Why is the time complexity of insertion sort not brought down even if we use binary search for the comparisons?
For the $j^{th}$ element, you would do ~ $\log j$ comparisons and (in the worst case) ~$j$ shifts.

Summing over $j$, you get

$$
\sum_{j = 1}^{n} (j + \log j) = \frac{n(n+1)}{2} + \log (n!) = O(n^2 + n \log n) = O(n^2)
$$

The idea is that the linear work of shifting trumps the logarithmic work of comparing. You end up doing less comparisons, but still a linear amount of work per iteration. So the complexity does not change.