You can Simplify
a bit your integrand. Then, first, I have tried with MonteCarlo
method to get a rough estimation of your integral. After a non-convergent result, the estimation is $\approx 0.3$. This gives me a clue about the result.
So following the suggestion of @MariuszIwaniuk:
NIntegrate[(R R1 (R Cos[s] - R1 Cos[s1])^4)/(R^2 + R1^2 + (z - z1)^2 -
2 R R1 Cos[s - s1])^(5/2), {s, 0, 2 Pi}, {z, -0.2705980500730985,0.2705980500730985},
{R,0, 1/2 Sqrt[1 - 16 z^2 + 32 z^4]}, {s1, 0, 2 Pi},
{z1, -0.7205904182426568, 0.7205904182426568},
{R1, 0, Sqrt[1 - 1.7313019390581716 z1^2 - 0.37467580052332317 z1^4]},
Method -> "LocalAdaptive"]
(* 0.300715 *)
After 200 runs by using MonteCarlo
, now the convergent result is:
Mean@Table[NIntegrate[(R R1 (R Cos[s] - R1 Cos[s1])^4)/
(R^2 + R1^2 + (z - z1)^2 - 2 R R1 Cos[s - s1])^(5/2),
{s, 0, \[Pi], 2 Pi}, {z, -0.2705980500730985, 0.2705980500730985},
{R, 0,1/2 Sqrt[1 - 16 z^2 + 32 z^4]}, {s1, 0, 2 Pi},
{z1, -0.7205904182426568, 0, 0.7205904182426568},
{R1, 0, Sqrt[1 - 1.7313019390581716 z1^2 - 0.37467580052332317 z1^4]},
Method -> {"MonteCarlo", "MaxPoints" -> 1000000}], {200}]
(* 0.300865 *)
which is faster and quite approximated.
NIntegrate[expr,Method -> "LocalAdaptive"]
$\endgroup$ – Mariusz Iwaniuk Apr 2 '18 at 18:00