Optimizing pow()

The implementation of x^a can be optimized for constant a.

It is obvious that for constant integer exponent you can just expand:

x³ = x¹⋅x²

x⁹ = x¹⋅x⁸

and since 13₁₀ = 1101₂ = 2⁰ + 2² + 2³:

x¹³ = x¹⋅x⁴⋅x⁸

Likewise for fractional a:

x^(3)/(4) = x^(1)/(2)⋅x^(1)/(4)

So for a = 2.4₁₀≃10.011001101₂:

x^2.4≃x²⋅x^(1)/(4)⋅x^(1)/(8)⋅x^(1)/(64)⋅x^(1)/(128)⋅x^(1)/(512)