Why is pyrrole more reactive than pyridine and benzene for an electrophilic substitution reaction?
From Peter Sykes:  


  Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be
  deformed by being attracted towards the nitrogen atom because of the latter's
  being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the negative end on N and the positive end on the nucleus:
  
  Pyridine is thus referred to as a $\pi$-deficient heterocycle and, by analogywith a benzene ring that carries an electron-withdrawing substituent, e.g.$\ce{NO2}$ (p. 6.22), one would expect it to be deactivated towards electrophilic
  attack. Substitution takes place, with difficulty, at the 3-position because this leads to the most stable Wheland intermediate (63); the intermediates
  for 2- and 4-attack (64 and 65, respectively) each has a canonical state in which the charge is located on divalent N a highly unstable,i.e. high energy, state.
  
  There are certain formal analogies here to m- attack or
  p. 6.24), byt pyridine is very much more difficult to substi
  Thus nitration, chlorination, bromination and FriedelCraf
  really be made to take place usefully, and sulphonation on
  ing with oleum for 24 hours at 230°C, with an $\ce{Hg^2+}$ catalyst attack is due partly to the fact that pyridine has an available nitrogen, and can thus protonate (66), or interact with an electrophile (67):
  
  
  
  The positive charge will clearly further destabilise any of the $\sigma$-com
  plexes for electrophilic substitution, as did a substituent such as $\ce{-NR3^+}$, on the
  benzene nucleus (p. 6.24), but the destabilisation will be much more marked
  than with $\ce{-NR3^+}$; as the charge is now on an atom of the ring itself and not merely on a substituent.
  Pyrrole (68) also has 6 electrons in delocalised orbitals, but here the
  nitrogen atom has to contribute two electrons to make up the six (thus be
  coming essentially non-basic in the process, cf. p. 3.26), and the dipole of
  pyrrole is found to be in the opposite direction to that of pyridine, i.e. with
  the positive end on nitrogen and the negative end on the nucleus:
  
  Pyrrole is thus referred to as a π- excessive heterocycle and behaves rather like a reactive benzene derivative, e.g. aniline (p. 6.25), undergoing very ready electrophilic attack. This may be complicated by the fact that in
  strongly acid solution protonation (69) is forced even on the weakly basic
  pyrrole (it takes place on the 2-carbon atom rather than on N. cf. p. 3.26)    


TLDR   


Pyridine is a $\pi$-deficient heterocycle owing to nitrogen's more electronegative nature than carbon. Pyrrole on the other hand is $\pi$ excessive since nitrogen donates its lone pair to the ring.  
The non-delocalized lone pair on pyridine tends to interact with the electrophile/$\ce{H+}$. This further destabilizes any of the $\sigma$ complexes formed in electrophilic aromatic substitution (EAS).