Integrand size = 31, antiderivative size = 125 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {5 x}{32 a^3}-\frac {1}{32 a^3 d (i-\cot (c+d x))}+\frac {i}{16 a^3 d (i+\cot (c+d x))^4}-\frac {1}{3 a^3 d (i+\cot (c+d x))^3}-\frac {23 i}{32 a^3 d (i+\cot (c+d x))^2}+\frac {13}{16 a^3 d (i+\cot (c+d x))} \]
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Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3167, 862, 90, 209} \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {1}{32 a^3 d (-\cot (c+d x)+i)}+\frac {13}{16 a^3 d (\cot (c+d x)+i)}-\frac {23 i}{32 a^3 d (\cot (c+d x)+i)^2}-\frac {1}{3 a^3 d (\cot (c+d x)+i)^3}+\frac {i}{16 a^3 d (\cot (c+d x)+i)^4}+\frac {5 x}{32 a^3} \]
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Rule 90
Rule 209
Rule 862
Rule 3167
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^5}{(i a+a x)^3 \left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^5}{\left (-\frac {i}{a}+\frac {x}{a}\right )^2 (i a+a x)^5} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{32 a^3 (-i+x)^2}+\frac {i}{4 a^3 (i+x)^5}-\frac {1}{a^3 (i+x)^4}-\frac {23 i}{16 a^3 (i+x)^3}+\frac {13}{16 a^3 (i+x)^2}+\frac {5}{32 a^3 \left (1+x^2\right )}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {1}{32 a^3 d (i-\cot (c+d x))}+\frac {i}{16 a^3 d (i+\cot (c+d x))^4}-\frac {1}{3 a^3 d (i+\cot (c+d x))^3}-\frac {23 i}{32 a^3 d (i+\cot (c+d x))^2}+\frac {13}{16 a^3 d (i+\cot (c+d x))}-\frac {5 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{32 a^3 d} \\ & = \frac {5 x}{32 a^3}-\frac {1}{32 a^3 d (i-\cot (c+d x))}+\frac {i}{16 a^3 d (i+\cot (c+d x))^4}-\frac {1}{3 a^3 d (i+\cot (c+d x))^3}-\frac {23 i}{32 a^3 d (i+\cot (c+d x))^2}+\frac {13}{16 a^3 d (i+\cot (c+d x))} \\ \end{align*}
Time = 0.97 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {120 c+120 d x+108 i \cos (2 (c+d x))+60 i \cos (4 (c+d x))+20 i \cos (6 (c+d x))+3 i \cos (8 (c+d x))+132 \sin (2 (c+d x))+60 \sin (4 (c+d x))+20 \sin (6 (c+d x))+3 \sin (8 (c+d x))}{768 a^3 d} \]
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Time = 0.96 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {5 x}{32 a^{3}}+\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 d \,a^{3}}+\frac {5 i {\mathrm e}^{-6 i \left (d x +c \right )}}{192 d \,a^{3}}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{256 d \,a^{3}}+\frac {9 i \cos \left (2 d x +2 c \right )}{64 d \,a^{3}}+\frac {11 \sin \left (2 d x +2 c \right )}{64 d \,a^{3}}\) | \(97\) |
derivativedivides | \(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{64}+\frac {1}{32 \tan \left (d x +c \right )+32 i}-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{64}+\frac {i}{16 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}}{d \,a^{3}}\) | \(102\) |
default | \(\frac {\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{64}+\frac {1}{32 \tan \left (d x +c \right )+32 i}-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{64}+\frac {i}{16 \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {3 i}{32 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}}{d \,a^{3}}\) | \(102\) |
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Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.61 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {{\left (120 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 12 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 60 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 20 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{768 \, a^{3} d} \]
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Time = 0.23 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 100663296 i a^{12} d^{4} e^{22 i c} e^{2 i d x} + 1006632960 i a^{12} d^{4} e^{18 i c} e^{- 2 i d x} + 503316480 i a^{12} d^{4} e^{16 i c} e^{- 4 i d x} + 167772160 i a^{12} d^{4} e^{14 i c} e^{- 6 i d x} + 25165824 i a^{12} d^{4} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{6442450944 a^{15} d^{5}} & \text {for}\: a^{15} d^{5} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 8 i c}}{32 a^{3}} - \frac {5}{32 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{32 a^{3}} \]
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Exception generated. \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.35 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {-\frac {60 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {60 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {12 \, {\left (5 \, \tan \left (d x + c\right ) + 7 i\right )}}{a^{3} {\left (i \, \tan \left (d x + c\right ) - 1\right )}} + \frac {-125 i \, \tan \left (d x + c\right )^{4} - 596 \, \tan \left (d x + c\right )^{3} + 1110 i \, \tan \left (d x + c\right )^{2} + 996 \, \tan \left (d x + c\right ) - 405 i}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{768 \, d} \]
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Time = 27.76 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {5\,x}{32\,a^3}+\frac {-\frac {27\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{16}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,33{}\mathrm {i}}{8}+\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,9{}\mathrm {i}}{8}+\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,9{}\mathrm {i}}{8}+\frac {31\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,33{}\mathrm {i}}{8}-\frac {27\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^8} \]
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