Integrand size = 15, antiderivative size = 33 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))} \]
[Out]
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3560, 8} \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))} \]
[In]
[Out]
Rule 8
Rule 3560
Rubi steps \begin{align*} \text {integral}& = \frac {i}{2 d (a+i a \tan (c+d x))}+\frac {\int 1 \, dx}{2 a} \\ & = \frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\frac {\frac {\arctan (\tan (c+d x))}{a}+\frac {1}{-i a+a \tan (c+d x)}}{2 d} \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x}{2 a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a d}\) | \(26\) |
derivativedivides | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {1}{2 d a \left (\tan \left (d x +c \right )-i\right )}\) | \(36\) |
default | \(\frac {\arctan \left (\tan \left (d x +c \right )\right )}{2 d a}+\frac {1}{2 d a \left (\tan \left (d x +c \right )-i\right )}\) | \(36\) |
norman | \(\frac {\frac {x}{2 a}+\frac {i}{2 a d}+\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}+\frac {\tan \left (d x +c \right )}{2 a d}}{1+\tan ^{2}\left (d x +c \right )}\) | \(58\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\frac {{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.82 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\begin {cases} \frac {i e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (\frac {\left (e^{2 i c} + 1\right ) e^{- 2 i c}}{2 a} - \frac {1}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x}{2 a} \]
[In]
[Out]
Exception generated. \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\text {Exception raised: RuntimeError} \]
[In]
[Out]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (25) = 50\).
Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=-\frac {-\frac {i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {-i \, \tan \left (d x + c\right ) - 3}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]
[In]
[Out]
Time = 4.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {1}{a+i a \tan (c+d x)} \, dx=\frac {x}{2\,a}+\frac {1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
[In]
[Out]