Integrand size = 21, antiderivative size = 65 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=-\frac {4}{5 a^3 d (a+a \sin (c+d x))^5}-\frac {1}{3 a^5 d (a+a \sin (c+d x))^3}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^4} \]
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Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2746, 45} \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=-\frac {1}{3 a^5 d (a \sin (c+d x)+a)^3}-\frac {4}{5 a^3 d (a \sin (c+d x)+a)^5}+\frac {1}{d \left (a^2 \sin (c+d x)+a^2\right )^4} \]
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Rule 45
Rule 2746
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2}{(a+x)^6} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {4 a^2}{(a+x)^6}-\frac {4 a}{(a+x)^5}+\frac {1}{(a+x)^4}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {4}{5 a^3 d (a+a \sin (c+d x))^5}-\frac {1}{3 a^5 d (a+a \sin (c+d x))^3}+\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )^4} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {\cos ^6(c+d x) \left (2-5 \sin (c+d x)+5 \sin ^2(c+d x)\right )}{15 a^8 d (-1+\sin (c+d x))^3 (1+\sin (c+d x))^8} \]
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Time = 0.79 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {4}{5 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}}{a^{8} d}\) | \(43\) |
default | \(\frac {\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {4}{5 \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}}{a^{8} d}\) | \(43\) |
risch | \(\frac {8 i \left (-10 i {\mathrm e}^{6 i \left (d x +c \right )}+5 \,{\mathrm e}^{7 i \left (d x +c \right )}+10 i {\mathrm e}^{4 i \left (d x +c \right )}-18 \,{\mathrm e}^{5 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{15 d \,a^{8} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{10}}\) | \(82\) |
parallelrisch | \(\frac {60 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+564 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+340 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+340 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+280 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+280 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d \,a^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{10}}\) | \(138\) |
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Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {5 \, \cos \left (d x + c\right )^{2} + 5 \, \sin \left (d x + c\right ) - 7}{15 \, {\left (5 \, a^{8} d \cos \left (d x + c\right )^{4} - 20 \, a^{8} d \cos \left (d x + c\right )^{2} + 16 \, a^{8} d + {\left (a^{8} d \cos \left (d x + c\right )^{4} - 12 \, a^{8} d \cos \left (d x + c\right )^{2} + 16 \, a^{8} d\right )} \sin \left (d x + c\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1120 vs. \(2 (58) = 116\).
Time = 9.26 (sec) , antiderivative size = 1120, normalized size of antiderivative = 17.23 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=\text {Too large to display} \]
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Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=-\frac {5 \, \sin \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) + 2}{15 \, {\left (a^{8} \sin \left (d x + c\right )^{5} + 5 \, a^{8} \sin \left (d x + c\right )^{4} + 10 \, a^{8} \sin \left (d x + c\right )^{3} + 10 \, a^{8} \sin \left (d x + c\right )^{2} + 5 \, a^{8} \sin \left (d x + c\right ) + a^{8}\right )} d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (61) = 122\).
Time = 0.38 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.11 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 170 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 282 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 170 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{10}} \]
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Time = 5.93 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^5(c+d x)}{(a+a \sin (c+d x))^8} \, dx=\frac {1}{a^8\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^4}-\frac {1}{3\,a^8\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3}-\frac {4}{5\,a^8\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^5} \]
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