Integrand size = 14, antiderivative size = 39 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3}-\frac {2 i \log (i+a x)}{a^3} \]
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Time = 0.02 (sec) , antiderivative size = 39, 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} \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {2 i \log (a x+i)}{a^3}+\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (1+i a x)}{1-i a x} \, dx \\ & = \int \left (\frac {2}{a^2}+\frac {2 i x}{a}-x^2-\frac {2 i}{a^2 (i+a x)}\right ) \, dx \\ & = \frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3}-\frac {2 i \log (i+a x)}{a^3} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=\frac {2 x}{a^2}+\frac {i x^2}{a}-\frac {x^3}{3}-\frac {2 i \log (i+a x)}{a^3} \]
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Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(-\frac {a^{3} x^{3}-3 i a^{2} x^{2}+6 i \ln \left (a x +i\right )-6 a x}{3 a^{3}}\) | \(37\) |
risch | \(\frac {2 x}{a^{2}}-\frac {x^{3}}{3}+\frac {i x^{2}}{a}-\frac {i \ln \left (a^{2} x^{2}+1\right )}{a^{3}}-\frac {2 \arctan \left (a x \right )}{a^{3}}\) | \(47\) |
default | \(\frac {2 x -\frac {1}{3} a^{2} x^{3}+i a \,x^{2}}{a^{2}}+\frac {-\frac {i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {2 \arctan \left (a x \right )}{a}}{a^{2}}\) | \(55\) |
meijerg | \(\frac {\frac {2 x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2}}-\frac {2 \left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}}{2 a^{2} \sqrt {a^{2}}}+\frac {i \left (a^{2} x^{2}-\ln \left (a^{2} x^{2}+1\right )\right )}{a^{3}}-\frac {-\frac {2 x \left (a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{15 a^{4}}+\frac {2 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}}{2 a^{2} \sqrt {a^{2}}}\) | \(110\) |
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 6 \, a x + 6 i \, \log \left (\frac {a x + i}{a}\right )}{3 \, a^{3}} \]
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Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=- \frac {x^{3}}{3} + \frac {i x^{2}}{a} + \frac {2 x}{a^{2}} - \frac {2 i \log {\left (a x + i \right )}}{a^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{2} x^{3} - 3 i \, a x^{2} - 6 \, x}{3 \, a^{2}} - \frac {2 \, \arctan \left (a x\right )}{a^{3}} - \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{a^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=-\frac {a^{3} x^{3} - 3 i \, a^{2} x^{2} - 6 \, a x}{3 \, a^{3}} - \frac {2 i \, \log \left (a x + i\right )}{a^{3}} \]
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Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int e^{2 i \arctan (a x)} x^2 \, dx=\frac {2\,x}{a^2}-\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,2{}\mathrm {i}}{a^3}-\frac {x^3}{3}+\frac {x^2\,1{}\mathrm {i}}{a} \]
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