Integrand size = 12, antiderivative size = 45 \[ \int e^{4 i \arctan (a x)} x \, dx=-\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2} \]
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Time = 0.02 (sec) , antiderivative size = 45, 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} \[ \int e^{4 i \arctan (a x)} x \, dx=-\frac {4 i}{a^2 (a x+i)}-\frac {8 \log (a x+i)}{a^2}-\frac {4 i x}{a}+\frac {x^2}{2} \]
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Rule 78
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x (1+i a x)^2}{(1-i a x)^2} \, dx \\ & = \int \left (-\frac {4 i}{a}+x+\frac {4 i}{a (i+a x)^2}-\frac {8}{a (i+a x)}\right ) \, dx \\ & = -\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int e^{4 i \arctan (a x)} x \, dx=-\frac {4 i x}{a}+\frac {x^2}{2}-\frac {4 i}{a^2 (i+a x)}-\frac {8 \log (i+a x)}{a^2} \]
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Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.11
method | result | size |
default | \(-\frac {-\frac {1}{2} a \,x^{2}+4 i x}{a}+\frac {-\frac {4 i}{a \left (a x +i\right )}-\frac {8 \ln \left (a x +i\right )}{a}}{a}\) | \(50\) |
risch | \(\frac {x^{2}}{2}-\frac {4 i x}{a}-\frac {4 i}{a^{2} \left (a x +i\right )}-\frac {4 \ln \left (a^{2} x^{2}+1\right )}{a^{2}}+\frac {8 i \arctan \left (a x \right )}{a^{2}}\) | \(53\) |
parallelrisch | \(-\frac {-a^{4} x^{4}+8 i a^{3} x^{3}+16 a^{2} \ln \left (a x +i\right ) x^{2}-9 a^{2} x^{2}+16 i a x +16 \ln \left (a x +i\right )}{2 a^{2} \left (a^{2} x^{2}+1\right )}\) | \(72\) |
meijerg | \(\frac {x^{2}}{2 a^{2} x^{2}+2}+\frac {2 i \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{a \sqrt {a^{2}}}-\frac {3 \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{a^{2}}-\frac {2 i \left (\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a \sqrt {a^{2}}}+\frac {\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{3 a^{2} x^{2}+3}-2 \ln \left (a^{2} x^{2}+1\right )}{2 a^{2}}\) | \(205\) |
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Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int e^{4 i \arctan (a x)} x \, dx=\frac {a^{3} x^{3} - 7 i \, a^{2} x^{2} + 8 \, a x - 16 \, {\left (a x + i\right )} \log \left (\frac {a x + i}{a}\right ) - 8 i}{2 \, {\left (a^{3} x + i \, a^{2}\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80 \[ \int e^{4 i \arctan (a x)} x \, dx=\frac {x^{2}}{2} - \frac {4 i}{a^{3} x + i a^{2}} - \frac {4 i x}{a} - \frac {8 \log {\left (a x + i \right )}}{a^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int e^{4 i \arctan (a x)} x \, dx=-\frac {4 \, {\left (i \, a x + 1\right )}}{a^{4} x^{2} + a^{2}} + \frac {a x^{2} - 8 i \, x}{2 \, a} + \frac {8 i \, \arctan \left (a x\right )}{a^{2}} - \frac {4 \, \log \left (a^{2} x^{2} + 1\right )}{a^{2}} \]
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Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int e^{4 i \arctan (a x)} x \, dx=-\frac {8 \, \log \left (a x + i\right )}{a^{2}} + \frac {a^{4} x^{2} - 8 i \, a^{3} x}{2 \, a^{4}} - \frac {4 i}{{\left (a x + i\right )} a^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int e^{4 i \arctan (a x)} x \, dx=\frac {x^2}{2}-\frac {8\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^2}-\frac {4{}\mathrm {i}}{a^3\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {x\,4{}\mathrm {i}}{a} \]
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