Optimal. Leaf size=174 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f} \]
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Rubi [A] time = 0.437845, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {2967, 2855, 2675, 2649, 206} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}+\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2855
Rule 2675
Rule 2649
Rule 206
Rubi steps
\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{2 a^3 c^2}\\ &=-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \sec ^2(e+f x) \sqrt{c-c \sin (e+f x)} \, dx}{4 a^3 c}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac{(A+B) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{8 a^3}\\ &=-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac{(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 )}{4 a^3 f}\\ &=\frac{(A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{4 \sqrt{2} a^3 \sqrt{c} f}-\frac{(A+B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{4 a^3 c f}-\frac{(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [C] time = 0.785356, size = 204, normalized size = 1.17 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-15 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4-10 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(-15-15 i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+12 (B-A)\right )}{60 a^3 f (\sin (e+f x)+1)^3 \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.209, size = 200, normalized size = 1.2 \begin{align*}{\frac{-1+\sin \left ( fx+e \right ) }{120\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f} \left ( 74\,{c}^{9/2}A+80\,A{c}^{9/2}\sin \left ( fx+e \right ) +30\,A{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{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 ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{2}A+26\,{c}^{9/2}B+80\,B{c}^{9/2}\sin \left ( fx+e \right ) +30\,B{c}^{9/2} \left ( \sin \left ( fx+e \right ) \right ) ^{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 ) \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}{c}^{2}B \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 = 1.76415, size = 729, normalized size = 4.19 \begin{align*} \frac{15 \, \sqrt{2}{\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \,{\left (A + B\right )} \cos \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 (15 \,{\left (A + B\right )} \cos \left (f x + e\right )^{2} - 40 \,{\left (A + B\right )} \sin \left (f x + e\right ) - 52 \, A - 28 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{240 \,{\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \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 [B] time = 2.14438, size = 1511, normalized size = 8.68 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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