Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$ $$c_j = \left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) ) = \left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) + \left(\frac{\partial}{\partial s}\right)^j \log(\beta-e^s).$$

Now, $$\log(\alpha-e^s) = \log \alpha + \log(1-\frac{e^s}{\alpha}) = \log \alpha - \sum_{k=1}^{\infty} \frac{e^{ks}}{k\alpha^k}$$ and hence $$ \left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) = - \sum_{k=1}^{\infty} \frac{k^{j-1}e^{ks}}{\alpha^k} = - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\left(\frac{e^s}{\alpha}\right)^{i+1}}{(1-\frac{e^s}{\alpha})^j},$$ and similarly for $\beta$. Here $\left\langle {\cdot\atop\cdot}\right\rangle$ are Eulerian numbers.

Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$ $$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left.\left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) + \left(\frac{\partial}{\partial s}\right)^j \log(\beta-e^s)\right|_{s=0}.$$

Now, $$\log(\alpha-e^s) = \log \alpha + \log(1-\frac{e^s}{\alpha}) = \log \alpha - \sum_{k=1}^{\infty} \frac{e^{ks}}{k\alpha^k}$$ and hence $$ \left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) = - \sum_{k=1}^{\infty} \frac{k^{j-1}e^{ks}}{\alpha^k} = - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\left(\frac{e^s}{\alpha}\right)^{i+1}}{(1-\frac{e^s}{\alpha})^j},$$ and similarly for $\beta$. Here $\left\langle {\cdot\atop\cdot}\right\rangle$ are Eulerian numbers.

Therefore, $$c_j = - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\alpha^{j-1-i}}{(\alpha-1)^j} - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\beta^{j-1-i}}{(\beta-1)^j}.$$