Integrand size = 38, antiderivative size = 145 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (b B+a C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d} \]
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Time = 0.24 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4082, 3872, 3852, 3853, 3855} \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (a C+b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 (a C+b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]
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Rule 3852
Rule 3853
Rule 3855
Rule 3872
Rule 4082
Rule 4157
Rubi steps \begin{align*} \text {integral}& = \int \sec ^4(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx \\ & = \frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^4(c+d x) (5 a B+4 b C+5 (b B+a C) \sec (c+d x)) \, dx \\ & = \frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+(b B+a C) \int \sec ^5(c+d x) \, dx+\frac {1}{5} (5 a B+4 b C) \int \sec ^4(c+d x) \, dx \\ & = \frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 (b B+a C)) \int \sec ^3(c+d x) \, dx-\frac {(5 a B+4 b C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (3 (b B+a C)) \int \sec (c+d x) \, dx \\ & = \frac {3 (b B+a C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d} \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.73 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 (b B+a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (45 (b B+a C) \sec (c+d x)+30 (b B+a C) \sec ^3(c+d x)+8 \left (15 (a B+b C)+5 (a B+2 b C) \tan ^2(c+d x)+3 b C \tan ^4(c+d x)\right )\right )}{120 d} \]
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Time = 0.89 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81
method | result | size |
parts | \(\frac {\left (B b +C a \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(117\) |
derivativedivides | \(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
default | \(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+B b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C b \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(154\) |
norman | \(\frac {-\frac {4 \left (25 a B +29 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (8 a B -5 B b -5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (8 a B +5 B b +5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (32 a B -3 B b -3 C a +16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (32 a B +3 B b +3 C a +16 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(222\) |
parallelrisch | \(\frac {-45 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+420 \left (B b +C a \right ) \sin \left (2 d x +2 c \right )+80 \left (5 a B +4 C b \right ) \sin \left (3 d x +3 c \right )+90 \left (B b +C a \right ) \sin \left (4 d x +4 c \right )+16 \left (5 a B +4 C b \right ) \sin \left (5 d x +5 c \right )+320 \sin \left (d x +c \right ) \left (a B +2 C b \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(230\) |
risch | \(-\frac {i \left (45 B b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a \,{\mathrm e}^{9 i \left (d x +c \right )}+210 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a \,{\mathrm e}^{7 i \left (d x +c \right )}-240 B a \,{\mathrm e}^{6 i \left (d x +c \right )}-560 a B \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-210 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a \,{\mathrm e}^{3 i \left (d x +c \right )}-400 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B b \,{\mathrm e}^{i \left (d x +c \right )}-45 C a \,{\mathrm e}^{i \left (d x +c \right )}-80 a B -64 C b \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) | \(279\) |
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Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{4} + 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{2} + 24 \, C b + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
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\[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
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Time = 0.23 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.38 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (133) = 266\).
Time = 0.31 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.28 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
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Time = 21.39 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.61 \[ \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,B\,b}{4}+\frac {3\,C\,a}{4}\right )}{d}-\frac {\left (2\,B\,a-\frac {5\,B\,b}{4}-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,b}{2}-\frac {16\,B\,a}{3}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,B\,a}{3}-\frac {B\,b}{2}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a+\frac {5\,B\,b}{4}+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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