Using Binary-Search in Insertion Sort we can only reduce number of Comparisons,however number of swaps remains same.

$ #Comparisons$ = $\mathcal{O}(n\log(n))$

$ #Swaps$ = $\mathcal{O} (n^{2})$

So,time complexity = $Comparion + Swaps$ = $\mathcal{O}(n\log(n))$ + $\mathcal{O} (n^{2})$ = $\mathcal{O} (n^{2})$

$ #Comparisons$ = $\mathcal{O}(n\log(n))$

$ #Swaps$ = $\mathcal{O} (n^{2})$

So,time complexity = $Comparion + Swaps$ = $\mathcal{O}(n\log(n))$ + $\mathcal{O} (n^{2})$ = $\mathcal{O} (n^{2})$