Integrand size = 31, antiderivative size = 131 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )} \]
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Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3161, 8} \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}+\frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2} \]
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Rule 8
Rule 3161
Rubi steps \begin{align*} \text {integral}& = \frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx}{2 a} \\ & = \frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {\int \frac {\cos (c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx}{4 a^2} \\ & = \frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3} \\ & = \frac {x}{8 a^3}+\frac {i \cos ^3(c+d x)}{6 d (a \cos (c+d x)+i a \sin (c+d x))^3}+\frac {i \cos ^2(c+d x)}{8 a d (a \cos (c+d x)+i a \sin (c+d x))^2}+\frac {i \cos (c+d x)}{8 d \left (a^3 \cos (c+d x)+i a^3 \sin (c+d x)\right )} \\ \end{align*}
Time = 0.80 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {12 c+12 d x+18 i \cos (2 (c+d x))+9 i \cos (4 (c+d x))+2 i \cos (6 (c+d x))+18 \sin (2 (c+d x))+9 \sin (4 (c+d x))+2 \sin (6 (c+d x))}{96 a^3 d} \]
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Time = 0.69 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.47
method | result | size |
risch | \(\frac {x}{8 a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 d \,a^{3}}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 d \,a^{3}}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 d \,a^{3}}\) | \(62\) |
derivativedivides | \(\frac {\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}}{d \,a^{3}}\) | \(75\) |
default | \(\frac {\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}}{d \,a^{3}}\) | \(75\) |
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Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.41 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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Time = 0.17 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {\left (4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{8 a^{3}} \]
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Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {-11 i \, \tan \left (d x + c\right )^{3} - 45 \, \tan \left (d x + c\right )^{2} + 69 i \, \tan \left (d x + c\right ) + 51}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
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Time = 25.84 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {x}{8\,a^3}+\frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,9{}\mathrm {i}}{2}-\frac {41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,9{}\mathrm {i}}{2}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a^3\,d\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^6} \]
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