To make the arithmetic easier, let's use a 12 W CFL. I'll assume it's plugged into in the US power grid, where it sees a constant 120 V potential from the power line. That means it is drawing sort of 100 milliamperes of current.
Most household breakers trip at about 15 amperes of current. Suppose you have a (defective --- they don't do this) bulb which draws this maximum current for one second while it warms up. That's a hundred fifty times more "warm-up" power than "steady state" power, in the worst case. But the energy consumed during that warm-up is overmatched by the steady-state operation energy after three minutes. So if you're going to be away for more than (in this extreme example) three minutes, you use less power by turning the bulbs off.
I have a family member who was a CFL early adopter: he has some early CFLs which he bought in 2000 or so that advertised "ten year lifespan" and are in a seldom-used room. And those bulbs are terrible. They take about five minutes to warm up, and during the warm-up period the light from the bulbs is too dim and the wrong color. Let's suppose that those terrible bulbs draw twice their rated current during those first time minutes (which I consider unlikely). It's still cheaper, in that case, to turn the bulbs off if you're going to be out of the room for more than twice the warm-up time. But the other feature of these terrible bulbs is that the warm-up is thermal: once the bulbs are hot, you can turn them off and back on and they return immediately to full brightness. So you still save power by turning the bulb off briefly.
For any realistic light bulb, I think you come out ahead turning the power off if you're going to be out of the room for more than a couple of minutes.