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(x^(1 + p)*(-(b*Cos[b*Log[x]]) + (1 + p)*Sin[b*Log[x]]))/(b^2 + (1 + p)^2)

Integrate[x^p*Sin[b*Log[x]], x] ==
(x^(1 + p)*(-(b*Cos[b*Log[x]]) + (1 + p)*Sin[b*Log[x]]))/(b^2 + (1 + p)^2) 

           p
Integrate[x  Sin[b Log[x]], x] == 

 1 + p
x      (-(b Cos[b Log[x]]) + (1 + p) Sin[b Log[x]])
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                    2          2
                   b  + (1 + p)
Time to compute: 0.08 second
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