Optimal. Leaf size=145 \[ -\frac{7 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{6 a^2 d}-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}+\frac{13 \cos (c+d x)}{3 a d \sqrt{a \sin (c+d x)+a}} \]
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Rubi [A] time = 0.252776, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2765, 2968, 3023, 2751, 2649, 206} \[ -\frac{7 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{6 a^2 d}-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}+\frac{13 \cos (c+d x)}{3 a d \sqrt{a \sin (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 2765
Rule 2968
Rule 3023
Rule 2751
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{\int \frac{\sin (c+d x) \left (2 a-\frac{7}{2} a \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{\int \frac{2 a \sin (c+d x)-\frac{7}{2} a \sin ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}-\frac{7 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{6 a^2 d}-\frac{\int \frac{-\frac{7 a^2}{4}+\frac{13}{2} a^2 \sin (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx}{3 a^3}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac{13 \cos (c+d x)}{3 a d \sqrt{a+a \sin (c+d x)}}-\frac{7 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{6 a^2 d}+\frac{11 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{4 a}\\ &=\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac{13 \cos (c+d x)}{3 a d \sqrt{a+a \sin (c+d x)}}-\frac{7 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{6 a^2 d}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{2 a d}\\ &=-\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac{13 \cos (c+d x)}{3 a d \sqrt{a+a \sin (c+d x)}}-\frac{7 \cos (c+d x) \sqrt{a+a \sin (c+d x)}}{6 a^2 d}\\ \end{align*}
Mathematica [C] time = 0.267868, size = 156, normalized size = 1.08 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (-11 \sin \left (\frac{1}{2} (c+d x)\right )+7 \sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{5}{2} (c+d x)\right )+11 \cos \left (\frac{1}{2} (c+d x)\right )+7 \cos \left (\frac{3}{2} (c+d x)\right )+\cos \left (\frac{5}{2} (c+d x)\right )+(33+33 i) (-1)^{3/4} (\sin (c+d x)+1) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{6 d (a (\sin (c+d x)+1))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.589, size = 183, normalized size = 1.3 \begin{align*} -{\frac{1}{12\,d\cos \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \left ( 33\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-24\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2}-8\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{a} \right ) +33\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}-30\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2}-8\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}\sqrt{a} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \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 \frac{\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.88381, size = 783, normalized size = 5.4 \begin{align*} \frac{33 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \,{\left (4 \, \cos \left (d x + c\right )^{3} + 16 \, \cos \left (d x + c\right )^{2} -{\left (4 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) + 15 \, \cos \left (d x + c\right ) + 3\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{24 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d -{\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \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 [B] time = 2.98985, size = 552, normalized size = 3.81 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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