Integrand size = 10, antiderivative size = 31 \[ \int e^{4 i \arctan (a x)} \, dx=x+\frac {4}{a (i+a x)}-\frac {4 i \log (i+a x)}{a} \]
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Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5169, 45} \[ \int e^{4 i \arctan (a x)} \, dx=\frac {4}{a (a x+i)}-\frac {4 i \log (a x+i)}{a}+x \]
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Rule 45
Rule 5169
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{(1-i a x)^2} \, dx \\ & = \int \left (1-\frac {4}{(i+a x)^2}-\frac {4 i}{i+a x}\right ) \, dx \\ & = x+\frac {4}{a (i+a x)}-\frac {4 i \log (i+a x)}{a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int e^{4 i \arctan (a x)} \, dx=x+\frac {4}{a (i+a x)}-\frac {4 \arctan (a x)}{a}-\frac {2 i \log \left (1+a^2 x^2\right )}{a} \]
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Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06
method | result | size |
default | \(x -4 a \left (-\frac {1}{a^{2} \left (a x +i\right )}+\frac {i \ln \left (a x +i\right )}{a^{2}}\right )\) | \(33\) |
risch | \(x +\frac {4}{a \left (a x +i\right )}-\frac {2 i \ln \left (a^{2} x^{2}+1\right )}{a}-\frac {4 \arctan \left (a x \right )}{a}\) | \(41\) |
parallelrisch | \(-\frac {4 i \ln \left (a x +i\right ) x^{2} a^{2}-a^{3} x^{3}-4 i a^{2} x^{2}+4 i \ln \left (a x +i\right )-5 a x}{\left (a^{2} x^{2}+1\right ) a}\) | \(65\) |
meijerg | \(\frac {\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}}{2 \sqrt {a^{2}}}+\frac {2 i a \,x^{2}}{a^{2} x^{2}+1}-\frac {3 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{\sqrt {a^{2}}}-\frac {2 i \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{a}+\frac {\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}}{2 \sqrt {a^{2}}}\) | \(194\) |
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Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int e^{4 i \arctan (a x)} \, dx=\frac {a^{2} x^{2} + i \, a x - 4 \, {\left (i \, a x - 1\right )} \log \left (\frac {a x + i}{a}\right ) + 4}{a^{2} x + i \, a} \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int e^{4 i \arctan (a x)} \, dx=x + \frac {4}{a^{2} x + i a} - \frac {4 i \log {\left (a x + i \right )}}{a} \]
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Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int e^{4 i \arctan (a x)} \, dx=x + \frac {4 \, {\left (a x - i\right )}}{a^{3} x^{2} + a} - \frac {4 \, \arctan \left (a x\right )}{a} - \frac {2 i \, \log \left (a^{2} x^{2} + 1\right )}{a} \]
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int e^{4 i \arctan (a x)} \, dx=x - \frac {4 i \, \log \left (a x + i\right )}{a} + \frac {4}{{\left (a x + i\right )} a} \]
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Time = 0.49 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int e^{4 i \arctan (a x)} \, dx=x+\frac {4}{a^2\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}-\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,4{}\mathrm {i}}{a} \]
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