Integrand size = 24, antiderivative size = 49 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-a^3 x+\frac {i a^3 \log (\cos (c+d x))}{d}-\frac {2 i a^4}{d (a-i a \tan (c+d x))} \]
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Time = 0.06 (sec) , antiderivative size = 49, 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} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 i a^4}{d (a-i a \tan (c+d x))}+\frac {i a^3 \log (\cos (c+d x))}{d}-a^3 x \]
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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}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \left (\frac {2 a}{(a-x)^2}+\frac {1}{-a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -a^3 x+\frac {i a^3 \log (\cos (c+d x))}{d}-\frac {2 i a^4}{d (a-i a \tan (c+d x))} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.82 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i a^3 \left (\log (i+\tan (c+d x))+\frac {2 a}{a-i a \tan (c+d x)}\right )}{d} \]
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Time = 2.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02
method | result | size |
risch | \(-\frac {i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{d}+\frac {2 a^{3} c}{d}+\frac {i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(50\) |
derivativedivides | \(\frac {-i a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {3 i a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(99\) |
default | \(\frac {-i a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {3 i a^{3} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(99\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {-i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d} \]
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Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {i a^{3} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \begin {cases} - \frac {i a^{3} e^{2 i c} e^{2 i d x}}{d} & \text {for}\: d \neq 0 \\2 a^{3} x e^{2 i c} & \text {otherwise} \end {cases} \]
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Time = 0.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.27 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (d x + c\right )} a^{3} + i \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {4 \, {\left (a^{3} \tan \left (d x + c\right ) - i \, a^{3}\right )}}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]
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Time = 0.61 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {-i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d} \]
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Time = 3.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2\,a^3}{d\,\left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}-\frac {a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{d} \]
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