Optimal. Leaf size=180 \[ \frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]
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Rubi [A] time = 0.42, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2967, 2859, 2687, 2650, 2649, 206} \[ -\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 2649
Rule 2650
Rule 2687
Rule 2859
Rule 2967
Rubi steps
\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(5 A-3 B) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a c^2}\\ &=\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 (5 A-3 B)) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{16 a c}\\ &=\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {(3 (5 A-3 B)) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{64 a c^2}\\ &=\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(3 (5 A-3 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 )}{32 a c^2 f}\\ &=\frac {3 (5 A-3 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{32 \sqrt {2} a c^{5/2} f}+\frac {3 (5 A-3 B) \cos (e+f x)}{32 a c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \sec (e+f x)}{4 a c f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A-3 B) \sec (e+f x)}{8 a c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 0.85, size = 404, normalized size = 2.24 \[ \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (8 (B-A) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+(7 A-B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 (7 A-B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+4 (A+B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+8 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )-(3+3 i) \sqrt [4]{-1} (5 A-3 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{32 a f (\sin (e+f x)+1) (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 282, normalized size = 1.57 \[ -\frac {3 \, \sqrt {2} {\left ({\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (3 \, {\left (5 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (5 \, A - 3 \, B\right )} \sin \left (f x + e\right ) - 12 \, A + 20 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{128 \, {\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} f \cos \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.78, size = 350, normalized size = 1.94 \[ -\frac {\sin \left (f x +e \right ) \left (-30 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+40 A \,c^{\frac {5}{2}}+18 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-24 B \,c^{\frac {5}{2}}\right )+\left (-15 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+30 A \,c^{\frac {5}{2}}+9 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-18 B \,c^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+30 A \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-24 A \,c^{\frac {5}{2}}-18 B \sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+40 B \,c^{\frac {5}{2}}}{64 c^{\frac {9}{2}} a \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]
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 \[ \int \frac {B \sin \left (f x + e\right ) + A}{{\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]
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 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
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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