Optimal. Leaf size=151 \[ \frac{(A+B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{(A+B) \cos (e+f x)}{4 a f (a \sin (e+f x)+a)^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.353968, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2972, 2743, 2741, 3770} \[ \frac{(A+B) \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{4 a^2 f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{(A+B) \cos (e+f x)}{4 a f (a \sin (e+f x)+a)^{3/2} \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2972
Rule 2743
Rule 2741
Rule 3770
Rubi steps
\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}} \, dx &=-\frac{(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \int \frac{1}{(a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac{(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}+\frac{((A+B) \cos (e+f x)) \int \sec (e+f x) \, dx}{4 a^2 \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{(A-B) \cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} \sqrt{c-c \sin (e+f x)}}-\frac{(A+B) \cos (e+f x)}{4 a f (a+a \sin (e+f x))^{3/2} \sqrt{c-c \sin (e+f x)}}+\frac{(A+B) \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{4 a^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.660088, size = 214, normalized size = 1.42 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-(A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-(A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+(A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-A+B\right )}{4 f (a (\sin (e+f x)+1))^{5/2} \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.336, size = 465, normalized size = 3.1 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{4\,f} \left ( -A \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +A\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,A\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,A\sin \left ( fx+e \right ) \ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,A \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,B\sin \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,B\sin \left ( fx+e \right ) \ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -3\,A\sin \left ( fx+e \right ) +2\,A\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,A\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +B\sin \left ( fx+e \right ) +2\,B\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,B\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -2\,A \right ) \left ( a \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \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{B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.55936, size = 1081, normalized size = 7.16 \begin{align*} \left [\frac{{\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{a c} \log \left (-\frac{a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt{a c} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \,{\left ({\left (A + B\right )} \sin \left (f x + e\right ) + 2 \, A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{8 \,{\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 )}}, -\frac{{\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{-a c} \arctan \left (\frac{\sqrt{-a c} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) -{\left ({\left (A + B\right )} \sin \left (f x + e\right ) + 2 \, A\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{4 \,{\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 )}}\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*} \int \frac{B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \sqrt{-c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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