Integrand size = 40, antiderivative size = 125 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^3 (5 B+7 C) x+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]
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Time = 0.37 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4157, 4102, 4081, 3855} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {(5 B+3 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {1}{2} a^3 x (5 B+7 C)+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]
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Rule 3855
Rule 4081
Rule 4102
Rule 4157
Rubi steps \begin{align*} \text {integral}& = \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 B+3 C)+3 a C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^2 (B+C)+6 a^2 C \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {1}{6} \int \left (-3 a^3 (5 B+7 C)-6 a^3 C \sec (c+d x)\right ) \, dx \\ & = \frac {1}{2} a^3 (5 B+7 C) x+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (5 B+7 C) x+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (30 B d x+42 C d x-12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 (5 B+4 C) \sin (c+d x)+3 (3 B+C) \sin (2 (c+d x))+B \sin (3 (c+d x))\right )}{12 d} \]
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Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73
method | result | size |
parallelrisch | \(\frac {5 a^{3} \left (-\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{5}+\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{5}+\frac {\left (3 B +C \right ) \sin \left (2 d x +2 c \right )}{10}+\frac {B \sin \left (3 d x +3 c \right )}{30}+3 \left (\frac {B}{2}+\frac {2 C}{5}\right ) \sin \left (d x +c \right )+d x \left (B +\frac {7 C}{5}\right )\right )}{2 d}\) | \(91\) |
derivativedivides | \(\frac {B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(147\) |
default | \(\frac {B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(147\) |
risch | \(\frac {5 a^{3} B x}{2}+\frac {7 a^{3} x C}{2}-\frac {15 i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {15 i B \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {B \,a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\) | \(189\) |
norman | \(\frac {\left (\frac {5}{2} B \,a^{3}+\frac {7}{2} a^{3} C \right ) x +\left (\frac {5}{2} B \,a^{3}+\frac {7}{2} a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (-10 B \,a^{3}-14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-10 B \,a^{3}-14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (15 B \,a^{3}+21 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a^{3} \left (11 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {19 a^{3} \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {5 a^{3} \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {a^{3} \left (5 B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {a^{3} \left (11 B +39 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a^{3} \left (13 B +17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (59 B +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {a^{3} \left (101 B +69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(391\) |
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Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (5 \, B + 7 \, C\right )} a^{3} d x + 3 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, {\left (11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]
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Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
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Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 12 \, {\left (d x + c\right )} B a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 36 \, {\left (d x + c\right )} C a^{3} - 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} \sin \left (d x + c\right ) - 36 \, C a^{3} \sin \left (d x + c\right )}{12 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.44 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (5 \, B a^{3} + 7 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
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Time = 16.57 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.42 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]
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