[Wiener's result][1] has been [improved][2] [several][3] [times][4], and it is hard to tell how big the private exponent must be to be safe from further progress.

Also, the proposed technique, assuming $d>n^{1/3}$, requires a minimum of $\lceil log_2(n)/3\rceil$ modular multiplications for the sparsest $d$ conceivable (a power of two), compared to say ${7\over6} \cdot log_2(n)$ for classical RSA using [sliding window exponentiation][5].

Thus, when not using the [Chinese Remainder Theorem][6], the technique allows a saving of a factor of $7\over2$ at best, which is less than the factor of nearly 4 allowed by the CRT; and when combining the technique with the CRT, one of the saving in the CRT (halving the size of the exponents) vanishes, thus the speedup is by a factor like $7\over4$ compared to [classical RSA with CRT][7]. That's not a huge speedup.

But this leaves the question unanswered.

  [1]: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.92.5261
  [2]: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.30.1345
  [3]: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.24.6366
  [4]: http://eprint.iacr.org/2008/315.pdf
  [5]: http://en.wikipedia.org/wiki/Exponentiation_by_squaring#Sliding_window_method
  [6]: http://en.wikipedia.org/wiki/Chinese_Remainder_Theorem
  [7]: http://www.di-mgt.com.au/crt_rsa.html