Integrand size = 24, antiderivative size = 16 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \log (i-a x)}{a} \]
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Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5181, 31} \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \log (-a x+i)}{a} \]
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Rule 31
Rule 5181
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{1+i a x} \, dx \\ & = -\frac {i \log (i-a x)}{a} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \log (i-a x)}{a} \]
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Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-\frac {i \ln \left (a x -i\right )}{a}\) | \(14\) |
default | \(-\frac {i \ln \left (i a x +1\right )}{a}\) | \(15\) |
meijerg | \(-\frac {i \ln \left (i a x +1\right )}{a}\) | \(15\) |
risch | \(-\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\) | \(26\) |
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none
Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \, \log \left (\frac {a x - i}{a}\right )}{a} \]
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Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=- \frac {i \log {\left (a x - i \right )}}{a} \]
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none
Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \, \log \left (i \, a x + 1\right )}{a} \]
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none
Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \, \log \left (i \, a x + 1\right )}{a} \]
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Time = 0.57 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )\,1{}\mathrm {i}}{a} \]
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