Optimal. Leaf size=183 \[ \frac {5 a^3 (3 A+4 (B+C)) \sin (c+d x)}{8 d}+\frac {(15 A+20 B+12 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{24 d}+\frac {1}{8} a^3 x (15 A+20 B+28 C)+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {(3 A+4 B) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 a d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
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Rubi [A] time = 0.44, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4086, 4017, 3996, 3770} \[ \frac {5 a^3 (3 A+4 (B+C)) \sin (c+d x)}{8 d}+\frac {(15 A+20 B+12 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{24 d}+\frac {(3 A+4 B) \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 a d}+\frac {1}{8} a^3 x (15 A+20 B+28 C)+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3770
Rule 3996
Rule 4017
Rule 4086
Rubi steps
\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (a (3 A+4 B)+4 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(3 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (a^2 (15 A+20 B+12 C)+12 a^2 C \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(3 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(15 A+20 B+12 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (3 A+4 (B+C))+24 a^3 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {5 a^3 (3 A+4 (B+C)) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(3 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(15 A+20 B+12 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{24 d}-\frac {\int \left (-3 a^4 (15 A+20 B+28 C)-24 a^4 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {1}{8} a^3 (15 A+20 B+28 C) x+\frac {5 a^3 (3 A+4 (B+C)) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(3 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(15 A+20 B+12 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{24 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a^3 (15 A+20 B+28 C) x+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (3 A+4 (B+C)) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {(3 A+4 B) \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{12 a d}+\frac {(15 A+20 B+12 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{24 d}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 147, normalized size = 0.80 \[ \frac {a^3 \left (24 (13 A+15 B+12 C) \sin (c+d x)+24 (4 A+3 B+C) \sin (2 (c+d x))+24 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))+180 A d x+8 B \sin (3 (c+d x))+240 B d x-96 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+96 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+336 C d x\right )}{96 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 131, normalized size = 0.72 \[ \frac {3 \, {\left (15 \, A + 20 \, B + 28 \, C\right )} a^{3} d x + 12 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 12 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (15 \, A + 12 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.34, size = 286, normalized size = 1.56 \[ \frac {24 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (15 \, A a^{3} + 20 \, B a^{3} + 28 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 204 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 228 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 147 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 84 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.44, size = 251, normalized size = 1.37 \[ \frac {A \,a^{3} \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4 d}+\frac {15 A \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{8 d}+\frac {15 a^{3} A x}{8}+\frac {15 A \,a^{3} c}{8 d}+\frac {B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{3}}{3 d}+\frac {11 a^{3} B \sin \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {7 a^{3} C x}{2}+\frac {7 C \,a^{3} c}{2 d}+\frac {A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{d}+\frac {3 a^{3} A \sin \left (d x +c \right )}{d}+\frac {3 a^{3} B \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {5 a^{3} B x}{2}+\frac {5 a^{3} B c}{2 d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 240, normalized size = 1.31 \[ -\frac {96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 96 \, {\left (d x + c\right )} B a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 288 \, {\left (d x + c\right )} C a^{3} - 48 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 96 \, A a^{3} \sin \left (d x + c\right ) - 288 \, B a^{3} \sin \left (d x + c\right ) - 288 \, C a^{3} \sin \left (d x + c\right )}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.71, size = 243, normalized size = 1.33 \[ \frac {45\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+60\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+84\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+24\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+12\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )+3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )+\frac {3\,A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{8}+9\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )+B\,a^3\,\sin \left (3\,c+3\,d\,x\right )+3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )+39\,A\,a^3\,\sin \left (c+d\,x\right )+45\,B\,a^3\,\sin \left (c+d\,x\right )+36\,C\,a^3\,\sin \left (c+d\,x\right )}{12\,d} \]
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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