Integrand size = 31, antiderivative size = 185 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^3 (13 A+15 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
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Time = 0.52 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3054, 3047, 3100, 2827, 3853, 3855, 3852, 8} \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^3 (13 A+15 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (43 A+45 B) \tan (c+d x) \sec ^2(c+d x)}{60 d}+\frac {a^3 (13 A+15 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {(7 A+5 B) \tan (c+d x) \sec ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a \cos (c+d x)+a)^2}{5 d} \]
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Rule 8
Rule 2827
Rule 3047
Rule 3054
Rule 3100
Rule 3852
Rule 3853
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+a \cos (c+d x))^2 (a (7 A+5 B)+a (2 A+5 B) \cos (c+d x)) \sec ^5(c+d x) \, dx \\ & = \frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+a \cos (c+d x)) \left (a^2 (43 A+45 B)+2 a^2 (11 A+15 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int \left (a^3 (43 A+45 B)+\left (2 a^3 (11 A+15 B)+a^3 (43 A+45 B)\right ) \cos (c+d x)+2 a^3 (11 A+15 B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{60} \int \left (15 a^3 (13 A+15 B)+4 a^3 (38 A+45 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} \left (a^3 (13 A+15 B)\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{15} \left (a^3 (38 A+45 B)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (a^3 (13 A+15 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^3 (38 A+45 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d} \\ & = \frac {a^3 (13 A+15 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (38 A+45 B) \tan (c+d x)}{15 d}+\frac {a^3 (13 A+15 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^3 (43 A+45 B) \sec ^2(c+d x) \tan (c+d x)}{60 d}+\frac {(7 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \\ \end{align*}
Time = 1.76 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.55 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^3 \left (15 (13 A+15 B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 (13 A+15 B) \sec (c+d x)+30 (3 A+B) \sec ^3(c+d x)+8 \left (60 (A+B)+5 (5 A+3 B) \tan ^2(c+d x)+3 A \tan ^4(c+d x)\right )\right )\right )}{120 d} \]
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Time = 5.86 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.06
method | result | size |
parts | \(-\frac {A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{3} \tan \left (d x +c \right )}{d}\) | \(196\) |
parallelrisch | \(\frac {40 \left (-\frac {39 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {15 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {39 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {15 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\left (\frac {15 A}{16}+\frac {57 B}{80}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {19 A}{20}+\frac {39 B}{40}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {39 A}{160}+\frac {9 B}{32}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {19 A}{100}+\frac {9 B}{40}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {3 B}{4}\right )\right ) a^{3}}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(217\) |
derivativedivides | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{3} \tan \left (d x +c \right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(271\) |
default | \(\frac {A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{3} \tan \left (d x +c \right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A \,a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(271\) |
risch | \(-\frac {i a^{3} \left (195 A \,{\mathrm e}^{9 i \left (d x +c \right )}+225 B \,{\mathrm e}^{9 i \left (d x +c \right )}-120 B \,{\mathrm e}^{8 i \left (d x +c \right )}+750 A \,{\mathrm e}^{7 i \left (d x +c \right )}+570 B \,{\mathrm e}^{7 i \left (d x +c \right )}-720 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1200 B \,{\mathrm e}^{6 i \left (d x +c \right )}-2320 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2400 B \,{\mathrm e}^{4 i \left (d x +c \right )}-750 A \,{\mathrm e}^{3 i \left (d x +c \right )}-570 B \,{\mathrm e}^{3 i \left (d x +c \right )}-1520 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1680 B \,{\mathrm e}^{2 i \left (d x +c \right )}-195 A \,{\mathrm e}^{i \left (d x +c \right )}-225 B \,{\mathrm e}^{i \left (d x +c \right )}-304 A -360 B \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {13 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}+\frac {13 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}\) | \(299\) |
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Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.89 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (38 \, A + 45 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \, {\left (13 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 24 \, A a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
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Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.82 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 45 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (13 \, A a^{3} + 15 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
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Time = 2.96 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.21 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A+15\,B\right )}{4\,d}-\frac {\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {91\,A\,a^3}{6}-\frac {35\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+32\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {133\,A\,a^3}{6}-\frac {61\,B\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {49\,B\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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