Divisor sum from prime-power factorisation
CJam, 15 bytes * 0.8 = 12

q~.{_@)#(\(/}:*


Try it online. The input order is exponent list first, then list of primes (-3 bytes thanks to @Dennis).

For each prime-exponent pair (p, e) find

(p^(e+1) - 1)/(p - 1)


then find the product of all of these. E.g. for 240 this would be

(1 + 2 + 4 + 8 + 16)(1 + 3)(1 + 5) = 31 * 4 * 6 = 744


Depending on how exponentiation is implemented, this can be better than O(sum of exponents).