Optimal. Leaf size=110 \[ \frac{5 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}+\frac{\sin ^{\frac{3}{2}}(2 a+2 b x) \cos (a+b x)}{8 b}-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac{5 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]
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Rubi [A] time = 0.075635, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4297, 4301, 4306} \[ \frac{5 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}+\frac{\sin ^{\frac{3}{2}}(2 a+2 b x) \cos (a+b x)}{8 b}-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac{5 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]
Antiderivative was successfully verified.
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Rule 4297
Rule 4301
Rule 4306
Rubi steps
\begin{align*} \int \cos ^3(a+b x) \sqrt{\sin (2 a+2 b x)} \, dx &=\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}+\frac{5}{8} \int \cos (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\\ &=\frac{5 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}+\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}+\frac{5}{16} \int \frac{\sin (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac{5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{32 b}+\frac{5 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{16 b}+\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{8 b}\\ \end{align*}
Mathematica [A] time = 0.181851, size = 84, normalized size = 0.76 \[ \frac{2 \sqrt{\sin (2 (a+b x))} (6 \sin (a+b x)+\sin (3 (a+b x)))-5 \left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x))+\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{32 b} \]
Antiderivative was successfully verified.
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Maple [B] time = 19.511, size = 88762396, normalized size = 806930.9 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (b x + a\right )^{3} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.550257, size = 774, normalized size = 7.04 \begin{align*} \frac{8 \, \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) + 10 \, \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 ) - 10 \, \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 ) + 5 \, \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 )}{128 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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