Optimal. Leaf size=193 \[ \frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}+\frac {4 (9 A-19 B) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(6 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {x (6 A-13 B)}{2 a^3}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(6 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.47, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2977, 2734} \[ \frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}+\frac {4 (9 A-19 B) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(6 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {x (6 A-13 B)}{2 a^3}+\frac {(A-B) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {(6 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2734
Rule 2977
Rubi steps
\begin {align*} \int \frac {\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {\cos ^3(c+d x) (4 a (A-B)-a (2 A-7 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^2(c+d x) \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \cos (c+d x) \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(6 A-13 B) x}{2 a^3}+\frac {8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}-\frac {(6 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 0.92, size = 435, normalized size = 2.25 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-600 d x (6 A-13 B) \cos \left (c+\frac {d x}{2}\right )-600 d x (6 A-13 B) \cos \left (\frac {d x}{2}\right )-4500 A \sin \left (c+\frac {d x}{2}\right )+4860 A \sin \left (c+\frac {3 d x}{2}\right )-900 A \sin \left (2 c+\frac {3 d x}{2}\right )+1452 A \sin \left (2 c+\frac {5 d x}{2}\right )+300 A \sin \left (3 c+\frac {5 d x}{2}\right )+60 A \sin \left (3 c+\frac {7 d x}{2}\right )+60 A \sin \left (4 c+\frac {7 d x}{2}\right )-1800 A d x \cos \left (c+\frac {3 d x}{2}\right )-1800 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-360 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-360 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+7020 A \sin \left (\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 190, normalized size = 0.98 \[ -\frac {15 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (6 \, A - 13 \, B\right )} d x - {\left (15 \, B \cos \left (d x + c\right )^{4} + 15 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right ) + 144 \, A - 304 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.94, size = 200, normalized size = 1.04 \[ -\frac {\frac {30 \, {\left (d x + c\right )} {\left (6 \, A - 13 \, B\right )}}{a^{3}} - \frac {60 \, {\left (2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 292, normalized size = 1.51 \[ \frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{2 d \,a^{3}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}+\frac {17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {7 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.00, size = 322, normalized size = 1.67 \[ -\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.27, size = 203, normalized size = 1.05 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (3\,A-5\,B\right )}{4\,a^3}+\frac {2\,A-10\,B}{4\,a^3}\right )}{d}-\frac {x\,\left (6\,A-13\,B\right )}{2\,a^3}+\frac {\left (2\,A-7\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-5\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {3\,A-5\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 15.91, size = 966, normalized size = 5.01 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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