Current divider and capacitor
Well \$I_c\$ will always be the sum of \$I_a\$ and \$I_b\$.

\$I_c = I_a + I_b \$

Assuming that the current supplied to the circuit is constant:

\$I_c\$ will be constant.

At the first instant, the capacitor is completely empty and acts like a short circuit. So the current is distributed like in a normal current divider. After the capacitor is fully charged, it acts as an open circuit and no current will flow through path B.

Now what happens between those two points:
The voltage across both paths must be the same (parallel circuit):

\$U(t) = I_a(t) * R_a\$

\$U(t) = I_b(t) * R_b + U_c(t)\$ 

\$U_c\$ is the voltage across the capacitor.

with that we end up with:

$$I_a(t) * R_a = I_b(t) * R_b + U_c(t)$$

Substituting the first statement this changes into:

$$(I_c(t)-I_b(t)) * R_a = I_b(t) * R_b + U_c(t)$$

Now we keep two things in mind: \$U_c(t) = \frac{Q_b(t)}{C}\$ and \$I(t)=\dot Q(t)\$ using this we end up with this lovely differential equation:

$$\frac{Q_b(t)}{C*(R_a+R_b)} + \dot Q_b(t) = I_c * R_a$$

Solved to (hopefully):

$$\large I_b(t) = I_c*\frac{R_a}{R_a+R_b}*e^{-\frac{t}{C*(R_a+R_b)}}$$

As for \$I_a(t)\$ well it has to add to \$I_b(t)\$ so that the result is constant.

$$\large I_a(t) = I_c (1-\frac{R_a}{R_a+R_b}*e^{-\frac{t}{C*(R_a+R_b)}})$$

Credits go mainly to this German Wikipedia article as I'm quite rusty with this kind of stuff.



Assuming that the voltage supplied to the circuit is constant:

\$I_a\$ will remain constant as it's just a simple resistor. 

\$I_b\$ will behave like the charging current of a RC circuit. That is:
$$
\large
I_b(t) = \frac{U}{R}  e^{\frac{-t}{RC}}
$$
So together:
$$
\large
I_c(t) = \frac{U}{R}e^{\frac{-t}{RC}} + I_a
$$