SF7

$\frac{6\Gamma(8/3)}{5\Gamma(2/3)}\,$

Since $\Gamma(n+1)=n\Gamma(n)\,$ ,

$\frac{6\frac{5}{3}\frac{2}{3}\Gamma(\frac{2}{3})}{5\Gamma(\frac{2}{3})} = \frac{6\frac{5}{3}\frac{2}{3}}{5} = \frac{4}{3}$