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Look closely at the circuit. If discharging is slower than charging, then there must be less current available for discharging. The equation of a capacitor is...

dv/dt = i/c

which means that the slope of the voltage is proportional to the current and inversely proportional to the capacitance. This does not matter if it is a charging or a discharging situation. For example, one amp of current into one farad of capacitance will be one volt per second, and negative one amp of current into (out of) that same one farad capacitance will be negative one volt per second.

You don't give a lot of information in the question, but it sounds like you are analyzing the filter in a rectifier. The charge path is through the diode, which is a low impedance, high current, circuit, while the discharge path is only through the load. This would explain, in one case, why the discharge time is greater than the charge time.

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Q: Why is discharging slower than charging in a capacitor filter?
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