Integrand size = 22, antiderivative size = 136 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {7 \arcsin (\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{64 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b} \]
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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 = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4382, 4386, 4387, 4390} \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {7 \arcsin (\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}-\frac {7 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{32 b}+\frac {7 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{64 b} \]
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Rule 4382
Rule 4386
Rule 4387
Rule 4390
Rubi steps \begin{align*} \text {integral}& = \frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{12} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx \\ & = \frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{16} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{32} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {7 \arcsin (\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{64 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b} \\ \end{align*}
Time = 0.63 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {-7 \arcsin (\cos (a+b x)-\sin (a+b x))+7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-\frac {2}{3} (10 \cos (a+b x)+9 \cos (3 (a+b x))+2 \cos (5 (a+b x))) \sqrt {\sin (2 (a+b x))}}{64 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 204.95 (sec) , antiderivative size = 523275790, normalized size of antiderivative = 3847616.10
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Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (118) = 236\).
Time = 0.27 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.14 \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 4 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 42 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 42 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 21 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{768 \, b} \]
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Timed out. \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
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\[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \]
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\[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}} \,d x } \]
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Timed out. \[ \int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\int {\cos \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \]
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