Integrand size = 21, antiderivative size = 135 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {3 a b \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d} \]
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Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2868, 3853, 3855, 3091, 3852} \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 a b \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d} \]
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Rule 2868
Rule 3091
Rule 3852
Rule 3853
Rule 3855
Rubi steps \begin{align*} \text {integral}& = (2 a b) \int \sec ^5(c+d x) \, dx+\int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx \\ & = \frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (3 a b) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (4 a^2+5 b^2\right ) \int \sec ^4(c+d x) \, dx \\ & = \frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 a b) \int \sec (c+d x) \, dx-\frac {\left (4 a^2+5 b^2\right ) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {3 a b \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d} \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.67 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {45 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (60 \left (a^2+b^2\right )+45 a b \sec (c+d x)+30 a b \sec ^3(c+d x)+20 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+12 a^2 \tan ^4(c+d x)\right )}{60 d} \]
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Time = 4.33 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
default | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
parts | \(-\frac {a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(115\) |
risch | \(-\frac {i \left (45 a b \,{\mathrm e}^{9 i \left (d x +c \right )}+210 a b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-320 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 a b \,{\mathrm e}^{3 i \left (d x +c \right )}-160 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-45 a b \,{\mathrm e}^{i \left (d x +c \right )}-32 a^{2}-40 b^{2}\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{4 d}\) | \(194\) |
parallelrisch | \(\frac {-450 b \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 b \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (160 a^{2}+200 b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (32 a^{2}+40 b^{2}\right ) \sin \left (5 d x +5 c \right )+420 a b \sin \left (2 d x +2 c \right )+90 a b \sin \left (4 d x +4 c \right )+\left (320 a^{2}+160 b^{2}\right ) \sin \left (d x +c \right )}{60 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(214\) |
norman | \(\frac {-\frac {8 \left (19 a^{2}+5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}-\frac {4 \left (a^{2}-3 a b -b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (a^{2}+3 a b -b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (4 a^{2}-5 a b +4 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {\left (4 a^{2}+5 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {\left (44 a^{2}-5 a b -20 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 d}-\frac {\left (44 a^{2}+5 a b -20 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {3 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {3 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\) | \(279\) |
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Time = 0.28 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (45 \, a b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 12 \, a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
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Timed out. \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\text {Timed out} \]
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Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 15 \, a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (123) = 246\).
Time = 0.33 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.01 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]
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Time = 18.54 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.64 \[ \int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx=\frac {3\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\left (2\,a^2-\frac {5\,a\,b}{2}+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^2}{3}+a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^2}{15}+\frac {20\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^2}{3}-a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {5\,a\,b}{2}+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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