Optimal. Leaf size=110 \[ \frac{3 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{8 b}-\frac{\sin ^{\frac{3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac{3 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{16 b} \]
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Rubi [A] time = 0.0675042, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4302, 4301, 4306} \[ \frac{3 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{8 b}-\frac{\sin ^{\frac{3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac{3 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{16 b} \]
Antiderivative was successfully verified.
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Rule 4302
Rule 4301
Rule 4306
Rubi steps
\begin{align*} \int \sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx &=-\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{4 b}+\frac{3}{4} \int \cos (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\\ &=\frac{3 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{8 b}-\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{4 b}+\frac{3}{8} \int \frac{\sin (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac{3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{16 b}+\frac{3 \sin (a+b x) \sqrt{\sin (2 a+2 b x)}}{8 b}-\frac{\cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{4 b}\\ \end{align*}
Mathematica [A] time = 0.194416, size = 86, normalized size = 0.78 \[ \frac{2 \sqrt{\sin (2 (a+b x))} (2 \sin (a+b x)-\sin (3 (a+b x)))-3 \left (\sin ^{-1}(\cos (a+b x)-\sin (a+b x))+\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{16 b} \]
Antiderivative was successfully verified.
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Maple [B] time = 15.386, size = 65166864, normalized size = 592426. \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 \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}} \sin \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.555751, size = 771, normalized size = 7.01 \begin{align*} -\frac{8 \, \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt{\cos \left (b x + a\right ) \sin \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 )}{64 \, 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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}} \sin \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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