Can you intuitively explain the decreasing time-period of oscillation with increasing pendulum length in some cases?

Consider a rigid body suspended about an axis of rotation which, in general does not pass through it’s center of mass (COM) and has a moment of inertia (MOI) $0 < I_{axis}$ about that axis. Let $I_C$ denote the moment of inertial of the object about an axis parallel to the one mentioned before and passing through the COM. The parallel axis theorem implies that $I_{axis} = I_C + ml^2$ where $0 leq l$ is the distance between the two axes. From the rotational dynamics of the object $I_{axis} ddot{theta} = -mgltheta$, where $0 leq theta rightarrow 0$ is the (small) angular displacement of the object, we can fine the time period of oscillation as $T = frac{2pi}{sqrt{frac{mgl}{I_{axis}}}}$ so that $T propto sqrt{frac{I_{axis}}{ml}}$.

In case of the simple pendulum, $I_C approx 0$ so that $I_{axis} = ml^2$, and thus $T propto sqrt{l}$ leading to the usual conclusion (matching our intuitive physical understanding) that the time-period of small oscillations increases as the length $l$ increases. However, in the case of general rigid bodies, i.e. not point-masses, the algebra results in $T propto sqrt{frac{I_C}{ml} + l}$. From the plot one can see that this expression explains the apparently counter-intuitive observation that for general rigid bodies (i.e. not point-masses), for small $l$, the time-period of small oscillations reduce as the length $l$ increases, in a mathematical sense.

In the case of a simple pendulum, it is physically intuitive that the time-period should increase with the increase in $l$ (distance traveled over an oscillation is increasing linearly with $l$, while the motivating force remains roughly of the same magnitude regardless of $l$). In the same sense what is the intuitive physical explanation of this apparently counter-intuitive behavior?