Integrand size = 22, antiderivative size = 71 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \]
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Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3640, 3677, 3612, 3556} \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {\log (\sin (c+d x))}{a^2 d}-\frac {3 i x}{4 a^2}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \]
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Rule 3556
Rule 3612
Rule 3640
Rule 3677
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot (c+d x) (4 a-2 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = \frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \left (8 a^2-6 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4} \\ & = -\frac {3 i x}{4 a^2}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \cot (c+d x) \, dx}{a^2} \\ & = -\frac {3 i x}{4 a^2}+\frac {\log (\sin (c+d x))}{a^2 d}+\frac {3}{4 a^2 d (1+i \tan (c+d x))}+\frac {1}{4 d (a+i a \tan (c+d x))^2} \\ \end{align*}
Time = 0.87 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {2}{(a+i a \tan (c+d x))^2}-\frac {7 \log (i-\tan (c+d x))-8 \log (\tan (c+d x))+\log (i+\tan (c+d x))+\frac {6 i}{-i+\tan (c+d x)}}{a^2}}{8 d} \]
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Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {7 i x}{4 a^{2}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{2 a^{2} d}+\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}-\frac {2 i c}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\) | \(72\) |
derivativedivides | \(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(90\) |
default | \(-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}-\frac {3 i \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {3 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(90\) |
norman | \(\frac {\frac {1}{a d}-\frac {3 i x}{4 a}-\frac {5 i \tan \left (d x +c \right )}{4 d a}-\frac {3 i \left (\tan ^{3}\left (d x +c \right )\right )}{4 d a}-\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{4 a}+\frac {\tan ^{2}\left (d x +c \right )}{2 a d}}{a \left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right )}{a^{2} d}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{2}}\) | \(144\) |
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none
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]
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Time = 0.22 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (- 7 i e^{4 i c} - 4 i e^{2 i c} - i\right ) e^{- 4 i c}}{4 a^{2}} + \frac {7 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} - \frac {7 i x}{4 a^{2}} + \frac {\log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{2} d} \]
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Exception generated. \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {16 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac {21 \, \tan \left (d x + c\right )^{2} - 54 i \, \tan \left (d x + c\right ) - 37}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
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Time = 4.86 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {7\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {1{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]
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