Integrand size = 24, antiderivative size = 83 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-12 a^5 x+\frac {12 i a^5 \log (\cos (c+d x))}{d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {i a^5 \tan ^2(c+d x)}{2 d}-\frac {8 i a^6}{d (a-i a \tan (c+d x))} \]
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Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {8 i a^6}{d (a-i a \tan (c+d x))}+\frac {i a^5 \tan ^2(c+d x)}{2 d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {12 i a^5 \log (\cos (c+d x))}{d}-12 a^5 x \]
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Rule 45
Rule 3568
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(a+x)^3}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \left (5 a+\frac {8 a^3}{(a-x)^2}-\frac {12 a^2}{a-x}+x\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -12 a^5 x+\frac {12 i a^5 \log (\cos (c+d x))}{d}+\frac {5 a^5 \tan (c+d x)}{d}+\frac {i a^5 \tan ^2(c+d x)}{2 d}-\frac {8 i a^6}{d (a-i a \tan (c+d x))} \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i a^5 \left (24 \log (i+\tan (c+d x))+10 i \tan (c+d x)-\tan ^2(c+d x)+\frac {16 i}{i+\tan (c+d x)}\right )}{2 d} \]
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Time = 14.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {4 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}}{d}+\frac {24 a^{5} c}{d}+\frac {2 i a^{5} \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+5\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {12 i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(85\) |
derivativedivides | \(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-10 i a^{5} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {5 i a^{5} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(205\) |
default | \(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-10 i a^{5} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {5 i a^{5} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{5} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(205\) |
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Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.51 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 \, {\left (2 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 i \, a^{5} + 6 \, {\left (-i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]
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Time = 0.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.58 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {12 i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {12 i a^{5} e^{2 i c} e^{2 i d x} + 10 i a^{5}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} + \begin {cases} - \frac {4 i a^{5} e^{2 i c} e^{2 i d x}}{d} & \text {for}\: d \neq 0 \\8 a^{5} x e^{2 i c} & \text {otherwise} \end {cases} \]
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Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-i \, a^{5} \tan \left (d x + c\right )^{2} + 24 \, {\left (d x + c\right )} a^{5} + 12 i \, a^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 10 \, a^{5} \tan \left (d x + c\right ) - \frac {16 \, {\left (a^{5} \tan \left (d x + c\right ) - i \, a^{5}\right )}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]
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Time = 0.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.76 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 \, {\left (-6 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 12 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 5 i \, a^{5}\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]
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Time = 3.72 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {8\,a^5}{d\,\left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}-\frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,12{}\mathrm {i}}{d}+\frac {5\,a^5\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,d} \]
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