Effects of Pulsed vs Direct Current on Battery Depletion
P = I^2R. 

So the power loss in the resistor goes up with the square of the current. Multiply this by time and you get energy (joules) lost to the resistor. 
Example:  Vbatt = 10 V and Rbatt = 1 ohm.  100 Joules required. 

Pulsed load of 4 ohm:


Ipulse = 2A,
load_power = 16 W,
Time to deliver energy = 6.25 seconds (total "on" time), 
Pbatt = 4 W, and  
Power lost in battery = 25 Joules


continous method Rload = 40 ohm:


Icont = 0.249 A,
load_power = 2.38 W,
time to deliver energy = 42 s,
Pbatt = 0.06 W, and
Power lost in battery = 2.5 J. 


The difference is the i^2 term!

Most importantly is the 0.06 W vs 4 W internal power dissipation.  The 4 W will generate signiciantly more heat than the 0.06 W due to the thermal resistance of the cell to ambient.  This heat will elevate the batteries temperature, and this elevated temperature will also reduce the batteries capacity.

So this is the secondary effect of using pulsed power, not only is the battery dissipating more energy due to the I^2 term, it is "losing" available capacity due to elevated temperature.