Comment #7 on issue 2607 by asmeurer: as_numer_denom() is too slow http://code.google.com/p/sympy/issues/detail?id=2607
Well, thanks to starblue on stackoverflow (see the above link), I have the efficient algorithm. The key is to compute it iteratively. So if you have a1/d1 + a2/d2 + … + an/dn then you compute (…(a1/d1 + a2/d2) + … ) + an/dn. If you work it out on paper, you'll see that this avoids all duplicate multiplications that can be avoided.
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