This answer is intended to provide additional insight into the [initial answer][1] posted by [Carlo Beenakker][2].

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The MeijerG function is defined as

$$\text{MeijerG}\left[\{\{a_1..a_n\},\{a_{n+1}..a_p\}\},\{\{b_1..b_m\},\{b_{m+1}..b_q\}\},z,r\right]$$ $$=\frac{r}{2 \pi i} \int \frac{\left(\left(\prod\limits_{i=1}^n \Gamma(1-a_i-r s)\right) \prod\limits_{i=1}^m \Gamma(b_i+r s)\right)}{\left(\prod\limits_{i=n+1}^p \Gamma(a_i+r s)\right) \prod\limits_{i=m+1}^q \Gamma(1-b_i-r s)} z^{-s} \, ds\tag{1}$$

and with the default $r=1$ this becomes

$$\text{MeijerG}\left[\{\{a_1..a_n\},\{a_{n+1}..a_p\}\},\{\{b_1..b_m\},\{b_{m+1}..b_q\}\},z\right]$$ $$=G_{p, q}^{m, n}\left(z\left|
\begin{array}{c}
 a_1 \text{..} a_p \\
 b_1 \text{..} b_q \\
\end{array}
\right.\right)$$ $$=\frac{1}{2 \pi i} \int \frac{\left(\left(\prod\limits_{i=1}^n \Gamma(1-a_i-s)\right) \prod\limits_{i=1}^m \Gamma(b_i+s)\right)}{\left(\prod\limits_{i=n+1}^p \Gamma(a_i+s)\right) \prod\limits_{i=m+1}^q \Gamma(1-b_i-s)} z^{-s} \, ds\tag{2}$$

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From the definition above it can be seen that

$$f(x)=\frac{1}{\pi}\ 2^{a+b-2}\ G_{0,5}^{5,0}\left(\frac{x}{16}|
\begin{array}{c}
 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\
\end{array}
\right)$$ $$=\frac{1}{\pi} 2^{a+b-2} \frac{1}{2 \pi i}\int \Gamma(s+0)\ \Gamma\left(\frac{a}{2}+s\right)\ \Gamma\left(\frac{a+1}{2}+s\right)\ \Gamma\left(\frac{b}{2}+s\right)\ \Gamma\left(\frac{b+1}{2}+s\right) \left(\frac{x}{16}\right)^{-s} \, ds$$ $$=\frac{1}{2 \pi i} \int \Gamma(s)\ \Gamma(a+2 s)\ \Gamma(b+2 s)\ x^{-s} \, ds\tag{3}$$

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Mathematica also gives the following result for the inverse Mellin transform.

$$f(x)=\mathcal{M}_s^{-1}[\Gamma(s)\ \Gamma(a+2 s)\ \Gamma(b+2 s)](x)=\frac{1}{2 \pi i} \int \Gamma(s)\ \Gamma(2s+a)\ \Gamma(2s+b)\ x^{-s}\,ds\tag{4}$$ $$=\pi ^3 2^{-a-b-4} \left(4^{a+b+2} \csc (\pi  a) \csc (\pi  b) \, _0\tilde{F}_4\left(;\frac{1}{2}-\frac{a}{2},1-\frac{a}{2},\frac{1}{2}-\frac{b}{2},1-\frac{b}{2};-\frac{x}{16}\right)-2 \csc (\pi  (b-a)) \left(4^{b+1} x^{a/2} \csc \left(\frac{\pi  a}{2}\right) \, _0\tilde{F}_4\left(;\frac{a+2}{2},\frac{1}{2},\frac{1}{2} (a-b+1),\frac{1}{2} (a-b+2);-\frac{x}{16}\right)-4^{a+1} x^{b/2} \csc \left(\frac{\pi  b}{2}\right) \, _0\tilde{F}_4\left(;\frac{b+2}{2},\frac{1}{2} (-a+b+1),\frac{1}{2} (-a+b+2),\frac{1}{2};-\frac{x}{16}\right)+4^b x^{\frac{a+1}{2}} \sec \left(\frac{\pi  a}{2}\right) \, _0\tilde{F}_4\left(;\frac{a+3}{2},\frac{3}{2},\frac{1}{2} (a-b+2),\frac{1}{2} (a-b+3);-\frac{x}{16}\right)-4^a x^{\frac{b+1}{2}} \sec \left(\frac{\pi  b}{2}\right) \, _0\tilde{F}_4\left(;\frac{b+3}{2},\frac{1}{2} (-a+b+2),\frac{1}{2} (-a+b+3),\frac{3}{2};-\frac{x}{16}\right)\right)\right)$$

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Mathematica also expands the MeijerG result as follows.

$$f(x)=\frac{1}{\pi }\ 2^{a+b-2}\ G_{0,5}^{5,0}\left(\frac{x}{16}|
\begin{array}{c}
 0,\frac{a}{2},\frac{a+1}{2},\frac{b}{2},\frac{b+1}{2} \\
\end{array}
\right)\tag{5}$$ $$=\frac{\pi ^2 x^{a/2} \csc \left(\frac{\pi  a}{2}\right) \csc \left(\frac{1}{2} \pi  (a-b)\right) \sec \left(\frac{1}{2} \pi  (a-b)\right) \, _0F_4\left(;\frac{1}{2},\frac{a}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{1}{2},\frac{a}{2}-\frac{b}{2}+1;-\frac{x}{16}\right)}{4 \Gamma \left(\frac{a}{2}+1\right) \Gamma (a-b+1)}-\frac{\pi ^2 x^{b/2} \csc \left(\frac{\pi  b}{2}\right) \csc \left(\frac{\pi  a}{2}-\frac{\pi  b}{2}\right) \sec \left(\frac{1}{2} \pi  (a-b)\right) \, _0F_4\left(;\frac{1}{2},\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{1}{2},-\frac{a}{2}+\frac{b}{2}+1;-\frac{x}{16}\right)}{4 \Gamma \left(\frac{b}{2}+1\right) \Gamma (-a+b+1)}+\frac{\pi ^2 x^{\frac{a+1}{2}} \sec \left(\frac{\pi  a}{2}\right) \csc \left(\frac{1}{2} \pi  (a-b)\right) \sec \left(\frac{1}{2} \pi  (a-b)\right) \, _0F_4\left(;\frac{3}{2},\frac{a}{2}+\frac{3}{2},\frac{a}{2}-\frac{b}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{4 \Gamma \left(\frac{a}{2}+\frac{3}{2}\right) \Gamma (a-b+2)}-\frac{\pi ^2 x^{\frac{b+1}{2}} \sec \left(\frac{\pi  b}{2}\right) \csc \left(\frac{\pi  a}{2}-\frac{\pi  b}{2}\right) \sec \left(\frac{1}{2} \pi  (a-b)\right) \, _0F_4\left(;\frac{3}{2},\frac{b}{2}+\frac{3}{2},-\frac{a}{2}+\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{4 \Gamma \left(\frac{b}{2}+\frac{3}{2}\right) \Gamma (-a+b+2)}+\frac{\pi ^2 \csc \left(\frac{\pi  a}{2}\right) \csc \left(\frac{1}{2} \pi  (a+1)\right) \csc \left(\frac{\pi  b}{2}\right) \csc \left(\frac{1}{2} \pi  (b+1)\right) \, _0F_4\left(;\frac{1}{2}-\frac{a}{2},1-\frac{a}{2},\frac{1}{2}-\frac{b}{2},1-\frac{b}{2};-\frac{x}{16}\right)}{4 \Gamma (1-a) \Gamma (1-b)}$$


  [1]: https://mathoverflow.net/q/420796
  [2]: https://mathoverflow.net/users/11260/carlo-beenakker