calculus and analysis - Faster code for calculat in Integral include derivative

I saw this question How I can get faster

I tried help that. But I can not understand. I want to write a faster code for:

m = 1; 
u[x_, t_] = (t^α*x*(2*t^2 + (1 + α)*(2 + α)))/Gamma[3 + α]; 
DUt = FullSimplify[(1/Gamma[m - α])*
   Integrate[(t - τ)^(m - α - 1)*
     D[u[x, τ], {τ, m}], 
         {τ, 0, t}], 
  Assumptions -> {m - 1 < α < m, t > 0}]

Any suggestion?

Answer

This is your definition:

 m = 1;
u[x_, t_] = (t^\[Alpha]*x*(2*t^2 + (1 + \[Alpha])*(2 + \[Alpha])))/
   Gamma[3 + \[Alpha]];

Your code makes 3.19 seconds on Mma10.4.1

 DUt = FullSimplify[(1/Gamma[m - \[Alpha]])*Integrate[(t - \[Tau])^(m - \[Alpha] - 1)*D[u[x, \[Tau]], {\[Tau], m}], {\[Tau], 0, t}], 
   Assumptions -> {m - 1 < \[Alpha] < m, t > 0}] // AbsoluteTiming

(* {3.18669, (1 + t^2) x}  *)

which is not bad at all. If one calculates the integral separately it takes 0.02 sec:

    int = Integrate[(t - \[Tau])^(m - \[Alpha] - 1)*D[u[x, \[Tau]], {\[Tau], m}], {\[Tau], 0, t}];
FullSimplify[(1/Gamma[m - \[Alpha]])*int, 
  Assumptions -> {m - 1 < \[Alpha] < m, t > 0}] // AbsoluteTiming

(*   {0.0244041, (1 + t^2) x}   *)

But I do not see a serious gain in it.

Have fun!