Optimal. Leaf size=182 \[ \frac{7 a c^4 (2 A-B) \cos ^3(e+f x)}{24 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{7 a c^4 (2 A-B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a c^4 x (2 A-B)-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f} \]
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Rubi [A] time = 0.295074, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2967, 2860, 2678, 2669, 2635, 8} \[ \frac{7 a c^4 (2 A-B) \cos ^3(e+f x)}{24 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{7 a c^4 (2 A-B) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a c^4 x (2 A-B)-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f} \]
Antiderivative was successfully verified.
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Rule 2967
Rule 2860
Rule 2678
Rule 2669
Rule 2635
Rule 8
Rubi steps
\begin{align*} \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx &=(a c) \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{1}{2} (a (2 A-B) c) \int \cos ^2(e+f x) (c-c \sin (e+f x))^3 \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{1}{10} \left (7 a (2 A-B) c^2\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{8} \left (7 a (2 A-B) c^3\right ) \int \cos ^2(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{8} \left (7 a (2 A-B) c^4\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}+\frac{7 a (2 A-B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}+\frac{1}{16} \left (7 a (2 A-B) c^4\right ) \int 1 \, dx\\ &=\frac{7}{16} a (2 A-B) c^4 x+\frac{7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}+\frac{7 a (2 A-B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}-\frac{a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac{a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac{7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f}\\ \end{align*}
Mathematica [A] time = 0.934998, size = 131, normalized size = 0.72 \[ \frac{a c^4 (120 (7 A-5 B) \cos (e+f x)+20 (13 A-7 B) \cos (3 (e+f x))+240 A \sin (2 (e+f x))-90 A \sin (4 (e+f x))-12 A \cos (5 (e+f x))+840 A f x+15 B \sin (2 (e+f x))+105 B \sin (4 (e+f x))-5 B \sin (6 (e+f x))+36 B \cos (5 (e+f x))-420 B f x)}{960 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.036, size = 342, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{A{c}^{4}a\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }-3\,A{c}^{4}a \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{2\,A{c}^{4}a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,A{c}^{4}a \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +3\,A{c}^{4}a\cos \left ( fx+e \right ) +B{c}^{4}a \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +{\frac{3\,B{c}^{4}a\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+2\,B{c}^{4}a \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{2\,B{c}^{4}a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-3\,B{c}^{4}a \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +A{c}^{4}a \left ( fx+e \right ) -B{c}^{4}a\cos \left ( fx+e \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.977028, size = 454, normalized size = 2.49 \begin{align*} -\frac{64 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} A a c^{4} - 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a c^{4} + 90 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{4} - 480 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{4} - 960 \,{\left (f x + e\right )} A a c^{4} - 192 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a c^{4} - 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a c^{4} - 5 \,{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 60 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} + 720 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 2880 \, A a c^{4} \cos \left (f x + e\right ) + 960 \, B a c^{4} \cos \left (f x + e\right )}{960 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.48433, size = 302, normalized size = 1.66 \begin{align*} -\frac{48 \,{\left (A - 3 \, B\right )} a c^{4} \cos \left (f x + e\right )^{5} - 320 \,{\left (A - B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 105 \,{\left (2 \, A - B\right )} a c^{4} f x + 5 \,{\left (8 \, B a c^{4} \cos \left (f x + e\right )^{5} + 2 \,{\left (18 \, A - 25 \, B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 21 \,{\left (2 \, A - B\right )} a c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 10.5282, size = 853, normalized size = 4.69 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.20789, size = 248, normalized size = 1.36 \begin{align*} -\frac{B a c^{4} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac{7}{16} \,{\left (2 \, A a c^{4} - B a c^{4}\right )} x - \frac{{\left (A a c^{4} - 3 \, B a c^{4}\right )} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{{\left (13 \, A a c^{4} - 7 \, B a c^{4}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac{{\left (7 \, A a c^{4} - 5 \, B a c^{4}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (6 \, A a c^{4} - 7 \, B a c^{4}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac{{\left (16 \, A a c^{4} + B a c^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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