Integrand size = 20, antiderivative size = 110 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b} \]
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Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4393, 4386, 4387, 4390} \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b}-\frac {3 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{4 b}+\frac {3 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{8 b} \]
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Rule 4386
Rule 4387
Rule 4390
Rule 4393
Rubi steps \begin{align*} \text {integral}& = 2 \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx \\ & = \frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b}+\frac {3}{2} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b}+\frac {3}{4} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = -\frac {3 \arcsin (\cos (a+b x)-\sin (a+b x))}{8 b}+\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{8 b}-\frac {3 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{4 b}+\frac {\sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{2 b} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.78 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {3 \left (-\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )-2 (2 \cos (a+b x)+\cos (3 (a+b x))) \sqrt {\sin (2 (a+b x))}}{8 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 12.12 (sec) , antiderivative size = 26835015, normalized size of antiderivative = 243954.68
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Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (96) = 192\).
Time = 0.27 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.55 \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \]
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Timed out. \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]
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\[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
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\[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int \csc (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{5/2}}{\sin \left (a+b\,x\right )} \,d x \]
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