Integrand size = 29, antiderivative size = 102 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4085, 3881, 3879} \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {(2 A+3 B) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {(2 A+3 B) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}+\frac {(A-B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rule 3879
Rule 3881
Rule 4085
Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a} \\ & = \frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2} \\ & = \frac {(A-B) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(2 A+3 B) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}
Time = 0.86 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (5 (8 A+3 B) \sin \left (\frac {d x}{2}\right )-15 (2 A+B) \sin \left (c+\frac {d x}{2}\right )+20 A \sin \left (c+\frac {3 d x}{2}\right )+15 B \sin \left (c+\frac {3 d x}{2}\right )-15 A \sin \left (2 c+\frac {3 d x}{2}\right )+7 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \]
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Time = 0.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
method | result | size |
parallelrisch | \(\frac {\left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {10 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+5 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a^{3} d}\) | \(56\) |
derivativedivides | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\) | \(64\) |
default | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{4 d \,a^{3}}\) | \(64\) |
risch | \(\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+30 A \,{\mathrm e}^{3 i \left (d x +c \right )}+15 B \,{\mathrm e}^{3 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+15 B \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )} A +15 B \,{\mathrm e}^{i \left (d x +c \right )}+7 A +3 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(114\) |
norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}-\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (5 A +3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (13 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2}}\) | \(117\) |
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Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {{\left ({\left (7 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A + 3 \, B\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {A {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
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Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]
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Time = 13.76 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,A+15\,B-10\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{60\,a^3\,d} \]
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