Initialize answer variable to zero
ans = 0 
Iterate over every possible bit -
for i from 0 to 30 
    Let count of numbers in [A,B] whose ith bit is set is - x 
    Let count of numbers in [C,D] whose ith bit is not set is - y 
    ans+=x*y*pow(2,i)
    Let count of numbers in [A,B] whose ith bit is not set is - a 
    Let count of numbers in [C,D] whose ith bit is set is - b
    ans+=a*b*pow(2,i)

To calculate count of numbers upto N, whose Kth bit is set, use this — Link. You can easily extend this idea for a range.