differential equation $y' = \frac{a}{(b+xy)^2}$ Solve the differential equation
$$y' = \frac{a}{(b+xy)^2}$$
My attempt: Substitution
$$y = \frac{z}{x} \implies y' =  \frac{xz' - z}{x^2} \\ 
\frac{xz'-z}{x^2} = \frac{a}{(b+z)^2}$$
But this doesn't seem to help much.
Wolfram Alpha says
$$-\frac{\sqrt a \tanh^{-1}\left(\frac{\sqrt b y}{\sqrt a}\right)}{b^{3/2}} + \frac{\sqrt a \tanh^{-1}\left(\frac{ b y(b+y)-ax}{\sqrt a b^{3/2}}\right)}{b^{3/2}} + \frac{y}{b} = c_1$$

Switch variables to make
$$x'=\frac 1a(b+x y)^2$$ which is quite ugly but doable.
You will have $x(y)$ but no way to inverse it. Using a CAS gives the nasty
$$x=\frac{b^{3/2} \left(-c_1 \cosh \left(\frac{\sqrt{b} y}{\sqrt{a}}\right)+i \sinh
   \left(\frac{\sqrt{b} y}{\sqrt{a}}\right)\right)}{\left(\sqrt{b} c_1 y+i
   \sqrt{a}\right) \cosh \left(\frac{\sqrt{b} y}{\sqrt{a}}\right)-\left(\sqrt{a}
   c_1+i \sqrt{b} y\right) \sinh \left(\frac{\sqrt{b} y}{\sqrt{a}}\right)}$$
