x(x+1)\dots(x+n-1)=\frac{\Gamma(x+n)}{\Gamma(x)}$$
Let $a,b,c,d$ be integers. Then, using the formula $\Gamma(1/4)\Gamma(3/4)=\dfrac{\pi}{\sin(\pi/4)}$:
$$\frac{\Gamma(a-1/4)}{\Gamma(b+1/4)}=\frac{\Gamma(a-1+3/4)}{\Gamma(b+1/4)}=\frac{\Gamma(3/4)\cdot(3/4)^{(a-1)}}{\Gamma(1/4)\cdot(1/4)^{(b)}}=\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}\cdot\frac{\pi}{\Gamma(1/4)^2\sin(\pi/4)}=\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}\cdot\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$$
Likewise for $c$ and $d$. Thus
$$\frac{\Gamma(a-1/4)}{\Gamma(b+1/4)}-\frac{\Gamma(c-1/4)}{\Gamma(d+1/4)}=\left(\frac{(3/4)^{(a-1)}}{(1/4)^{(b)}}-\frac{(3/4)^{(c-1)}}{(1/4)^{(d)}}\right)\cdot\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$$
It's possible to rewrite this with the falling factorial, using the formula $x^{(n)}=(x+n-1)_n$. The factor before $\frac{\pi\sqrt{2}}{\Gamma(1/4)^2}$ only involves finite products of rational numbers.