Inverse Mellin transform of $f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}$ Given the function
$$f(s)= 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)},$$
can we find and inverse Mellin transform for $f(s)$? That is, 
$$\frac{1}{2 \pi i}\int_{- i\infty}^{ i\infty} 2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}   x^{-s-1 }  ds$$
for $x>0$. 
I was wondering if the integral can be expressed in terms of hypergeometric functions?   For example, this is very similar to the  Mellin–Barnes integral 
\begin{align}
{}_2F_1(a,b;c;z) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)} \frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^s\,ds
\end{align}
However, I am not sure how to connect it to my problem!
Thanks. 

Let me offer a partial solution, which I might finish later. Repeatedly using the Gauss multiplication formula yields the following identity:
$$2^{ \frac{s}{6} }\frac{\Gamma \left( \frac{s+1}{3/2} \right)}{ \Gamma \left( \frac{s+1}{2} \right)}=\frac1{\sqrt{\pi } \sqrt[6]{2}}\frac{\Gamma \left(\frac{s}{6}+\frac{11}{12}\right) \Gamma \left(\frac{s}{6}+\frac{2}{3}\right) \Gamma \left(\frac{s}{6}+\frac{5}{12}\right)}{\Gamma\left(\frac{s}{6}+\frac{1}{2}\right) \Gamma \left(\frac{s}{6}+\frac{5}{6}\right)}\left(\sqrt{\frac38}\right)^{-s}$$
and now the inverse Mellin transform can be directly converted into a Meijer $G$-function, yielding
$$\frac{3\sqrt[6]{32}}{\sqrt{\pi}}G_{2,3}^{3,0}\left(\frac{27z^6}{512}\middle|
{{\frac12,\frac56}\atop{\frac{5}{12},\frac23,\frac{11}{12}}}\right)$$
This can then be further expanded into a sum of $3$ ${}_2 F_2$ hypergeometric functions, using this formula.
