Dijkstra's algorithm runtime for dense graphs
The runtime of Dijkstra's algorithm (with Fibonacci Heaps) is $O(|E|+|V|\log|V|)$, which is different from what you were posting.

If $|E|\in \Theta(|V|^2)$, that is your graph is very dense, then this gives you runtime of $O(|V|^2+|V|\log|V|)=O(|V|^2)$. A better runtime would be "surprising", since you have to look at every edge at least once.

When using binary heaps, you get a runtime of $O((|E|+|V|)\log|V|)$ which for $|E|\in \Theta(|V|^2)$ gives $O(|V|^2\log |V|)$.