This is found in [1] $\S 82$, Satz 5  

If I plugged in correctly, we get for your continued fraction:  Let
$a, c, d$ be complex numbers satisfying: $c \ne 0, a \ne 0, a^2 \ne 4c$, and
$(a^2-4c)/a^2$ is not a negative real.  Then the value is
$$
{\frac { \left( d+c \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d
 \right) }{2c} 
\;{\mbox{$_2$F$_1$}\left(1,{\frac {d+c}{c}};{\frac { \left( 3\,c+d \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d \right) }{2c\sqrt {{a}^{2}-4\,c}}};{\frac {\sqrt {{a}^{2}-4\,c}-a}{2\sqrt {{a}^{2}-4\,c}}}\right)}^{-1}}
$$


The sign of the square-root is chosen so that $\displaystyle\frac{a}{\sqrt{a^2-4c}}$ has positive real part.




[1] Oskar Perron, *Die lehre von den Kettenbrüchen*, 2 Auflage 1929