If you restrict the values of $c$ and $d$ to real values where $c<0$ and $d\neq 0$, then all values of the desired random variable are real and a symbolic solution exists. (With more work maybe other symbolic solutions exist if the conditions on $c$ and $d$ are relaxed - I won't tackle that.)
We see that with $c<0$ and $d\neq 0$
FullSimplify[1/2 + Sqrt[c^2 d^2 (d^2 - c x^2)]/(2 c d^2), Assumptions -> {c < 0, d != 0}]
$$\frac{1}{2}-\frac{\sqrt{d^4-c d^2 x^2}}{2 d^2}$$
Because of the restrictions we can simplify this to
$$\frac{1}{2}-\frac{1}{2} \sqrt{1-\frac{c x^2}{d^2}}$$
Because we know that $-c/d^2$ is positive, we can further simplify to
1/2 - Sqrt[1 + e x^2]/2
Now a symbolic solution can be obtained:
a = TransformedDistribution[Sqrt[x^2], x \[Distributed] NormalDistribution[μ, σ]];
b = TransformedDistribution[1/2 - Sqrt[1 + e x^2]/2, x \[Distributed] a,
Assumptions -> {e > 0}];
PDF[b, z]
$$\begin{array}{cc}
\{ &
\begin{array}{cc}
\frac{\frac{(1-2 z) e^{-\frac{\left(2 \sqrt{e} \sqrt{(z-1) z}+e\right)^2}{8 e^2}}}{\sqrt{e} \sqrt{(z-1) z}}+\frac{(1-2 z) e^{-\frac{1}{8} \left(1-\frac{2 \sqrt{(z-1) z}}{\sqrt{e}}\right)^2}}{\sqrt{e} \sqrt{(z-1) z}}}{2 \sqrt{2 \pi }} & z\leq 0 \\
0 & \text{True} \\
\end{array}
\\
\end{array}$$
x
$\endgroup$