Optimal. Leaf size=157 \[ \frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}+\frac{5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{7/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}} \]
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Rubi [A] time = 0.323023, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2736, 2680, 2649, 206} \[ \frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}+\frac{5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{7/2} f}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}} \]
Antiderivative was successfully verified.
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Rule 2736
Rule 2680
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac{1}{6} \left (5 a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac{\left (5 a^3\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx}{8 c}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}-\frac{\left (5 a^3\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 c^3}\\ &=\frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 c-x^2} \, dx,x,-\frac{c \cos (e+f x)}{\sqrt{c-c \sin (e+f x)}}\right )}{8 c^3 f}\\ &=-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{7/2} f}+\frac{a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac{5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac{5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 1.56934, size = 307, normalized size = 1.96 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 \sin \left (\frac{1}{2} (e+f x)\right )+33 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+66 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-52 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-104 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+32 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6\right )}{24 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.645, size = 245, normalized size = 1.6 \begin{align*}{\frac{{a}^{3}}{48\, \left ( -1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}{c}^{3}-45\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{3}+66\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{c}+45\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{3}-160\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{c}^{3/2}-15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{3}+120\,\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{5/2} \right ) \sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }{c}^{-{\frac{13}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \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{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.12637, size = 1121, normalized size = 7.14 \begin{align*} \frac{15 \, \sqrt{2}{\left (a^{3} \cos \left (f x + e\right )^{4} - 3 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} + 4 \, a^{3} \cos \left (f x + e\right ) + 8 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{3} + 4 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) - 8 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{c} \log \left (-\frac{c \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{-c \sin \left (f x + e\right ) + c} \sqrt{c}{\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) +{\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (33 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 46 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3} +{\left (33 \, a^{3} \cos \left (f x + e\right )^{2} + 14 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{96 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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