Integrand size = 14, antiderivative size = 37 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \]
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Time = 0.02 (sec) , antiderivative size = 37, 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} \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-2 a^2 \log (x)+2 a^2 \log (-a x+i)+\frac {2 i a}{x}-\frac {1}{2 x^2} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {1-i a x}{x^3 (1+i a x)} \, dx \\ & = \int \left (\frac {1}{x^3}-\frac {2 i a}{x^2}-\frac {2 a^2}{x}+\frac {2 a^3}{-i+a x}\right ) \, dx \\ & = -\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}+\frac {2 i a}{x}-2 a^2 \log (x)+2 a^2 \log (i-a x) \]
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92
method | result | size |
default | \(-\frac {1}{2 x^{2}}+\frac {2 i a}{x}-2 a^{2} \ln \left (x \right )+2 a^{2} \ln \left (-a x +i\right )\) | \(34\) |
risch | \(\frac {2 i a x -\frac {1}{2}}{x^{2}}-2 a^{2} \ln \left (-x \right )+2 i a^{2} \arctan \left (a x \right )+a^{2} \ln \left (a^{2} x^{2}+1\right )\) | \(46\) |
parallelrisch | \(-\frac {-4 \ln \left (x \right ) x^{3} a^{3}+4 \ln \left (a x -i\right ) x^{3} a^{3}+4 i \ln \left (x \right ) x^{2} a^{2}-4 i \ln \left (a x -i\right ) x^{2} a^{2}+i+4 a^{3} x^{3}+3 a x}{2 \left (-a x +i\right ) x^{2}}\) | \(82\) |
meijerg | \(a^{2} \left (-\frac {2 i a x}{2 i a x +2}-\ln \left (i a x +1\right )+1+\ln \left (x \right )+\ln \left (i a \right )\right )-a^{2} \left (-\frac {4 i a x}{4 i a x +4}-3 \ln \left (i a x +1\right )+1+3 \ln \left (x \right )+3 \ln \left (i a \right )+\frac {1}{2 a^{2} x^{2}}-\frac {2 i}{x a}\right )\) | \(98\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-\frac {4 \, a^{2} x^{2} \log \left (x\right ) - 4 \, a^{2} x^{2} \log \left (\frac {a x - i}{a}\right ) - 4 i \, a x + 1}{2 \, x^{2}} \]
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Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=- 2 a^{2} \left (\log {\left (4 a^{3} x \right )} - \log {\left (4 a^{3} x - 4 i a^{2} \right )}\right ) - \frac {- 4 i a x + 1}{2 x^{2}} \]
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Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.35 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=2 \, a^{2} \log \left (i \, a x + 1\right ) - 2 \, a^{2} \log \left (x\right ) - \frac {4 \, a^{2} x^{2} - 3 i \, a x + 1}{2 i \, a x^{3} + 2 \, x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=-2 \, a^{2} \log \left (\frac {i}{i \, a x + 1} - i\right ) + \frac {5 \, a^{2} - \frac {6 \, a^{2}}{i \, a x + 1}}{2 \, {\left (\frac {i}{i \, a x + 1} - i\right )}^{2}} \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-2 i \arctan (a x)}}{x^3} \, dx=a^2\,\mathrm {atan}\left (2\,a\,x-\mathrm {i}\right )\,4{}\mathrm {i}+\frac {-\frac {1}{2}+a\,x\,2{}\mathrm {i}}{x^2} \]
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