How to derive the solution of $F_S(x)=P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right)$? [EDIT]
I came across a received signal-to-interference-plus-noise-ratio (SINR), $S$, of a wireless communication system as
\begin{align*}
 S = \frac{\phi|h|^2\rho_1}{1+|g|^{2} \rho _{2} }, \tag{1}
\end{align*}
where $\phi$ is the power allocation coefficient, $\rho_1$ is the SNR (signal-to-noise-ratio) of channel 1, $\rho_2$ is the SNR of channel 2, $h \sim\mathcal{CN}(0,\sigma^2)$ is channel 1's gain, and $g \sim\mathcal{CN}(0,\sigma^2)$ is channel 2's gain. Furthermore, $H= \mathbb{E}[|h|^2]$ and $G= \mathbb{E}[|g|^2]$.
The authors state that the CDF of the received SINR is
\begin{align*} F_S(x)=&P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right), \end{align*}
where the RHS is found to be
\begin{align*} P \left ({|h|^{2} \le \frac { x \left ({1 + |g|^{2} \rho _{2} }\right)}{\phi \rho _{1}} }\right) = 1 - \frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}},\tag{2} \end{align*}
where  $a=\frac{G \rho_2}{\phi H\rho_1}$.
I understand that $z=h_r^2+h_i^2=\frac{2}{\sigma^2} |h|^2$ is a chi squared distributed random variable with $k=2$ degrees of freedom. Then, the CDF for $k=2$ is
\begin{equation}
F(z;k=2)=1-e^{-z/2}, \tag{3}
\end{equation}
which allows us to obtain the structure of (2).
However, I am unable to figure out how the $\frac {1}{1 + a x}$ in $\frac {1}{1 + a x} e^{-\frac {x}{\phi H \rho _{1}}}$ was derived.
Does the fact that $g$ is a complex random variable also affect the outcome?
Could someone help me understand how the expression in (2) was derived?
Thanks in advance.

The integral turns out to be easier than it looks.
Let's simplify notation a little.  By choosing suitable units for the variables, we may make $\sigma^2=1/2,$ entailing $H=G=1.$  Consequently everything depends on the values of $$\frac{1}{\phi\rho_1} = \beta \gt 0,$$ say, and $$a = \rho_2 \beta \gt 0.$$ $Y=|h|^2$ and $X=|g|^2$ are independent Exponential variables, which means that for any number $\lambda \ge 0,$
$$\Pr(X \le \lambda) = \Pr(Y \le \lambda) = 1 - e^{-\lambda}.$$
The problem is for any $\lambda\ge 0$ to compute
$$\begin{aligned}
\Pr(Y \le \beta \lambda(1 + \rho_2 X)) &= E\left[1 - e^{-( \beta\lambda(1 + \rho_2 X))}\right] \\
&= \int_0^\infty \left(1 - e^{-( \beta\lambda(1 + \rho_2 x))}\right) \,\mathrm{d}\left(1-e^{-x}\right) \\
&= \int_0^\infty e^{-x}\,\mathrm{d}x - e^{-\lambda}\int_0^\infty e^{- \beta\lambda \rho_2 x - x}\,\mathrm{d}x \\
&= 1 - e^{-\beta\lambda} \frac{1}{\beta\lambda\rho_1+1} \\
&= 1 - \frac{1}{1 + a\lambda}e^{-\lambda/(\phi \rho_1)}.
\end{aligned}$$
In the question, "$x$" is used instead of "$\lambda:$" because $H=1,$ this result obviously agrees with the quoted formula in the question, QED.
