SE = Stored energy
KE = Kinetic energy
v = velocity
m1 = mass of arrow
m2 = virtual mass

SE = 1/2(m1 + m2) v^2
Therefore:
v=((2SE/(m1+m2))^1/2

and:
KE = 1/2m1v^2
Therefore substituting ((2SE/(m1+m2))^1/2 for v and reducing:
KE=m1SE/(m1+m2)

Now as anyone can see, as m1 is multiplied by SE in the numerator and m1 is only added by m2 in the denominator it isclear that as m1 increases KE increases as well. Now as SE is a constant for a given bow setup as KE increases with arrow mass then so does efficiency.