Integrand size = 21, antiderivative size = 146 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \]
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Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2746, 46, 212} \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))}{64 a^2 d}-\frac {a^2}{32 d (a \sin (c+d x)+a)^4}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}-\frac {a}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {3}{32 d (a \sin (c+d x)+a)^2} \]
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Rule 46
Rule 212
Rule 2746
Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^5 \text {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^3}+\frac {5}{64 a^6 (a-x)^2}+\frac {1}{8 a^3 (a+x)^5}+\frac {3}{16 a^4 (a+x)^4}+\frac {3}{16 a^5 (a+x)^3}+\frac {5}{32 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {15 \text {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d} \\ & = \frac {15 \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^4(c+d x) (1-\sin (c+d x))^2 (1+\sin (c+d x))^2 \left (\frac {15}{64} \text {arctanh}(\sin (c+d x))+\frac {1}{64 (1-\sin (c+d x))^2}+\frac {5}{64 (1-\sin (c+d x))}-\frac {1}{32 (1+\sin (c+d x))^4}-\frac {1}{16 (1+\sin (c+d x))^3}-\frac {3}{32 (1+\sin (c+d x))^2}-\frac {5}{32 (1+\sin (c+d x))}\right )}{a^2 d} \]
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Time = 1.38 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {-\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128}}{d \,a^{2}}\) | \(103\) |
default | \(\frac {-\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128}}{d \,a^{2}}\) | \(103\) |
risch | \(-\frac {i \left (60 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}+160 i {\mathrm e}^{8 i \left (d x +c \right )}-35 \,{\mathrm e}^{9 i \left (d x +c \right )}+72 i {\mathrm e}^{6 i \left (d x +c \right )}-242 \,{\mathrm e}^{7 i \left (d x +c \right )}+160 i {\mathrm e}^{4 i \left (d x +c \right )}+242 \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i {\mathrm e}^{2 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{32 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} a^{2} d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{64 a^{2} d}-\frac {15 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 a^{2} d}\) | \(208\) |
norman | \(\frac {\frac {17 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {17 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {23 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d a}+\frac {49 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {57 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {57 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {29 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {29 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 a^{2} d}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 a^{2} d}\) | \(280\) |
parallelrisch | \(\frac {\left (-60 \sin \left (5 d x +5 c \right )-255 \cos \left (2 d x +2 c \right )-30 \cos \left (4 d x +4 c \right )+15 \cos \left (6 d x +6 c \right )-120 \sin \left (d x +c \right )-180 \sin \left (3 d x +3 c \right )-210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (60 \sin \left (5 d x +5 c \right )+255 \cos \left (2 d x +2 c \right )+30 \cos \left (4 d x +4 c \right )-15 \cos \left (6 d x +6 c \right )+120 \sin \left (d x +c \right )+180 \sin \left (3 d x +3 c \right )+210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+34 \sin \left (5 d x +5 c \right )-48 \cos \left (2 d x +2 c \right )-88 \cos \left (4 d x +4 c \right )-16 \cos \left (6 d x +6 c \right )+612 \sin \left (d x +c \right )+262 \sin \left (3 d x +3 c \right )+152}{64 a^{2} d \left (17 \cos \left (2 d x +2 c \right )-\cos \left (6 d x +6 c \right )+4 \sin \left (5 d x +5 c \right )+12 \sin \left (3 d x +3 c \right )+2 \cos \left (4 d x +4 c \right )+14+8 \sin \left (d x +c \right )\right )}\) | \(297\) |
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Time = 0.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {60 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) - 8}{128 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]
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\[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} + 30 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{3} - 50 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{128 \, d} \]
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Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{2} - 110 \, \sin \left (d x + c\right ) + 69\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 580 \, \sin \left (d x + c\right )^{3} + 1038 \, \sin \left (d x + c\right )^{2} + 868 \, \sin \left (d x + c\right ) + 301}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{512 \, d} \]
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Time = 0.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{64\,a^2\,d}+\frac {-\frac {15\,{\sin \left (c+d\,x\right )}^5}{64}-\frac {15\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {25\,{\sin \left (c+d\,x\right )}^2}{32}+\frac {17\,\sin \left (c+d\,x\right )}{64}-\frac {1}{4}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]
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