Integrand size = 12, antiderivative size = 29 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i+a x)}{a^2} \]
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Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5170, 78} \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 \log (a x+i)}{a^2}+\frac {2 i x}{a}-\frac {x^2}{2} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1+i a x)}{1-i a x} \, dx \\ & = \int \left (\frac {2 i}{a}-x+\frac {2}{a (i+a x)}\right ) \, dx \\ & = \frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i+a x)}{a^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2 i x}{a}-\frac {x^2}{2}+\frac {2 \log (i+a x)}{a^2} \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {-a^{2} x^{2}+4 i a x +4 \ln \left (a x +i\right )}{2 a^{2}}\) | \(29\) |
risch | \(-\frac {x^{2}}{2}+\frac {2 i x}{a}+\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{2}}-\frac {2 i \arctan \left (a x \right )}{a^{2}}\) | \(38\) |
default | \(\frac {-\frac {1}{2} a \,x^{2}+2 i x}{a}+\frac {\frac {\ln \left (a^{2} x^{2}+1\right )}{a}-\frac {2 i \arctan \left (a x \right )}{a}}{a}\) | \(46\) |
meijerg | \(\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}+\frac {i \left (\frac {2 x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2}}-\frac {2 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{a \sqrt {a^{2}}}-\frac {a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}\) | \(79\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a^{2} x^{2} - 4 i \, a x - 4 \, \log \left (\frac {a x + i}{a}\right )}{2 \, a^{2}} \]
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Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int e^{2 i \arctan (a x)} x \, dx=- \frac {x^{2}}{2} + \frac {2 i x}{a} + \frac {2 \log {\left (a x + i \right )}}{a^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a x^{2} - 4 i \, x}{2 \, a} - \frac {2 i \, \arctan \left (a x\right )}{a^{2}} + \frac {\log \left (a^{2} x^{2} + 1\right )}{a^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x \, dx=-\frac {a^{2} x^{2} - 4 i \, a x}{2 \, a^{2}} + \frac {2 \, \log \left (a x + i\right )}{a^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int e^{2 i \arctan (a x)} x \, dx=\frac {2\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {x^2}{2}+\frac {x\,2{}\mathrm {i}}{a} \]
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