Integrand size = 14, antiderivative size = 26 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \]
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Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5170, 78} \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=2 i a \log (x)-2 i a \log (a x+i)-\frac {1}{x} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+i a x}{x^2 (1-i a x)} \, dx \\ & = \int \left (\frac {1}{x^2}+\frac {2 i a}{x}-\frac {2 i a^2}{i+a x}\right ) \, dx \\ & = -\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}+2 i a \log (x)-2 i a \log (i+a x) \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {2 i a \ln \left (x \right ) x -2 i a \ln \left (a x +i\right ) x -1}{x}\) | \(26\) |
risch | \(-\frac {1}{x}-2 a \arctan \left (a x \right )-i a \ln \left (a^{2} x^{2}+1\right )+2 i a \ln \left (x \right )\) | \(34\) |
default | \(-\frac {1}{x}+2 i a \ln \left (x \right )-2 a^{2} \left (\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\right )\) | \(43\) |
meijerg | \(\frac {a^{2} \left (-\frac {2}{x \sqrt {a^{2}}}-\frac {2 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{2 \sqrt {a^{2}}}+i a \left (-\ln \left (a^{2} x^{2}+1\right )+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-a \arctan \left (a x \right )\) | \(67\) |
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none
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=\frac {2 i \, a x \log \left (x\right ) - 2 i \, a x \log \left (\frac {a x + i}{a}\right ) - 1}{x} \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=- 2 a \left (- i \log {\left (4 a^{2} x \right )} + i \log {\left (4 a^{2} x + 4 i a \right )}\right ) - \frac {1}{x} \]
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-2 \, a \arctan \left (a x\right ) - i \, a \log \left (a^{2} x^{2} + 1\right ) + 2 i \, a \log \left (x\right ) - \frac {1}{x} \]
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Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-2 i \, a \log \left (a x + i\right ) + 2 i \, a \log \left ({\left | x \right |}\right ) - \frac {1}{x} \]
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Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^{2 i \arctan (a x)}}{x^2} \, dx=-4\,a\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )-\frac {1}{x} \]
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