solution of a ODE $dy/dx = ((y+a)(y+b) + x)/(y+b)^2$ In my work, I derived an ODE of the following form

$\frac{dy}{dx} = \frac{(y+a)(y+b) + x}{(y+b)^2}$

where $a > 0$ and $b > 0$ are constants. I wonder if there is an analytical solution of this equation? Anyone can give a clue on how to solve it? 
 A: Let $r=y+b$ ,
Then $\dfrac{dr}{dx}=\dfrac{dy}{dx}$
$\therefore\dfrac{dr}{dx}=\dfrac{r(r+a-b)+x}{r^2}$
$(x+r(r+a-b))\dfrac{dx}{dr}=r^2$
This belongs to an Abel equation of the second kind.
Let $u=x+r(r+a-b)$ ,
Then $x=u-r(r+a-b)$
$\dfrac{dx}{dr}=\dfrac{du}{dr}-2r-a+b$
$\therefore u\left(\dfrac{du}{dr}-2r-a+b\right)=r^2$
$u\dfrac{du}{dr}-(2r+a-b)u=r^2$
$u\dfrac{du}{dr}=(2r+a-b)u+r^2$
Let $s=r+\dfrac{a-b}{2}$ ,
Then $\dfrac{du}{dr}=\dfrac{du}{ds}\dfrac{ds}{dr}=\dfrac{du}{ds}$
$\therefore u\dfrac{du}{ds}=2su+\left(s-\dfrac{a-b}{2}\right)^2$
$u\dfrac{du}{ds}=2su+s^2-(a-b)s+\dfrac{(a-b)^2}{4}$
Let $t=s^2$ ,
Then $\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=2s\dfrac{du}{dt}$
$\therefore2su\dfrac{du}{dt}=2su+s^2-(a-b)s+\dfrac{(a-b)^2}{4}$
$u\dfrac{du}{dt}=u+\dfrac{s}{2}-\dfrac{a-b}{2}+\dfrac{(a-b)^2}{8s}$
$u\dfrac{du}{dt}=u\pm\dfrac{\sqrt t}{2}-\dfrac{a-b}{2}\pm\dfrac{(a-b)^2}{8\sqrt t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
For the special case $a-b=-1$ , this exactly belongs to the ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=137.
WLOG, just consider $u\dfrac{du}{dt}=u+\dfrac{\sqrt t}{2}+\dfrac{1}{2}+\dfrac{1}{8\sqrt t}$ ,
The general solution is $\begin{cases}t=\dfrac{(C_1\tau J_1(\tau)+C_2\tau Y_1(\tau)+C_1J_0(\tau)+C_2Y_0(\tau))^2}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\\u=\dfrac{\tau^2((C_1J_1(\tau)+C_2Y_1(\tau))^2\pm(C_1J_0(\tau)+C_2Y_0(\tau))^2)}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\end{cases}$
$\begin{cases}\left(r-\dfrac{1}{2}\right)^2=\dfrac{(C_1\tau J_1(\tau)+C_2\tau Y_1(\tau)+C_1J_0(\tau)+C_2Y_0(\tau))^2}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\\x+r(r-1)=\dfrac{\tau^2((C_1J_1(\tau)+C_2Y_1(\tau))^2\pm(C_1J_0(\tau)+C_2Y_0(\tau))^2)}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\end{cases}$
$\begin{cases}\left(y+b-\dfrac{1}{2}\right)^2=\dfrac{(C_1\tau J_1(\tau)+C_2\tau Y_1(\tau)+C_1J_0(\tau)+C_2Y_0(\tau))^2}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\\x+(y+b)(y+b-1)=\dfrac{\tau^2((C_1J_1(\tau)+C_2Y_1(\tau))^2\pm(C_1J_0(\tau)+C_2Y_0(\tau))^2)}{4(C_1J_0(\tau)+C_2Y_0(\tau))^2}\end{cases}$
A: $$\frac{dy}{dx} = \frac{(y+a)(y+b) + x}{(y+b)^2}$$
Let $\quad Y(x)=\frac{1}{y+b}$
$$Y'=-Y^2+(b-a)Y^3-xY^4$$
This is an Abel's differential equation of the first kind :
http://mathworld.wolfram.com/AbelsDifferentialEquation.html
where $f_0(x)=0\:,\:f_1(x)=0\:,\:f_2(x)=-1\:,\:f_3(x)=b-a\:,\:f_4(x)=-x$
In the general case, the Abel's ODEs are unsolvable differential equations. The most likely, there is no closed form for the solutions. So, my answer is to recommend numerical methods for solving this kind of ODE.
