Integrand size = 29, antiderivative size = 122 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+5 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 A+5 C) \tan ^3(c+d x)}{15 d} \]
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Time = 0.14 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3100, 2827, 3853, 3855, 3852} \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {(4 A+5 C) \tan ^3(c+d x)}{15 d}+\frac {(4 A+5 C) \tan (c+d x)}{5 d}+\frac {A \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 B \tan (c+d x) \sec (c+d x)}{8 d} \]
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Rule 2827
Rule 3100
Rule 3852
Rule 3853
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (5 B+(4 A+5 C) \cos (c+d x)) \sec ^5(c+d x) \, dx \\ & = \frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+B \int \sec ^5(c+d x) \, dx+\frac {1}{5} (4 A+5 C) \int \sec ^4(c+d x) \, dx \\ & = \frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 B) \int \sec ^3(c+d x) \, dx-\frac {(4 A+5 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {(4 A+5 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 A+5 C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (3 B) \int \sec (c+d x) \, dx \\ & = \frac {3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(4 A+5 C) \tan (c+d x)}{5 d}+\frac {3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {A \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(4 A+5 C) \tan ^3(c+d x)}{15 d} \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {45 B \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 B \sec (c+d x)+30 B \sec ^3(c+d x)+8 \left (15 (A+C)+5 (2 A+C) \tan ^2(c+d x)+3 A \tan ^4(c+d x)\right )\right )}{120 d} \]
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Time = 8.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {-A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(104\) |
default | \(\frac {-A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(104\) |
parts | \(-\frac {A \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {B \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(109\) |
risch | \(-\frac {i \left (45 B \,{\mathrm e}^{9 i \left (d x +c \right )}+210 B \,{\mathrm e}^{7 i \left (d x +c \right )}-240 C \,{\mathrm e}^{6 i \left (d x +c \right )}-640 A \,{\mathrm e}^{4 i \left (d x +c \right )}-560 C \,{\mathrm e}^{4 i \left (d x +c \right )}-210 B \,{\mathrm e}^{3 i \left (d x +c \right )}-320 A \,{\mathrm e}^{2 i \left (d x +c \right )}-400 C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B \,{\mathrm e}^{i \left (d x +c \right )}-64 A -80 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(174\) |
parallelrisch | \(\frac {-450 B \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 B \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (320 A +400 C \right ) \sin \left (3 d x +3 c \right )+\left (64 A +80 C \right ) \sin \left (5 d x +5 c \right )+420 \sin \left (2 d x +2 c \right ) B +90 \sin \left (4 d x +4 c \right ) B +640 \left (A +\frac {C}{2}\right ) \sin \left (d x +c \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(197\) |
norman | \(\frac {-\frac {8 \left (19 A +5 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {2 \left (2 A -3 B -2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (2 A +3 B -2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (8 A -5 B +8 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (8 A +5 B +8 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (88 A -5 B -40 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}-\frac {\left (88 A +5 B -40 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 B \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(247\) |
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Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {45 \, B \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, B \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 45 \, B \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 30 \, B \cos \left (d x + c\right ) + 24 \, A\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
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Timed out. \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C - 15 \, B {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (110) = 220\).
Time = 0.34 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.02 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 160 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
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Time = 3.85 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.61 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\left (2\,A-\frac {5\,B}{4}+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B}{2}-\frac {8\,A}{3}-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A}{15}+\frac {20\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A}{3}-\frac {B}{2}-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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