Integrand size = 24, antiderivative size = 30 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{a (i+a x)}-\frac {i \log (i+a x)}{a} \]
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Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5181, 45} \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{a (a x+i)}-\frac {i \log (a x+i)}{a} \]
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Rule 45
Rule 5181
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+i a x}{(1-i a x)^2} \, dx \\ & = \int \left (-\frac {2}{(i+a x)^2}-\frac {i}{i+a x}\right ) \, dx \\ & = \frac {2}{a (i+a x)}-\frac {i \log (i+a x)}{a} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{a (i+a x)}-\frac {i \log (i+a x)}{a} \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {2}{a \left (a x +i\right )}-\frac {i \ln \left (a x +i\right )}{a}\) | \(28\) |
risch | \(\frac {2}{a \left (a x +i\right )}-\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}-\frac {\arctan \left (a x \right )}{a}\) | \(40\) |
parallelrisch | \(-\frac {i \ln \left (a x +i\right ) x^{2} a^{2}-2 i x^{2} a^{2}+i \ln \left (a x +i\right )-2 a x}{\left (a^{2} x^{2}+1\right ) a}\) | \(57\) |
meijerg | \(\frac {\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}}{2 \sqrt {a^{2}}}+\frac {3 i a \,x^{2}}{2 \left (a^{2} x^{2}+1\right )}-\frac {3 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{2 \sqrt {a^{2}}}-\frac {i \left (-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )\right )}{2 a}\) | \(140\) |
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none
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {{\left (-i \, a x + 1\right )} \log \left (\frac {a x + i}{a}\right ) + 2}{a^{2} x + i \, a} \]
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Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{a^{2} x + i a} - \frac {i \log {\left (a x + i \right )}}{a} \]
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none
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2 \, {\left (a x - i\right )}}{a^{3} x^{2} + a} - \frac {\arctan \left (a x\right )}{a} - \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]
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none
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {i \, \log \left (a x + i\right )}{a} + \frac {2}{{\left (a x + i\right )} a} \]
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Time = 0.58 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{3 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {2}{x\,a^2+a\,1{}\mathrm {i}}-\frac {\ln \left (a\,x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \]
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