BJT switching circuit with a capacitor as a load
The first expression gives how the capacitor is charging from 0V to \$E_C\$ through the resistor \$R_C\$. So time constant = \$R_CC_L\$. hence the expression:
$$u_{CL}\left ( t \right )=E_c(1-e^{\frac{-t}{R_C C_L}})$$

After the capacitor is charged to \$E_C\$ the input changed and transistor becomes 'on'. Now the capacitor will discharge from \$E_C\$ to \$U_{CE_{sat}}\$ through the transistor, equivalent resistance of which is \$R_{CE}\$. Where \$U_{CE_{sat}}\$ is the collector-emitter voltage when transistor is in saturation. So the discharging of capacitor can be expressed as:

$$u_{CL}\left ( t \right )=U_{CE_{sat}} +(E_c-U_{CE_{sat}}) e^{\frac{-t}{R_{CE} C_L}}$$


  1st expression is the charging equation.


\$U_{CE_{sat}}\$ is usually a small voltage (0.2-0.3V) compared to \$E_C\$, many a people approximates \$E_c-U_{CE_{sat}}\$ to \$E_C\$. And in that case this becomes your second expression. 


  2nd expression is the discharging equation.


From the two expressions above, one can see that the charging and discharging time depends on the capacitance and resistance values through which it charge/discharges. Now if the value of \$R_{CE}\$ is less than that of \$R_C\$, the time taken for the capacitor to discharge fully will be less than that of the time required to charge. I think in your case, the discharging time few times smaller than charging time and hence in the plot, it looks like an abrupt transition.