Integrand size = 14, antiderivative size = 49 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \]
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Time = 0.03 (sec) , antiderivative size = 49, 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} \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {2 \log (-a x+i)}{a^4}+\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 (1-i a x)}{1+i a x} \, dx \\ & = \int \left (\frac {2 i}{a^3}+\frac {2 x}{a^2}-\frac {2 i x^2}{a}-x^3-\frac {2}{a^3 (-i+a x)}\right ) \, dx \\ & = \frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {2 i x}{a^3}+\frac {x^2}{a^2}-\frac {2 i x^3}{3 a}-\frac {x^4}{4}-\frac {2 \log (i-a x)}{a^4} \]
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Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {\frac {1}{4} a^{3} x^{4}+\frac {2}{3} i a^{2} x^{3}-a \,x^{2}-2 i x}{a^{3}}-\frac {2 \ln \left (-a x +i\right )}{a^{4}}\) | \(48\) |
risch | \(-\frac {x^{4}}{4}-\frac {2 i x^{3}}{3 a}+\frac {x^{2}}{a^{2}}+\frac {2 i x}{a^{3}}-\frac {\ln \left (a^{2} x^{2}+1\right )}{a^{4}}-\frac {2 i \arctan \left (a x \right )}{a^{4}}\) | \(55\) |
parallelrisch | \(-\frac {-3 a^{5} x^{5}-5 i a^{4} x^{4}+4 a^{3} x^{3}+24 i+12 i a^{2} x^{2}-24 \ln \left (a x -i\right ) x a +24 i \ln \left (a x -i\right )}{12 a^{4} \left (-a x +i\right )}\) | \(73\) |
meijerg | \(-\frac {-\frac {i x a \left (-3 a^{4} x^{4}-5 i a^{3} x^{3}+10 a^{2} x^{2}+30 i a x +60\right )}{12 \left (i a x +1\right )}+5 \ln \left (i a x +1\right )}{a^{4}}+\frac {-\frac {i a x \left (2 a^{2} x^{2}+6 i a x +12\right )}{4 \left (i a x +1\right )}+3 \ln \left (i a x +1\right )}{a^{4}}\) | \(108\) |
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Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {3 \, a^{4} x^{4} + 8 i \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 24 i \, a x + 24 \, \log \left (\frac {a x - i}{a}\right )}{12 \, a^{4}} \]
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Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=- \frac {x^{4}}{4} - \frac {2 i x^{3}}{3 a} + \frac {x^{2}}{a^{2}} + \frac {2 i x}{a^{3}} - \frac {2 \log {\left (a x - i \right )}}{a^{4}} \]
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Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=-\frac {i \, {\left (-3 i \, a^{3} x^{4} + 8 \, a^{2} x^{3} + 12 i \, a x^{2} - 24 \, x\right )}}{12 \, a^{3}} - \frac {2 \, \log \left (i \, a x + 1\right )}{a^{4}} \]
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Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.39 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {{\left (i \, a x + 1\right )}^{4} {\left (\frac {20}{i \, a x + 1} - \frac {54}{{\left (i \, a x + 1\right )}^{2}} + \frac {84}{{\left (i \, a x + 1\right )}^{3}} - 3\right )}}{12 \, a^{4}} + \frac {2 \, \log \left (\frac {1}{\sqrt {a^{2} x^{2} + 1} {\left | a \right |}}\right )}{a^{4}} \]
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Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int e^{-2 i \arctan (a x)} x^3 \, dx=\frac {x^2}{a^2}-\frac {x^4}{4}-\frac {2\,\ln \left (x-\frac {1{}\mathrm {i}}{a}\right )}{a^4}+\frac {x\,2{}\mathrm {i}}{a^3}-\frac {x^3\,2{}\mathrm {i}}{3\,a} \]
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