Optimal. Leaf size=217 \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}+\frac{5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt{c-c \sin (e+f x)}}-\frac{5 a^3 (A+13 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{7/2} f}-\frac{a^3 c (A+13 B) \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.549385, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2967, 2859, 2680, 2679, 2649, 206} \[ \frac{a^3 c^3 (A+B) \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}+\frac{5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt{c-c \sin (e+f x)}}-\frac{5 a^3 (A+13 B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} c^{7/2} f}-\frac{a^3 c (A+13 B) \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2859
Rule 2680
Rule 2679
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{1}{12} \left (a^3 (A+13 B) c^2\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{1}{48} \left (5 a^3 (A+13 B)\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}-\frac{\left (5 a^3 (A+13 B)\right ) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{32 c^2}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac{5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt{c-c \sin (e+f x)}}-\frac{\left (5 a^3 (A+13 B)\right ) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 c^3}\\ &=\frac{a^3 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac{5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt{c-c \sin (e+f x)}}+\frac{\left (5 a^3 (A+13 B)\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 (A+13 B) \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 (A+B) c^3 \cos ^7(e+f x)}{6 f (c-c \sin (e+f x))^{13/2}}-\frac{a^3 (A+13 B) c \cos ^5(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac{5 a^3 (A+13 B) \cos ^3(e+f x)}{48 c f (c-c \sin (e+f x))^{5/2}}+\frac{5 a^3 (A+13 B) \cos (e+f x)}{16 c^3 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 3.27079, size = 422, normalized size = 1.94 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (64 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )+3 (11 A+47 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+6 (11 A+47 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-4 (13 A+25 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-8 (13 A+25 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+32 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} (A+13 B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+48 B \cos \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6+48 B \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6\right )}{24 f (c-c \sin (e+f x))^{7/2} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.516, size = 524, normalized size = 2.4 \begin{align*} \text{result 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 \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\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 = 1.69733, size = 1434, normalized size = 6.61 \begin{align*} \text{result too large to display} \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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