Integrand size = 31, antiderivative size = 179 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {5 (3 A+B) \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{32 d (a+a \sin (c+d x))^4}-\frac {a (3 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {3 A}{32 d (a+a \sin (c+d x))^2}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \]
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Time = 0.16 (sec) , antiderivative size = 179, 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} \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {5 (3 A+B) \text {arctanh}(\sin (c+d x))}{64 a^2 d}-\frac {a^2 (A-B)}{32 d (a \sin (c+d x)+a)^4}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {a (3 A-B)}{48 d (a \sin (c+d x)+a)^3}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {3 A}{32 d (a \sin (c+d x)+a)^2} \]
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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)^5} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {A+B}{32 a^5 (a-x)^3}+\frac {5 A+3 B}{64 a^6 (a-x)^2}+\frac {A-B}{8 a^3 (a+x)^5}+\frac {3 A-B}{16 a^4 (a+x)^4}+\frac {3 A}{16 a^5 (a+x)^3}+\frac {5 A+B}{32 a^6 (a+x)^2}+\frac {5 (3 A+B)}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{32 d (a+a \sin (c+d x))^4}-\frac {a (3 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {3 A}{32 d (a+a \sin (c+d x))^2}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {(5 (3 A+B)) \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d} \\ & = \frac {5 (3 A+B) \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {A+B}{64 d (a-a \sin (c+d x))^2}-\frac {a^2 (A-B)}{32 d (a+a \sin (c+d x))^4}-\frac {a (3 A-B)}{48 d (a+a \sin (c+d x))^3}-\frac {3 A}{32 d (a+a \sin (c+d x))^2}+\frac {5 A+3 B}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5 A+B}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.69 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {15 (3 A+B) \text {arctanh}(\sin (c+d x))+\frac {3 (A+B)}{(-1+\sin (c+d x))^2}-\frac {3 (5 A+3 B)}{-1+\sin (c+d x)}-\frac {6 (A-B)}{(1+\sin (c+d x))^4}+\frac {4 (-3 A+B)}{(1+\sin (c+d x))^3}-\frac {18 A}{(1+\sin (c+d x))^2}-\frac {6 (5 A+B)}{1+\sin (c+d x)}}{192 a^2 d} \]
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Time = 1.39 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\left (-\frac {15 A}{128}-\frac {5 B}{128}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{32}-\frac {B}{32}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {5 A}{64}+\frac {3 B}{64}}{\sin \left (d x +c \right )-1}-\frac {3 A}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {\frac {A}{8}-\frac {B}{8}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}+\frac {B}{32}}{1+\sin \left (d x +c \right )}+\left (\frac {15 A}{128}+\frac {5 B}{128}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(151\) |
default | \(\frac {\left (-\frac {15 A}{128}-\frac {5 B}{128}\right ) \ln \left (\sin \left (d x +c \right )-1\right )-\frac {-\frac {A}{32}-\frac {B}{32}}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {\frac {5 A}{64}+\frac {3 B}{64}}{\sin \left (d x +c \right )-1}-\frac {3 A}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {\frac {A}{8}-\frac {B}{8}}{4 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {\frac {3 A}{16}-\frac {B}{16}}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {\frac {5 A}{32}+\frac {B}{32}}{1+\sin \left (d x +c \right )}+\left (\frac {15 A}{128}+\frac {5 B}{128}\right ) \ln \left (1+\sin \left (d x +c \right )\right )}{d \,a^{2}}\) | \(151\) |
parallelrisch | \(\frac {-360 \left (\frac {7}{4}+\frac {17 \cos \left (2 d x +2 c \right )}{8}-\frac {\cos \left (6 d x +6 c \right )}{8}+\frac {\sin \left (5 d x +5 c \right )}{2}+\frac {3 \sin \left (3 d x +3 c \right )}{2}+\frac {\cos \left (4 d x +4 c \right )}{4}+\sin \left (d x +c \right )\right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+360 \left (\frac {7}{4}+\frac {17 \cos \left (2 d x +2 c \right )}{8}-\frac {\cos \left (6 d x +6 c \right )}{8}+\frac {\sin \left (5 d x +5 c \right )}{2}+\frac {3 \sin \left (3 d x +3 c \right )}{2}+\frac {\cos \left (4 d x +4 c \right )}{4}+\sin \left (d x +c \right )\right ) \left (A +\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-144 A -592 B \right ) \cos \left (2 d x +2 c \right )+\left (-264 A -152 B \right ) \cos \left (4 d x +4 c \right )+\left (-48 A +16 B \right ) \cos \left (6 d x +6 c \right )+\left (786 A -122 B \right ) \sin \left (3 d x +3 c \right )+\left (102 A -94 B \right ) \sin \left (5 d x +5 c \right )+\left (1836 A +356 B \right ) \sin \left (d x +c \right )+456 A +728 B}{192 d \,a^{2} \left (14+2 \cos \left (4 d x +4 c \right )+8 \sin \left (d x +c \right )+12 \sin \left (3 d x +3 c \right )+17 \cos \left (2 d x +2 c \right )-\cos \left (6 d x +6 c \right )+4 \sin \left (5 d x +5 c \right )\right )}\) | \(346\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (160 i B \,{\mathrm e}^{7 i \left (d x +c \right )}+45 A \,{\mathrm e}^{10 i \left (d x +c \right )}+216 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+15 B \,{\mathrm e}^{10 i \left (d x +c \right )}+180 i A \,{\mathrm e}^{i \left (d x +c \right )}-105 A \,{\mathrm e}^{8 i \left (d x +c \right )}+60 i B \,{\mathrm e}^{i \left (d x +c \right )}-35 B \,{\mathrm e}^{8 i \left (d x +c \right )}+160 i B \,{\mathrm e}^{3 i \left (d x +c \right )}-726 A \,{\mathrm e}^{6 i \left (d x +c \right )}+480 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-242 B \,{\mathrm e}^{6 i \left (d x +c \right )}+60 i B \,{\mathrm e}^{9 i \left (d x +c \right )}+726 A \,{\mathrm e}^{4 i \left (d x +c \right )}+180 i A \,{\mathrm e}^{9 i \left (d x +c \right )}+242 B \,{\mathrm e}^{4 i \left (d x +c \right )}+480 i A \,{\mathrm e}^{7 i \left (d x +c \right )}+105 A \,{\mathrm e}^{2 i \left (d x +c \right )}-952 i B \,{\mathrm e}^{5 i \left (d x +c \right )}+35 B \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A -15 B \right )}{96 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d \,a^{2}}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{64 a^{2} d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{64 a^{2} d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{64 a^{2} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{64 a^{2} d}\) | \(429\) |
norman | \(\frac {\frac {\left (117 A +103 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}+\frac {\left (117 A +103 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{48 a d}+\frac {\left (143 A +69 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a d}+\frac {\left (143 A +69 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a d}+\frac {\left (17 A +11 B \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}+\frac {\left (17 A +11 B \right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}+\frac {\left (49 A -5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 a d}+\frac {\left (49 A -5 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 a d}+\frac {\left (33 A +59 B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}+\frac {\left (33 A +59 B \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}-\frac {\left (21 A -137 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}-\frac {\left (21 A -137 B \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (171 A -7 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {5 \left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 a^{2} d}+\frac {5 \left (3 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 a^{2} d}\) | \(434\) |
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Time = 0.32 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {60 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A + B\right )} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{4} - 20 \, {\left (3 \, A + B\right )} \cos \left (d x + c\right )^{2} - 36 \, A - 12 \, B\right )} \sin \left (d x + c\right ) - 24 \, A - 72 \, B}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]
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Timed out. \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
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Time = 0.22 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (15 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{5} + 30 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{4} - 10 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{3} - 50 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right )^{2} - 17 \, {\left (3 \, A + B\right )} \sin \left (d x + c\right ) + 48 \, A - 16 \, B\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, {\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, {\left (3 \, A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \]
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Time = 0.39 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {60 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, {\left (3 \, A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (45 \, A \sin \left (d x + c\right )^{2} + 15 \, B \sin \left (d x + c\right )^{2} - 110 \, A \sin \left (d x + c\right ) - 42 \, B \sin \left (d x + c\right ) + 69 \, A + 31 \, B\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {375 \, A \sin \left (d x + c\right )^{4} + 125 \, B \sin \left (d x + c\right )^{4} + 1740 \, A \sin \left (d x + c\right )^{3} + 548 \, B \sin \left (d x + c\right )^{3} + 3114 \, A \sin \left (d x + c\right )^{2} + 894 \, B \sin \left (d x + c\right )^{2} + 2604 \, A \sin \left (d x + c\right ) + 612 \, B \sin \left (d x + c\right ) + 903 \, A + 93 \, B}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{1536 \, d} \]
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Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx=\frac {\left (-\frac {15\,A}{64}-\frac {5\,B}{64}\right )\,{\sin \left (c+d\,x\right )}^5+\left (-\frac {15\,A}{32}-\frac {5\,B}{32}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {5\,A}{32}+\frac {5\,B}{96}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {25\,A}{32}+\frac {25\,B}{96}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {17\,A}{64}+\frac {17\,B}{192}\right )\,\sin \left (c+d\,x\right )-\frac {A}{4}+\frac {B}{12}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (3\,A+B\right )}{64\,a^2\,d} \]
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