The mass change of electron after absorbing photon If a photon is absorbed by an electron, will the rest mass of the electron increase? Or will the relativistic mass of the electron increase?
 A: Relativistic mass is an archaic concept, and I believe, it is not used by professional physicists any longer. No. The rest mass won't increase only its momentum will change and it will scatter.
A: I would guess the sort of process you thinking about is Compton scattering where a photon hits a free electron and transfers energy to it. The scattered electron ends up with an increased energy, but this does not produce an increased mass.
The relativistic expression for the total energy of a particle is:
$$ E^2 = p^2 c^2 + m^2c^4 $$
where the $m$ in this equation is the rest mass, which is a constant. The increased energy of the electron after the photon has hit it comes from an increase in the momentum not an increase in the mass. The momentum is in turn related to the velocity by:
$$ p = \frac{mv}{\sqrt{1 - v^2/c^2}} $$
where again $m$ is the (constant) rest mass.
However there is a situation in which absorption of a photon does lead to an increase in mass. Suppose we have a bound state like the electron in a hydrogen atom, and the photon is absorbed and excites the electron from the $1s$ to the $2s$ or some higher energy state. In this case the mass of the bound state increases by $E/c^2$, where $E$ is the photon energy. This is simply Einstein's famous equation $E=mc^2$ in action.
However in cases like this it is the mass of the whole bound state that increases, and not the mass of the electron or indeed any individual component of the bound state.
A: Relativistic mass is a concept barely used. The idea behind it is that you can keep the classical formula,
$$ p = m_{rel}v  ,$$
where you define,
$$ m_{rel} = \frac{m_0}{\sqrt{1-v^2/c^2}}.$$
However this yields the correct formula,
$$ p = \frac{m_0v}{\sqrt{1-v^2/c^2}} ,$$
it causes interpretation issues, since in your example the relativistic mass would increase but this doesn't have a real physical interpretation, it is just a concept introduced to correct the classical formula $ p = mv$. Therefore this relativistic mass is not useful in most of the cases and is barely used nowadays.
A: There are good answers, but I want to introduce the representation of Feynman diagrams, because that is what is being used when studying the behavior of elementary particles, and both the photon and the electron are elementary particles. These diagrams are used to calculate the probability of interaction , a strict mathematical operation implied by it.

The squiggly line represents the four momentum of a photon , and the dark the four momentum of the electron. The interaction happens at the points called vertices. What happens is that for a tiny interval of the variable the photon is absorbed completely in a summed four vector in the internal dark line , and the electron and photon reappear with the scattered momenta/energies at a second vertex. Between the two vertices An integral is applied over the variables and the boundary conditions of the problem. 
The incoming and outgoing electron have the invariant mass given by the four vector they are carrying. Only energy and momentum change with the absorption of part of the energy of the incoming photon.
It has to be stressed that for a free electron there will always be a scattered off photon, even of very low energy, because of momentum conservation at the center of mass, there cannot be two particles coming in and one going out.  

If a photon is absorbed by an electron, will the rest mass of the electron increase?

The answer is no, it is not called an invariant mass in vain.

Or will the relativistic mass of the electron increase?

Yes, the relativistic mass will increase. The concept is not really useful at the particle level. It is useful where newtonian mechanics is assumed, and newtonian forces are estimated, as with starships and the velocity of light, but it is confusing terminology at the particle level.
The case of electrons bound in an atom is covered by other answers.
A: A situation where an electron could absorb a photon is when the electron is in an atom and the photon has the correct energy to allow the electron to move up to a higher quantized energy level. The modern view is that the rest mass of a fundamental particle, like the electron, is always the same. However, one can talk of the rest mass of a composite system, such as an atom, changing. 
Before the absorption, choose a frame where the atom is stationary and say its mass is $m_i$ and the photons energy is $E_\gamma$. After the absorption call atom's mass  $m_f$ and it's momentum $p_f$. Then from conservation of energy and the relativistic energy momentum formula:
$$m_i+E_\gamma=\sqrt{m_f^2+p_f^2}$$
where we are setting the speed of light $c=1$. From conservation of momentum $p_f=E_\gamma$. Substituting this into the above equation and solving for $m_f$ gives
$$
m_f= \sqrt{m_i} \sqrt{2 E_{\gamma }+m_i}.
$$
In general, $E_\gamma\ll m_i$ and so we can expand the above expression to first order in $E_\gamma$ to get
$$
m_f\approx m_i+E_\gamma/c^2
$$
where we have restored the factors of $c$ using dimensional analysis.
