Find the analytic form of expression for the following integral $$
\hspace{0.3cm} \large {\int_{0} ^{\infty}  \frac{\frac{1}{x^4} \hspace{0.1cm} e^{- \frac{r}{x^2}}\hspace{0.1cm}e^{- \frac{r}{z^2}} }{ \frac{1}{x^2} \hspace{0.1cm} e^{- \frac{r}{x^2}}+  \frac{1}{y^2} \hspace{0.1cm} e^{- \frac{r}{y^2}}}}  dr   \hspace{.2cm}   ; \hspace{1cm}            x>0,y>0,z>0   $$ where $ x $ ,$ y $  and $z $ are constants independent of $ r $.

The integral is pretty awkward as written. Let's rename the parameters:
$$a=\frac{1}{x^2} \\ b= \frac{1}{y^2} \\ c= \frac{1}{z^2}$$
Then we have:
$$a^2 \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}$$
So we need to find the following integral:
$$I(a,b,c)= \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}$$
I'd really prefer the integral to be symmetric in $(a,b)$, so let's change the parameter:
$$c=b+s$$
Now we have:
$$J(a,b,s)=J(b,a,s)= \int_0^\infty \frac{e^{-(a+b+s)r} ~dr}{ae^{-ar}+be^{-br}}$$
We notice that the integrand is in the form of harmonic mean:
$$2 a bJ(a,b,s)= \int_0^\infty \frac{2a e^{-ar} b e^{-br} }{ae^{-ar}+be^{-br}}~e^{-sr} ~dr \leq \int_0^\infty \sqrt{ab} e^{-\frac{a+b}{2}r} e^{-sr} ~dr$$
Using the harmonic mean - geometric mean inequality, we have obtained a very nice bound, assuming:
$$\frac{a+b}{2}+s>0  \qquad \Rightarrow \qquad \frac{a-b}{2}+c >0$$
$$2 a bJ(a,b,s) \leq \frac{2\sqrt{ab}}{a+b+2s}  \\  J(a,b,s) \leq \frac{1}{(a+b+2s)\sqrt{ab}}$$
The equality holds for the case $a=b$, where we get:
$$J(a,a,s) = \frac{1}{2a(a+s)}=\frac{1}{2ac}$$
Which is confirmed by considering the original integral.

However, what about the general expression? Let's transform the integrand a little:
$$J(a,b,s)=\frac{1}{a} \int_0^\infty \frac{e^{-(b+s)r} ~dr}{1+\frac{b}{a} e^{-(b-a)r}}$$
Let's introduce a substitution:
$$e^{-(b-a)r}=t \\ -(b-a) dr= \frac{dt}{t}$$
We assume $b>a$, without the loss of generality, since the integral is symmetric in $a,b$.
We obtain:
$$J(a,b,s)=\frac{1}{a(b-a)} \int_0^1 \frac{t^{(b+s)/(b-a)-1} ~dt}{1+\frac{b}{a} t}$$
$$J(a,b,s)=\frac{1}{a(b-a)} \int_0^1 t^{\frac{b+s}{b-a}-1} \left( 1+\frac{b}{a} t\right)^{-1} ~dt$$
This integral can be expressed in terms of the hypergeometric function:
$${_2 F_1} (\alpha, \beta; \gamma; p)$$
$$\alpha=1 \\ \beta=\frac{b+s}{b-a} \\  \gamma=\beta+1=\frac{2b-a+s}{b-a} \\ p=-\frac{b}{a}$$

$$J(a,b,s)=\frac{1}{a(b+s)} {_2 F_1} \left(1,\frac{b+s}{b-a};\frac{2b-a+s}{b-a};-\frac{b}{a} \right)$$
For the original integral we get:

$$a^2 \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}=\frac{a}{c} ~{_2 F_1} \left(1,\frac{c}{b-a};1+\frac{c}{b-a};-\frac{b}{a} \right)$$

