Here's an extrapolatory approach based on Richardson extrapolation of the integrals over each "period." (`Richardson[]` code from [Anton Antonov's answer here](http://mathematica.stackexchange.com/a/95166/4999)):

    Clear[Richardson]
    Richardson[A_, n_, N_] := 
     Total@Table[(A[n + k]*(n + k)^N*If[OddQ[k + N], -1, 1])/(k! (N - k)!), {k, 0, N}]

    (* the integral from 2 n Pi to 2 (n+1) Pi *)
    Clear[aa];
    aa[n_Integer /; n >= 0] := 
      aa[n] = NIntegrate[ArcCot[x]*Sin[x]/(5/4 + Cos[x]), {x, 2 n Pi, 2 (n + 1) Pi}, 
        WorkingPrecision -> 50];
    
We get around machine precision with 100 terms plus 6 extrapolation terms:

    sf[n_] := Sum[aa[i], {i, 0, n}]; (* partial sum = integral over {0, 2 (n+1) Pi} *)
    Richardson[sf, 100, 6]
    % - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
    (*
      0.743355751361227474780881457479018423835    
      -2.97535713733265290609713*10^-16
    *)

Without extrapolation:

    sf[100]
    % - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
    (*
      0.74207790286093203962531860751097243182349263171074
      -0.00127784850029573269127658323333660172377741318898
    *)

----

*Update*

More extrapolatory terms yields an approximation accurate to machine precision with many fewer terms overall:

    Richardson[sf, 1, 16]
    % - Pi*Log[3/2/(1 + 1/2*Exp[-1])]
    (*
      0.74335575136122767678505611095516483469029
      -9.553153907978914419885698*10^-17
    *)