Optimal. Leaf size=212 \[ \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}} \]
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Rubi [A]
time = 0.33, 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 = {3051, 2818,
2819, 2816, 2746, 31} \begin {gather*} -\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}} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 2746
Rule 2816
Rule 2818
Rule 2819
Rule 3051
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 ) \text {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 = 0.77, size = 207, normalized size = 0.98 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2} \left (2 (A+B)-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) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \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-B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (c-c \sin (e+f x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1091\) vs.
\(2(190)=380\).
time = 0.31, size = 1092, normalized size = 5.15
method | result | size |
default | \(\text {Expression too large to display}\) | \(1092\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 546 vs.
\(2 (203) = 406\).
time = 0.70, size = 546, normalized size = 2.58 \begin {gather*} -\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} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.56, size = 311, normalized size = 1.47 \begin {gather*} \frac {\sqrt {2} {\left (\frac {4 \, \sqrt {2} B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \sqrt {2} {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-32 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 32\right )}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (3 \, A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {a}}{4 \, f} \end {gather*}
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 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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