Integrand size = 14, antiderivative size = 38 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}-\frac {4 a}{i+a x}+4 i a \log (x)-4 i a \log (i+a x) \]
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Time = 0.02 (sec) , antiderivative size = 38, 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, 90} \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-\frac {4 a}{a x+i}+4 i a \log (x)-4 i a \log (a x+i)-\frac {1}{x} \]
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Rule 90
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{x^2 (1-i a x)^2} \, dx \\ & = \int \left (\frac {1}{x^2}+\frac {4 i a}{x}+\frac {4 a^2}{(i+a x)^2}-\frac {4 i a^2}{i+a x}\right ) \, dx \\ & = -\frac {1}{x}-\frac {4 a}{i+a x}+4 i a \log (x)-4 i a \log (i+a x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-\frac {1}{x}-\frac {4 a}{i+a x}+4 i a \log (x)-4 i a \log (i+a x) \]
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Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {1}{x}+4 i a \ln \left (x \right )-4 a^{2} \left (\frac {1}{a \left (a x +i\right )}+\frac {i \ln \left (a x +i\right )}{a}\right )\) | \(44\) |
risch | \(\frac {-5 a x -i}{\left (a x +i\right ) x}+4 i a \ln \left (x \right )-4 a \arctan \left (a x \right )-2 i a \ln \left (a^{2} x^{2}+1\right )\) | \(48\) |
parallelrisch | \(\frac {4 i a^{3} \ln \left (x \right ) x^{3}-4 i a^{3} \ln \left (a x +i\right ) x^{3}-4 i a^{3} x^{3}-1+4 i a \ln \left (x \right ) x -4 i a \ln \left (a x +i\right ) x -5 a^{2} x^{2}}{\left (a^{2} x^{2}+1\right ) x}\) | \(81\) |
meijerg | \(\frac {a^{2} \left (-\frac {2 \left (3 a^{2} x^{2}+2\right )}{x \sqrt {a^{2}}\, \left (2 a^{2} x^{2}+2\right )}-\frac {3 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{2 \sqrt {a^{2}}}+2 i a \left (-\frac {2 a^{2} x^{2}}{2 a^{2} x^{2}+2}-\ln \left (a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-\frac {3 a^{2} \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{\sqrt {a^{2}}}-\frac {2 i a^{3} x^{2}}{a^{2} x^{2}+1}+\frac {a^{2} \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}}}{a^{2} \left (a^{2} x^{2}+1\right )}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{a^{3}}\right )}{2 \sqrt {a^{2}}}\) | \(213\) |
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Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.58 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-\frac {5 \, a x + 4 \, {\left (-i \, a^{2} x^{2} + a x\right )} \log \left (x\right ) + 4 \, {\left (i \, a^{2} x^{2} - a x\right )} \log \left (\frac {a x + i}{a}\right ) + i}{a x^{2} + i \, x} \]
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Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=4 a \left (i \log {\left (8 a^{2} x \right )} - i \log {\left (8 a^{2} x + 8 i a \right )}\right ) + \frac {- 5 a x - i}{a x^{2} + i x} \]
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Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-4 \, a \arctan \left (a x\right ) - 2 i \, a \log \left (a^{2} x^{2} + 1\right ) + 4 i \, a \log \left (x\right ) - \frac {5 \, a^{2} x^{2} - 4 i \, a x + 1}{a^{2} x^{3} + x} \]
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Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-4 i \, a \log \left (a x + i\right ) + 4 i \, a \log \left ({\left | x \right |}\right ) - \frac {5 \, a x + i}{a x^{2} + i \, x} \]
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Time = 0.50 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {e^{4 i \arctan (a x)}}{x^2} \, dx=-8\,a\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )-\frac {5\,x+\frac {1{}\mathrm {i}}{a}}{x^2+\frac {x\,1{}\mathrm {i}}{a}} \]
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