Integrand size = 26, antiderivative size = 80 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\frac {(c-i d) x}{4 a^2}+\frac {i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac {i c+d}{4 f \left (a^2+i a^2 \tan (e+f x)\right )} \]
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Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3607, 3560, 8} \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\frac {d+i c}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac {x (c-i d)}{4 a^2}+\frac {-d+i c}{4 f (a+i a \tan (e+f x))^2} \]
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Rule 8
Rule 3560
Rule 3607
Rubi steps \begin{align*} \text {integral}& = \frac {i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac {(c-i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{2 a} \\ & = \frac {i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac {i c+d}{4 f \left (a^2+i a^2 \tan (e+f x)\right )}+\frac {(c-i d) \int 1 \, dx}{4 a^2} \\ & = \frac {(c-i d) x}{4 a^2}+\frac {i c-d}{4 f (a+i a \tan (e+f x))^2}+\frac {i c+d}{4 f \left (a^2+i a^2 \tan (e+f x)\right )} \\ \end{align*}
Time = 0.86 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\frac {(c-i d) \arctan (\tan (e+f x))}{4 a^2 f}-\frac {i c-d}{4 a^2 f (i-\tan (e+f x))^2}-\frac {c-i d}{4 a^2 f (i-\tan (e+f x))} \]
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Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {i x d}{4 a^{2}}+\frac {x c}{4 a^{2}}+\frac {i c \,{\mathrm e}^{-2 i \left (f x +e \right )}}{4 a^{2} f}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} d}{16 f \,a^{2}}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} c}{16 f \,a^{2}}\) | \(73\) |
derivativedivides | \(-\frac {i d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f \,a^{2}}+\frac {c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}\) | \(117\) |
default | \(-\frac {i d \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f \,a^{2}}+\frac {c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}+\frac {d}{4 f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}\) | \(117\) |
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Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (4 \, {\left (c - i \, d\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 4 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]
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Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.02 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (16 i a^{2} c f e^{4 i e} e^{- 2 i f x} + \left (4 i a^{2} c f e^{2 i e} - 4 a^{2} d f e^{2 i e}\right ) e^{- 4 i f x}\right ) e^{- 6 i e}}{64 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {c - i d}{4 a^{2}} + \frac {\left (c e^{4 i e} + 2 c e^{2 i e} + c - i d e^{4 i e} + i d\right ) e^{- 4 i e}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (c - i d\right )}{4 a^{2}} \]
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Exception generated. \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.43 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {2 \, {\left (-i \, c - d\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2}} - \frac {2 \, {\left (-i \, c - d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2}} - \frac {3 i \, c \tan \left (f x + e\right )^{2} + 3 \, d \tan \left (f x + e\right )^{2} + 10 \, c \tan \left (f x + e\right ) - 10 i \, d \tan \left (f x + e\right ) - 11 i \, c - 3 \, d}{a^{2} {\left (\tan \left (f x + e\right ) - i\right )}^{2}}}{16 \, f} \]
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Time = 5.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {c+d \tan (e+f x)}{(a+i a \tan (e+f x))^2} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {d}{4\,a^2}+\frac {c\,1{}\mathrm {i}}{4\,a^2}\right )+\frac {c}{2\,a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {x\,\left (d+c\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^2} \]
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