Integrand size = 20, antiderivative size = 84 \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b} \]
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Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4386, 4391} \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{4 b} \]
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Rule 4386
Rule 4391
Rubi steps \begin{align*} \text {integral}& = \frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}+\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 {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )-2 \sin (a+b x) \sqrt {\sin (2 (a+b x))}}{4 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.02 (sec) , antiderivative size = 5537888, normalized size of antiderivative = 65927.24
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Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (74) = 148\).
Time = 0.26 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.17 \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) + 2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{16 \, b} \]
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Timed out. \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \]
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\[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right ) \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]
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\[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right ) \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]
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Timed out. \[ \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int \cos \left (a+b\,x\right )\,\sqrt {\sin \left (2\,a+2\,b\,x\right )} \,d x \]
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