Integrand size = 33, antiderivative size = 115 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
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Time = 0.23 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4137, 4007, 4004, 3879} \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {A x}{a^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]
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Rule 3879
Rule 4004
Rule 4007
Rule 4137
Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {-5 a A+a (2 A-2 B-3 C) \sec (c+d x)}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = \frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2} \\ & = \frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(115)=230\).
Time = 2.02 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 A d x \cos \left (\frac {d x}{2}\right )+150 A d x \cos \left (c+\frac {d x}{2}\right )+75 A d x \cos \left (c+\frac {3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 A \sin \left (\frac {d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )+30 C \sin \left (\frac {d x}{2}\right )+270 A \sin \left (c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )-230 A \sin \left (c+\frac {3 d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )+30 C \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )-64 A \sin \left (2 c+\frac {5 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )+6 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \]
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Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63
method | result | size |
parallelrisch | \(\frac {3 \left (-A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \left (-7 A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x A}{60 a^{3} d}\) | \(73\) |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(127\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(127\) |
norman | \(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {A x}{a}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}+\frac {\left (7 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (23 A -13 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}-\frac {\left (25 A -5 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2}}\) | \(155\) |
risch | \(\frac {A x}{a^{3}}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}-15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}-30 B \,{\mathrm e}^{3 i \left (d x +c \right )}-15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}-40 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+115 A \,{\mathrm e}^{i \left (d x +c \right )}-20 B \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+32 A -7 B -3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(172\) |
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Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.28 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x - {\left ({\left (32 \, A - 7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
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\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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Time = 0.32 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.78 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
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Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} A}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
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Time = 16.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.37 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]
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