Integral of $\ln\left(\sum_{k=0}^N a_kx^k\right)$ Given the series $$\displaystyle S(x,N)=\sum_{k=0}^N a_kx^k$$
I would like to know if exists a closed formula to calculate the undefined integral of:
$$G(x,N)=\ln\left(\sum_{k=0}^N a_kx^k\right)$$
It's easy to calculate:
$$\int G(x,N)dx$$ for $N=2$ and also for $N=3$ even if in this last case the result is more complicated. What happens if $N\gt 3$?
Thanks.
 A: I  think that in the case $N>3$ one would need to impose a certain number of conditions on the $a_{k}$. For example, a little sketching out on my notepad gave me a quick solution to the case $N=4$ subject to $a_{1} = a_{3}=0$; yielding the so-called "biquadratic case", namely,
\begin{equation}
S(x,4) = a_{0}+a_{2}x^{2}+a_{4}x^{4}
\end{equation}  
From which we make a simple substitution $s=x^{2}$, viz
\begin{equation}
S(s,4) =a_{0}+a_{2}s +a_{4}s^{2}
\end{equation}
From which we obtain, (wlog) the roots to this quadratic as
\begin{equation}
\Delta^{\pm} = \frac{a_{2} \pm \sqrt{a_{2}^{2} - 4a_{0}a_{4}}}{2a_{4}}
\end{equation}
This then brings about the four original roots of the biquadratic as
\begin{eqnarray}
x_{1} &=& - \sqrt{\Delta^{+}} \\
x_{2} &=& + \sqrt{\Delta^{+}} \\
x_{3} &=& - \sqrt{\Delta^{-}} \\
x_{4} &=& + \sqrt{\Delta^{-}} \\
\end{eqnarray}
Thus (wlog)
\begin{equation}
S(x, 4) = (x_{1} + \sqrt{\Delta^{+}})(x_{2} - \sqrt{\Delta^{+}})(x_{3} + \sqrt{\Delta^{-}})(x_{4} - \sqrt{\Delta^{-}}).
\end{equation}
Now,by defintion of $G(x, N)$ we have;
\begin{eqnarray}
G(x, 4) &=& \ln S(x, 4) \\
        &=& \ln \left( (x_{1} + \sqrt{\Delta^{+}})(x_{2} - \sqrt{\Delta^{+}})(x_{3} + \sqrt{\Delta^{-}})(x_{4} - \sqrt{\Delta^{-}})\right) \\
        &=& \sum_{i=1}^{4} \ln \left(x_{i} +(-1)^{i+1} \Delta_{i} \right).
\end{eqnarray}
Where I have defined,
\begin{equation}
\left( \begin{array}{c}
\Delta_{1}  \\
\Delta_{2}  \\
\Delta_{3}  \\
\Delta_{4}    \end{array} \right)
=
\left( \begin{array}{c}
\sqrt{\Delta^{+}} \\
\sqrt{\Delta^{+}} \\
\sqrt{\Delta^{-}} \\
\sqrt{\Delta^{-}} \end{array} \right)
\end{equation}
Forming the integral of $\int G(x, 4)dx$ is now a trivial task. Of course, one can apply a "general" formula for the solution of a quartic in the case where all $a_{k}$ are non-zero, but the application of the "formula" is hugely cumbersome.
