Integrand size = 20, antiderivative size = 136 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b} \]
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Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4386, 4387, 4391} \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {5 \sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{24 b}-\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]
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Rule 4386
Rule 4387
Rule 4391
Rubi steps \begin{align*} \text {integral}& = \frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{6} \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx \\ & = -\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{8} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = \frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{16} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b} \\ \end{align*}
Time = 0.60 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {-5 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+\frac {2}{3} \sqrt {\sin (2 (a+b x))} (14 \sin (a+b x)-3 \sin (3 (a+b x))-2 \sin (5 (a+b x)))}{32 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 71.55 (sec) , antiderivative size = 221760772, normalized size of antiderivative = 1630593.91
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Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (118) = 236\).
Time = 0.28 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.13 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) - 30 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 30 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 15 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{384 \, b} \]
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Timed out. \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
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\[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
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\[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2} \,d x \]
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