Is there a (possibly hypergeometric-type) explicit evaluation of the
continued fraction
$$a-\dfrac{1.(c+d)}{2a-\dfrac{2.(2c+d)}{3a-\dfrac{3.(3c+d)}{4a-\ddots}}}$$
Even the special case $d=0$, $a=1$ would be interesting. Note that
$$\tanh^{-1}(z)=\dfrac{z}{1-\dfrac{1^2z^2}{3-\dfrac{2^2z^2}{5-\ddots}}}$$
but that does not seem to help.



Evaluation of hypergeometric type continued fraction
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


