Is there any simple way to find the polynomials satisfying the functional relation \begin{align*} P\left(-c + \frac{K}{u+c}\right) \frac{(u+c)^2}{K} = P(u) \tag{*} \end{align*} Where $K = (b+c)(a+c)$?
I have a hunch that one can restrain oneself into looking at polynomials of degree 2, else the left handside is not a polynomial. I tried inserting and assuming that $$ P(x) = A x^2 + B x + C $$ When comparing coefficients I got something silly and strange. The problem rises when trying to evaluate. $$ \int_a^b \frac{\log(x+c)}{P(x)}\,\mathrm{d}x $$ If $P(x)$ satifies $(*)$ then, the integral can be rewritten into $$ \int_a^b \frac{\log K}{P(x)}\,\mathrm{d}x $$ Which is much easier to evalute. Alas finding such polynomials seems somewhat cumbersome. If $c=0$ then $P(x)=(x+b)(x+a)$ satifies the relation.
It seems the following.
When $u$ tends to infinity then the modulus of the left side of the equality (*) grows not faster than a polynomial of degree 2. Therefore the polynomial from the right side of the equality has a degree at most two. Now, assuming that $P(x) = A x^2 + B x + C$ and comparing coefficients from both sides, we obtain a system
$$\begin{cases} A=\frac 1K\left(Ac^2-Bc+C\right)\\ B=\frac {2c}K\left(Ac^2-Bc+C\right)-2Ac+B\\ C=\frac {c^2}K\left(Ac^2-Bc+C\right)+c(-2Ac+B) \end{cases} $$
Solving it, we obtain that $P(u)\equiv B(u+c)$ or a degenerated solution $K=0$ and $P(-c)=0$.