Integrand size = 29, antiderivative size = 136 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(3 A-B) x}{a^3}+\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Time = 0.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {4105, 3872, 2717, 8} \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (3 A-B)}{a^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rule 8
Rule 2717
Rule 3872
Rule 4105
Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos (c+d x) (a (6 A-B)-3 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos (c+d x) \left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \cos (c+d x) \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(2 (36 A-11 B)) \int \cos (c+d x) \, dx}{15 a^3}-\frac {(3 A-B) \int 1 \, dx}{a^3} \\ & = -\frac {(3 A-B) x}{a^3}+\frac {2 (36 A-11 B) \sin (c+d x)}{15 a^3 d}-\frac {(A-B) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(9 A-4 B) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(3 A-B) \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}
Time = 2.46 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\sin (c+d x) \left (27 A-7 B-\frac {3 (A-B)}{(1+\sec (c+d x))^3}+\frac {-9 A+4 B}{(1+\sec (c+d x))^2}+\frac {15 (3 A-B) \left (\arcsin (\cos (c+d x)) (1+\cos (c+d x))+\sqrt {\sin ^2(c+d x)}\right )}{\sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{3/2}}\right )}{15 a^3 d} \]
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Time = 1.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65
method | result | size |
parallelrisch | \(\frac {729 \left (\left (\frac {26 A}{81}-\frac {64 B}{729}\right ) \cos \left (2 d x +2 c \right )+\frac {5 A \cos \left (3 d x +3 c \right )}{243}+\left (A -\frac {68 B}{243}\right ) \cos \left (d x +c \right )+\frac {58 A}{81}-\frac {152 B}{729}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-720 \left (A -\frac {B}{3}\right ) d x}{240 a^{3} d}\) | \(89\) |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {8 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-8 \left (3 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(136\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A +\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {8 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-8 \left (3 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(136\) |
norman | \(\frac {-\frac {\left (3 A -B \right ) x}{a}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}-\frac {\left (3 A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {\left (25 A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (27 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (45 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2}}\) | \(158\) |
risch | \(-\frac {3 A x}{a^{3}}+\frac {x B}{a^{3}}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {2 i \left (90 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 B \,{\mathrm e}^{4 i \left (d x +c \right )}+300 A \,{\mathrm e}^{3 i \left (d x +c \right )}-135 B \,{\mathrm e}^{3 i \left (d x +c \right )}+420 A \,{\mathrm e}^{2 i \left (d x +c \right )}-185 B \,{\mathrm e}^{2 i \left (d x +c \right )}+270 \,{\mathrm e}^{i \left (d x +c \right )} A -115 B \,{\mathrm e}^{i \left (d x +c \right )}+72 A -32 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(178\) |
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Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.27 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (3 \, A - B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (3 \, A - B\right )} d x - {\left (15 \, A \cos \left (d x + c\right )^{3} + {\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right ) + 72 \, A - 22 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \cos {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.31 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.70 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]
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Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.15 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (d x + c\right )} {\left (3 \, A - B\right )}}{a^{3}} - \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
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Time = 14.01 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2\,a^3}+\frac {3\,\left (A-B\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}-\frac {x\,\left (3\,A-B\right )}{a^3}+\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]
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