Is the impossibility of integrating the reciprocal of a function connected to the limitations of algebra? I will do two examples:
Let $$f(x)=\sum _{n=0}^{11} x^n$$
and then integrate in Mathematica:
$$\int \frac{1}{f(x)} \, dx$$
and one sees that Mathematica gives a solution.
However, let
$$f(x)=\sum _{n=0}^{12} x^n$$
and then try to integrate in Mathematica:
$$\int \frac{1}{f(x)} \, dx$$
then Mathematica gives the answer as a RootSum

So therefore I ask if the integrability of the reciprocal of a
  function is limited by the limitations of algebra?
$$\int \frac{1}{f(x)} \, dx=?$$

I know almost nothing about algebra or integrals.
 A: For the first case $$f(x)=\sum _{n=0}^{11} x^n$$ this can be factorized as
$$f(x)=(x+1) \left(x^2+1\right) \left(x^2-x+1\right) \left(x^2+x+1\right)
   \left(x^4-x^2+1\right)$$ and partial fraction decomposition leads to
$$\frac 1{f(x)}=\frac{1-2 x}{12 \left(x^2-x+1\right)}+\frac{1-x}{6 \left(x^2+1\right)}+\frac{1}{4
   \left(x^2+x+1\right)}+\frac{x^3-x^2-2 x+2}{6 \left(x^4-x^2+1\right)}+\frac{1}{6
   (x+1)}$$ which is workable.
This is not at all the case with the second one which does not show any real root. So, let
$$f(x)=\sum _{n=0}^{12} x^n=\prod_{i=1}^{12}(x-ri)$$ and partial fraction decomposition will lead to
$$\frac 1{f(x)}=\sum_{i=1}^{12}\frac{a_i}{x-r_i}\implies \int \frac {dx}{f(x)}=\sum_{i=1}^{12}a_i\log(x-ri)$$
Edit
If you note $$f_n(x)=\sum _{k=0}^{n} x^k$$ and try to compute (using a CAS)
$$\int \frac{dx}{f_n(x)}$$ you should notice that, for $0 \leq n \leq 20$, you would get explicit results for $n=0,1,2,3,5,7,11,15$. For all other values, you will get the answer in terms of RootSum as you called it.
