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you want to integrate along the line $\int_{\gamma-i\infty}^{\gamma+i\infty}ds$ where $\gamma>\max(0,-a/2,-b/2)$, then Mathematica says that _{$$F\left ( s \right )=\frac{1}{2i\pi}\int_{\gamma-i\infty}^{\gamma+i\infty} \Gamma(s)\Gamma (2s+a)\Gamma( 2s+b)x^{-s}\,ds=\frac{\pi ^2 x^{a/2} \, _0F_4\left(;\frac{1}{2},\frac{a}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{1}{2},\frac{a}{2}-\frac{b}{2}+1;-\frac{x}{16}\right)}{a \Gamma \left(\frac{a}{2}\right) \Gamma (a-b+1)\sin \left(\frac{\pi a}{2}\right) \sin (\pi (a-b))}-\frac{\pi ^2 x^{b/2} \, _0F_4\left(;\frac{1}{2},\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{1}{2},-\frac{a}{2}+\frac{b}{2}+1;-\frac{x}{16}\right)}{b \Gamma \left(\frac{b}{2}\right) \Gamma (-a+b+1)\sin \left(\frac{\pi b}{2}\right) \sin (\pi (a-b))}+\frac{\pi ^2 x^{\frac{a+1}{2}} \, _0F_4\left(;\frac{3}{2},\frac{a}{2}+\frac{3}{2},\frac{a}{2}-\frac{b}{2}+1,\frac{a}{2}-\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{2 \Gamma \left(\frac{a}{2}+\frac{3}{2}\right) \Gamma (a-b+2)\cos \left(\frac{\pi a}{2}\right) \sin (\pi (a-b))}-\frac{\pi ^2 x^{\frac{b+1}{2}} \, _0F_4\left(;\frac{3}{2},\frac{b}{2}+\frac{3}{2},-\frac{a}{2}+\frac{b}{2}+1,-\frac{a}{2}+\frac{b}{2}+\frac{3}{2};-\frac{x}{16}\right)}{2 \Gamma \left(\frac{b}{2}+\frac{3}{2}\right) \Gamma (-a+b+2)\cos \left(\frac{\pi b}{2}\right) \sin (\pi (a-b))}+\frac{\pi ^2 \, _0F_4\left(;\frac{1}{2}-\frac{a}{2},1-\frac{a}{2},\frac{1}{2}-\frac{b}{2},1-\frac{b}{2};-\frac{x}{16}\right)}{\Gamma (1-a) \Gamma (1-b)\sin (\pi a) \sin (\pi b)}.$$}