Integrand size = 31, antiderivative size = 146 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {(5 A+B) \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{24 d (a+a \sin (c+d x))^3}-\frac {a (3 A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {3 A}{16 d (a+a \sin (c+d x))} \]
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Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2915, 78, 212} \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {a^2 (A-B)}{24 d (a \sin (c+d x)+a)^3}+\frac {(5 A+B) \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a (3 A-B)}{32 d (a \sin (c+d x)+a)^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {3 A}{16 d (a \sin (c+d x)+a)} \]
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Rule 78
Rule 212
Rule 2915
Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {A+B}{16 a^4 (a-x)^3}+\frac {2 A+B}{16 a^5 (a-x)^2}+\frac {A-B}{8 a^3 (a+x)^4}+\frac {3 A-B}{16 a^4 (a+x)^3}+\frac {3 A}{16 a^5 (a+x)^2}+\frac {5 A+B}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{24 d (a+a \sin (c+d x))^3}-\frac {a (3 A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {3 A}{16 d (a+a \sin (c+d x))}+\frac {(5 A+B) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d} \\ & = \frac {(5 A+B) \text {arctanh}(\sin (c+d x))}{16 a d}+\frac {a (A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {2 A+B}{16 d (a-a \sin (c+d x))}-\frac {a^2 (A-B)}{24 d (a+a \sin (c+d x))^3}-\frac {a (3 A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {3 A}{16 d (a+a \sin (c+d x))} \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {6 (5 A+B) \text {arctanh}(\sin (c+d x))+\frac {3 (A+B)}{(-1+\sin (c+d x))^2}-\frac {6 (2 A+B)}{-1+\sin (c+d x)}-\frac {4 (A-B)}{(1+\sin (c+d x))^3}+\frac {-9 A+3 B}{(1+\sin (c+d x))^2}-\frac {18 A}{1+\sin (c+d x)}}{96 a d} \]
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Time = 0.97 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {\left (-\frac {5 A}{32}-\frac {B}{32}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{16}-\frac {B}{16}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{8}+\frac {B}{16}}{\sin \left (d x +c \right )-1}-\frac {3 A}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{8}-\frac {B}{8}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {5 A}{32}+\frac {B}{32}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(132\) |
default | \(\frac {\left (-\frac {5 A}{32}-\frac {B}{32}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{16}-\frac {B}{16}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {A}{8}+\frac {B}{16}}{\sin \left (d x +c \right )-1}-\frac {3 A}{16 \left (1+\sin \left (d x +c \right )\right )}-\frac {\frac {A}{8}-\frac {B}{8}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}+\left (\frac {5 A}{32}+\frac {B}{32}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d a}\) | \(132\) |
parallelrisch | \(\frac {-45 \left (A +\frac {B}{5}\right ) \left (2+\frac {\sin \left (5 d x +5 c \right )}{3}+\sin \left (3 d x +3 c \right )+\frac {2 \sin \left (d x +c \right )}{3}+\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {8 \cos \left (2 d x +2 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \left (A +\frac {B}{5}\right ) \left (2+\frac {\sin \left (5 d x +5 c \right )}{3}+\sin \left (3 d x +3 c \right )+\frac {2 \sin \left (d x +c \right )}{3}+\frac {2 \cos \left (4 d x +4 c \right )}{3}+\frac {8 \cos \left (2 d x +2 c \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-16 A -80 B \right ) \cos \left (2 d x +2 c \right )+\left (-14 A -22 B \right ) \cos \left (4 d x +4 c \right )+\left (84 A -12 B \right ) \sin \left (3 d x +3 c \right )+\left (8 A -8 B \right ) \sin \left (5 d x +5 c \right )+\left (236 A +28 B \right ) \sin \left (d x +c \right )+30 A +102 B}{48 a d \left (6+\sin \left (5 d x +5 c \right )+3 \sin \left (3 d x +3 c \right )+2 \sin \left (d x +c \right )+2 \cos \left (4 d x +4 c \right )+8 \cos \left (2 d x +2 c \right )\right )}\) | \(294\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (6 i B \,{\mathrm e}^{7 i \left (d x +c \right )}+15 A \,{\mathrm e}^{8 i \left (d x +c \right )}-30 i A \,{\mathrm e}^{i \left (d x +c \right )}+3 B \,{\mathrm e}^{8 i \left (d x +c \right )}-6 i B \,{\mathrm e}^{i \left (d x +c \right )}+40 A \,{\mathrm e}^{6 i \left (d x +c \right )}-110 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+8 B \,{\mathrm e}^{6 i \left (d x +c \right )}-22 i B \,{\mathrm e}^{3 i \left (d x +c \right )}+18 A \,{\mathrm e}^{4 i \left (d x +c \right )}+30 i A \,{\mathrm e}^{7 i \left (d x +c \right )}-150 B \,{\mathrm e}^{4 i \left (d x +c \right )}+22 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+40 A \,{\mathrm e}^{2 i \left (d x +c \right )}+110 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+8 B \,{\mathrm e}^{2 i \left (d x +c \right )}+15 A +3 B \right )}{24 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d a}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{16 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{16 a d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{16 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{16 a d}\) | \(371\) |
norman | \(\frac {\frac {\left (41 A +37 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (41 A +37 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}-\frac {\left (A -19 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (3 A +7 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {\left (3 A +7 B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {\left (2 A +10 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (2 A +10 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (11 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (11 A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}+\frac {\left (43 A -B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (43 A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {\left (5 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a d}+\frac {\left (5 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a d}\) | \(377\) |
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Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {6 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (5 \, A + B\right )} \cos \left (d x + c\right )^{2} + 10 \, A + 2 \, B\right )} \sin \left (d x + c\right ) - 4 \, A - 20 \, B}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]
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\[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {A \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
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Time = 0.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {3 \, {\left (5 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {2 \, {\left (3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{4} + 3 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right )^{2} - 5 \, {\left (5 \, A + B\right )} \sin \left (d x + c\right ) + 8 \, A - 8 \, B\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a}}{96 \, d} \]
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Time = 0.46 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, {\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {6 \, {\left (5 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, A \sin \left (d x + c\right )^{2} + 3 \, B \sin \left (d x + c\right )^{2} - 38 \, A \sin \left (d x + c\right ) - 10 \, B \sin \left (d x + c\right ) + 25 \, A + 9 \, B\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, A \sin \left (d x + c\right )^{3} + 11 \, B \sin \left (d x + c\right )^{3} + 201 \, A \sin \left (d x + c\right )^{2} + 33 \, B \sin \left (d x + c\right )^{2} + 255 \, A \sin \left (d x + c\right ) + 27 \, B \sin \left (d x + c\right ) + 117 \, A - 3 \, B}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \]
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Time = 10.67 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A+B\right )}{16\,a\,d}-\frac {\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {5\,A}{16}+\frac {B}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (-\frac {25\,A}{48}-\frac {5\,B}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A}{6}-\frac {B}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]
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