Word factorization in $O(n^2 \log n)$ time
If I'm not misunderstanding you, I think the minimum cost factorization can be calculated in $O(n^2)$ time as follows.

For each index i, we will calculate a bunch of values $(p_i^\ell, r_i^\ell)$ for $\ell=1,2,\ldots$ as follows. Let $p_i^1\ge 1$ be the smallest integer such that there is an integer $r\ge 2$ satisfying $$S[i-rp_i^1+1, i-p_i^1] = S[i-(r-1)p_i^1+1, i].$$ For this particular $p_i^1$, let $r_i^1$ be the largest $r$ with this property.
If no such $p_i$ exists, set $L_i=0$ so we know there are zero $(p_i^\ell,r_i^\ell)$ values for this index.

Let $p_i^2$ be the smallest integer strictly bigger than $(r_i^1-1)p_i^1$ satisfying, likewise,
$$S[i-r_i^2p_i^2+1, i-p_i^2] = S[i-(r_i^2-1)p_i^2+1, i]$$
for some $r_i^2\ge 2$. Like before, take $r_i^2$ to be the maximal one having fixed $p_i^2$.
In general $p_i^\ell$ is the smallest such number strictly bigger than $(r_i^{\ell-1}-1)p_i^{\ell-1}$. If no such $p_i^\ell$ exists, then $L_i=\ell-1$.

Note that for each index i, we have $L_i=O(\log (i+1))$ due to $p_i^\ell$ values increasing geometrically with $\ell$. (if $p_i^{\ell+1}$ exists, it's not just strictly bigger than $(r_i^\ell-1)p_i^\ell$ but bigger than that by at least $p_i^\ell/2$. This establishes the geometric increase.)


  Suppose now all $(p_i^\ell,r_i^\ell)$ values are given to us. The minimum cost is given by the recurrence
  $$\mathrm{dp}(i,j) = \min\left\{\mathrm{dp}(i, j-1) + 1, \min_\ell \left(\mathrm{dp}\left(i,j - r_j^\ell p_j^\ell\right) + \mathrm{dp}(j-r_j^\ell p_j^\ell+1,j-p_j^\ell)\right)\right\}$$
  with the understanding that for $i&gt;j$ we set $\mathrm{dp}(i,j) = +\infty$.
  The table can be filled in $O(n^2 + n\sum_j L_j)$ time.


We already observed above that $\sum_j L_j = O(\sum_j \log (j+1)) = \Theta(n\log n)$ by bounding the sum term by term. But actually if we look at the whole sum, we can prove something sharper.

Consider the suffix tree $T(\overleftarrow{S})$ of the reverse of $S$ (i.e., the prefix tree of S). We will charge each contribution to the sum $\sum_i L_i$ to an edge of $T(\overleftarrow{S})$ so that each edge will be charged at most once. Charge each $p_i^j$ to the edge emanating from $\mathrm{nca}(v(i), v(i-p_i^j))$ and going towards $v(i-p_i^j)$. Here $v(i)$ is the leaf of the prefix tree corresponding to $S[1..i]$ and nca denotes the nearest common ancestor.

This shows that $O(\sum_i L_i)=O(n)$. The values $(p_i^j,r_i^j)$ can be calculated in time $O(n+\sum_i L_i)$ by a traversal of the suffix tree but  I will leave the details to a later edit if anyone is interested.

Let me know if this makes sense.