Optimal. Leaf size=193 \[ \frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}} \]
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Rubi [A] time = 0.66, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2841, 2739, 2737, 2667, 31} \[ \frac {a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}} \]
Antiderivative was successfully verified.
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Rule 31
Rule 2667
Rule 2737
Rule 2739
Rule 2841
Rubi steps
\begin {align*} \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{9/2}} \, dx &=\frac {\int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx}{a c}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c^2}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^3}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^4}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^3 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^3 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^3 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^4 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c^4 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 4.13, size = 234, normalized size = 1.21 \[ \frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (3 \sin (3 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+30 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-18 \cos (2 (e+f x)) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+1\right )-9 \sin (e+f x) \left (5 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+4\right )+34\right )}{6 c^4 f (\sin (e+f x)-1)^4 \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 2 \, a^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{5 \, c^{5} \cos \left (f x + e\right )^{4} - 20 \, c^{5} \cos \left (f x + e\right )^{2} + 16 \, c^{5} - {\left (c^{5} \cos \left (f x + e\right )^{4} - 12 \, c^{5} \cos \left (f x + e\right )^{2} + 16 \, c^{5}\right )} \sin \left (f x + e\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [B] time = 0.38, size = 397, normalized size = 2.06 \[ \frac {\left (6 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-3 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-18 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-8 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+9 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+6 \left (\cos ^{2}\left (f x +e \right )\right )-24 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+12 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+24 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+14 \sin \left (f x +e \right )-12 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-6\right ) \left (\cos ^{2}\left (f x +e \right )-\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}}{3 f \left (\cos ^{3}\left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 \left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \cos \left (f x +e \right )-2 \cos \left (f x +e \right )+4 \sin \left (f x +e \right )+4\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {9}{2}}} \]
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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]
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 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]
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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