f(n) is

  int result = 0;
  for (int i = 1; i <= n; i *= 2)
     for (int j = 0; j <= n; j++)
        result++;

which can be simplified to

  int result = 0;
  for (int i = 1; i <= n; i *= 2)
     result += n+1;

which gives

  sum j=0..⌊log2 n⌋ of n+1  =
  (⌊log2 n⌋ + 1) (n + 1)

  (log2 n) (n + 1)  ≤  f(n)  ≤  (log2 n + 1) (n + 1)

so is O(n log n)