Integrand size = 22, antiderivative size = 40 \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b} \]
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Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4382, 2719} \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}+\frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{2 b} \]
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Rule 2719
Rule 4382
Rubi steps \begin{align*} \text {integral}& = \frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b}+\frac {1}{2} \int \sqrt {\sin (2 a+2 b x)} \, dx \\ & = \frac {E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{2 b}+\frac {\sin ^{\frac {3}{2}}(2 a+2 b x)}{6 b} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sin ^{\frac {3}{2}}(2 (a+b x))}{6 b} \]
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result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 5.16 (sec) , antiderivative size = 26159851, normalized size of antiderivative = 653996.28
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\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]
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Timed out. \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\text {Timed out} \]
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\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]
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\[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int { \cos \left (b x + a\right )^{2} \sqrt {\sin \left (2 \, b x + 2 \, a\right )} \,d x } \]
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Timed out. \[ \int \cos ^2(a+b x) \sqrt {\sin (2 a+2 b x)} \, dx=\int {\cos \left (a+b\,x\right )}^2\,\sqrt {\sin \left (2\,a+2\,b\,x\right )} \,d x \]
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