This is found in [1] $\S 82$, Satz 5. It covers the case where the numerator $a_n$ is polynomial of degree $2$ in $n$ and the denominator $b_n$ is degree $1$.
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.
(The numerator ${{}_2F_1}$ has a zero in there, so it turns out to be constant.)
[1] Oskar Perron, Die lehre von den Kettenbrüchen, 2 Auflage 1929