how to find explicit formula for $f_{n+1}=af_n+\frac{b}{f_n}$?

I tried to find an explicit formula for the recurrence relation$$f_{n+1}=af_n+\frac{b}{f_n} , f_0 \ne0$$I will show what I got in five cases

Case1

for $f_0=-\frac{\sqrt{b}}{\sqrt{1-a}}\ne0$ I got that $f_n=f_0$ where $a\ne1$

Case2

if $b=0$ then it's easy to show that $f_n=a^n f_0$

Case3

if $a=\frac{1}{2}$ then by putting $u_n=\frac{f_n-\sqrt{2b}}{f_n+\sqrt{2b}}$ then$$ u_{n+1}=(u_n)^2\to u_n=(u_0)^{2^n}$$So$$ f_n=-\sqrt{2b} \frac{x^{2^n}+1}{x^{2^n}-1} , x=\frac{f_0-\sqrt{2b}}{f_0+\sqrt{2b}}$$

Case4

for $a=1 , b\ne0$ I didn't get any information for $f_{n+1}=f_n+\frac{b}{f_n}$

Case5

for $a\notin\{1,\frac{1}{2}\} , b\ne0,f_0\ne-\frac{\sqrt{b}}{\sqrt{1-a}}$ then by putting$$ u_n=\frac{f_n-\frac{\sqrt{b}}{\sqrt{1-a}}}{f_n+\frac{\sqrt{b}}{\sqrt{1-a}}}$$then$$ u_{n+1}=(u_n)^2+(2a-1)u_n$$now my QUESTION how to find the solution in case 4 and 5 ?

is there another special values for $a,b,f_0$ to find the solution ?

I looked here and here and didn't get any ideas.