Integrand size = 26, antiderivative size = 80 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {(A-i B) x}{4 a^2}+\frac {i A-B}{4 d (a+i a \tan (c+d x))^2}+\frac {i A+B}{4 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Time = 0.07 (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} \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {B+i A}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {x (A-i B)}{4 a^2}+\frac {-B+i A}{4 d (a+i a \tan (c+d x))^2} \]
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Rule 8
Rule 3560
Rule 3607
Rubi steps \begin{align*} \text {integral}& = \frac {i A-B}{4 d (a+i a \tan (c+d x))^2}+\frac {(A-i B) \int \frac {1}{a+i a \tan (c+d x)} \, dx}{2 a} \\ & = \frac {i A-B}{4 d (a+i a \tan (c+d x))^2}+\frac {i A+B}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {(A-i B) \int 1 \, dx}{4 a^2} \\ & = \frac {(A-i B) x}{4 a^2}+\frac {i A-B}{4 d (a+i a \tan (c+d x))^2}+\frac {i A+B}{4 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {(A-i B) \arctan (\tan (c+d x))}{4 a^2 d}-\frac {i A-B}{4 a^2 d (i-\tan (c+d x))^2}-\frac {A-i B}{4 a^2 d (i-\tan (c+d x))} \]
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Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {i x B}{4 a^{2}}+\frac {x A}{4 a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{4 a^{2} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )} B}{16 a^{2} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )} A}{16 a^{2} d}\) | \(73\) |
derivativedivides | \(\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(117\) |
default | \(\frac {A \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {i B \arctan \left (\tan \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}-\frac {i A}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {B}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}\) | \(117\) |
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none
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\left (4 \, {\left (A - i \, B\right )} d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, A e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]
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Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.02 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\begin {cases} \frac {\left (16 i A a^{2} d e^{4 i c} e^{- 2 i d x} + \left (4 i A a^{2} d e^{2 i c} - 4 B a^{2} d e^{2 i c}\right ) e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac {A - i B}{4 a^{2}} + \frac {\left (A e^{4 i c} + 2 A e^{2 i c} + A - i B e^{4 i c} + i B\right ) e^{- 4 i c}}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (A - i B\right )}{4 a^{2}} \]
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Exception generated. \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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none
Time = 0.40 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac {2 \, {\left (-i \, A - B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {3 i \, A \tan \left (d x + c\right )^{2} + 3 \, B \tan \left (d x + c\right )^{2} + 10 \, A \tan \left (d x + c\right ) - 10 i \, B \tan \left (d x + c\right ) - 11 i \, A - 3 \, B}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
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Time = 7.44 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {\frac {A}{2\,a^2}+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B}{4\,a^2}+\frac {A\,1{}\mathrm {i}}{4\,a^2}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}-\frac {x\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^2} \]
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