Optimal. Leaf size=191 \[ -\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}+\frac{7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.328142, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2736, 2675, 2687, 2650, 2649, 206} \[ -\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}+\frac{7 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}+\frac{7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2736
Rule 2675
Rule 2687
Rule 2650
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac{\int \sec ^6(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{a^3 c^3}\\ &=-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac{7 \int \sec ^4(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{10 a^3 c^2}\\ &=-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac{7 \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{12 a^3 c}\\ &=-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac{7 \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 a^3}\\ &=\frac{7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}+\frac{7 \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{32 a^3 c}\\ &=\frac{7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}-\frac{7 \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 )}{16 a^3 c f}\\ &=\frac{7 \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{16 \sqrt{2} a^3 c^{3/2} f}+\frac{7 \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac{7 \sec (e+f x)}{12 a^3 c f \sqrt{c-c \sin (e+f x)}}-\frac{7 \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac{\sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [C] time = 1.2306, size = 174, normalized size = 0.91 \[ \frac{\left (\frac{1}{1920}+\frac{i}{1920}\right ) \cos (e+f x) \left ((1-i) (-231 \sin (e+f x)+105 \sin (3 (e+f x))+350 \cos (2 (e+f x))+206)+840 \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5\right )}{a^3 c f (\sin (e+f x)-1) (\sin (e+f x)+1)^3 \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.603, size = 170, normalized size = 0.9 \begin{align*} -{\frac{1}{480\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 105\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\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-105\, \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c+42\,{c}^{7/2}\sin \left ( fx+e \right ) -350\,{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-210\,{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}+278\,{c}^{7/2} \right ){c}^{-{\frac{9}{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(-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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Fricas [A] time = 1.21606, size = 645, normalized size = 3.38 \begin{align*} \frac{105 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\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 (175 \, \cos \left (f x + e\right )^{2} + 21 \,{\left (5 \, \cos \left (f x + e\right )^{2} - 4\right )} \sin \left (f x + e\right ) - 36\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{960 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\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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