Optimal. Leaf size=175 \[ -\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{(5 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.391888, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2967, 2855, 2687, 2650, 2649, 206} \[ -\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{(5 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2855
Rule 2687
Rule 2650
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx &=\frac{\int \sec ^4(e+f x) (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{(5 A+B) \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{6 a^2 c}\\ &=-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{(5 A+B) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}+\frac{(5 A+B) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{16 a^2 c}\\ &=\frac{(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}-\frac{(5 A+B) \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 a^2 c f}\\ &=\frac{(5 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{8 \sqrt{2} a^2 c^{3/2} f}+\frac{(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac{(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 c^2 f}\\ \end{align*}
Mathematica [C] time = 0.881429, size = 300, normalized size = 1.71 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+6 (A+B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+4 (B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+(-3-3 i) \sqrt [4]{-1} (5 A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-12 A \cos ^2(e+f x)\right )}{24 a^2 f (\sin (e+f x)+1)^2 (c-c \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.145, size = 258, normalized size = 1.5 \begin{align*} -{\frac{1}{48\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( \sin \left ( fx+e \right ) \left ( 15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}cA+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}cB-20\,A{c}^{5/2}-4\,B{c}^{5/2} \right ) + \left ( 30\,A{c}^{5/2}+6\,B{c}^{5/2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}cA-3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{3/2}cB-4\,A{c}^{5/2}-20\,B{c}^{5/2} \right ){c}^{-{\frac{7}{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 = 1.81056, size = 560, normalized size = 3.2 \begin{align*} \frac{3 \, \sqrt{2}{\left (5 \, A + B\right )} \sqrt{c} \cos \left (f x + e\right )^{3} \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 (3 \,{\left (5 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (5 \, A + B\right )} \sin \left (f x + e\right ) - 2 \, A - 10 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \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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