Optimal. Leaf size=212 \[ -\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a (A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.49, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2972, 2739, 2740, 2737, 2667, 31} \[ -\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a (A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 31
Rule 2667
Rule 2737
Rule 2739
Rule 2740
Rule 2972
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {(A+5 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {(a (A+5 B)) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (a^2 (A+5 B)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (a^3 (A+5 B) \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (a^3 (A+5 B) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 1.16, size = 207, normalized size = 0.98 \[ \frac {(a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-4 (A+2 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (A+B)-B \sin (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{f (c-c \sin (e+f x))^{5/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2} + {\left (B a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]
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 = 0.65, size = 1092, normalized size = 5.15 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 506, normalized size = 2.39 \[ -\frac {{\left (\frac {8 \, a^{\frac {5}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (c^{3} - \frac {4 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} + \frac {a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}}\right )} A - B {\left (\frac {10 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} - \frac {5 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left (\frac {5 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {16 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {14 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {16 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {5 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{c^{\frac {5}{2}} - \frac {4 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {7 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {8 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {7 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]
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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