Optimal. Leaf size=141 \[ -\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}}-\frac{4 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}} \]
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Rubi [A] time = 0.279105, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2740, 2737, 2667, 31} \[ -\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{f \sqrt{c-c \sin (e+f x)}}-\frac{4 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2740
Rule 2737
Rule 2667
Rule 31
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^{5/2}}{\sqrt{c-c \sin (e+f x)}} \, dx &=-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+(2 a) \int \frac{(a+a \sin (e+f x))^{3/2}}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+\left (4 a^2\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c-c \sin (e+f x)}} \, dx\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}+\frac{\left (4 a^3 c \cos (e+f x)\right ) \int \frac{\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}-\frac{\left (4 a^3 \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=-\frac{4 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{f \sqrt{c-c \sin (e+f x)}}-\frac{a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.577391, size = 127, normalized size = 0.9 \[ -\frac{(a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (12 \sin (e+f x)-\cos (2 (e+f x))+32 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{4 f \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.184, size = 315, normalized size = 2.2 \begin{align*} -{\frac{1}{2\,f \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) - \left ( \cos \left ( fx+e \right ) \right ) ^{3}+2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4\,\sin \left ( fx+e \right ) +2\,\cos \left ( fx+e \right ) -4 \right ) } \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{3}-6\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) -16\,\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 ) +8\,\sin \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-16\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +8\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) +5\,\sin \left ( fx+e \right ) -\cos \left ( fx+e \right ) +16\,\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -8\,\ln \left ( 2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1} \right ) -5 \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{{\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 [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{c \sin \left (f x + e\right ) - c}, x\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{{\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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