Optimal. Leaf size=191 \[ \frac {3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 \cos (e+f x)}{8 a c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.37, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2743, 2741, 3770} \[ \frac {3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 \cos (e+f x)}{8 a c f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2741
Rule 2743
Rule 3770
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}} \, dx}{2 a}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx}{4 a c}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 \int \frac {1}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {(3 \cos (e+f x)) \int \sec (e+f x) \, dx}{8 a c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {\cos (e+f x)}{2 f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{8 a c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 287, normalized size = 1.50 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 \cos ^2(e+f x)-\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+3 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{8 f (a (\sin (e+f x)+1))^{3/2} (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.53, size = 377, normalized size = 1.97 \[ \left [\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - \cos \left (f x + e\right )^{3}\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} + 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}}, -\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - \cos \left (f x + e\right )^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (3 \, \cos \left (f x + e\right )^{2} + 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 229, normalized size = 1.20 \[ -\frac {\left (3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+3 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\left (\cos ^{2}\left (f x +e \right )\right )-3 \sin \left (f x +e \right )+1\right ) \cos \left (f x +e \right )}{8 f \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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