Integrand size = 22, antiderivative size = 82 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]
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Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4378, 4392, 4391} \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}+\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{4 b} \]
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Rule 4378
Rule 4391
Rule 4392
Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{4} \int \sec (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx \\ & = -\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}-\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-2 \csc (a+b x) \sqrt {\sin (2 (a+b x))}}{4 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 35.44 (sec) , antiderivative size = 211562444, normalized size of antiderivative = 2580029.80
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Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (74) = 148\).
Time = 0.27 (sec) , antiderivative size = 295, normalized size of antiderivative = 3.60 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) \sin \left (b x + a\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) \sin \left (b x + a\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 8 \, \sin \left (b x + a\right )}{16 \, b \sin \left (b x + a\right )} \]
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Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]
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\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^3}{{\sin \left (2\,a+2\,b\,x\right )}^{3/2}} \,d x \]
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