Integrand size = 14, antiderivative size = 52 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}-\frac {4 i a}{x}-\frac {4 i a^2}{i+a x}-8 a^2 \log (x)+8 a^2 \log (i+a x) \]
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Time = 0.02 (sec) , antiderivative size = 52, 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} \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=-\frac {4 i a^2}{a x+i}-8 a^2 \log (x)+8 a^2 \log (a x+i)-\frac {4 i a}{x}-\frac {1}{2 x^2} \]
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Rule 90
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{x^3 (1-i a x)^2} \, dx \\ & = \int \left (\frac {1}{x^3}+\frac {4 i a}{x^2}-\frac {8 a^2}{x}+\frac {4 i a^3}{(i+a x)^2}+\frac {8 a^3}{i+a x}\right ) \, dx \\ & = -\frac {1}{2 x^2}-\frac {4 i a}{x}-\frac {4 i a^2}{i+a x}-8 a^2 \log (x)+8 a^2 \log (i+a x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=-\frac {1}{2 x^2}-\frac {4 i a}{x}-\frac {4 i a^2}{i+a x}-8 a^2 \log (x)+8 a^2 \log (i+a x) \]
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Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {1}{2 x^{2}}-\frac {4 i a}{x}-8 a^{2} \ln \left (x \right )-4 a^{3} \left (\frac {i}{a \left (a x +i\right )}-\frac {2 \ln \left (a x +i\right )}{a}\right )\) | \(53\) |
risch | \(\frac {-8 i a^{2} x^{2}+\frac {7}{2} a x -\frac {1}{2} i}{\left (a x +i\right ) x^{2}}-8 i a^{2} \arctan \left (a x \right )+4 a^{2} \ln \left (a^{2} x^{2}+1\right )-8 a^{2} \ln \left (x \right )\) | \(62\) |
parallelrisch | \(-\frac {16 \ln \left (x \right ) x^{4} a^{4}-16 \ln \left (a x +i\right ) x^{4} a^{4}-9 a^{4} x^{4}+16 i a^{3} x^{3}+1+16 a^{2} \ln \left (x \right ) x^{2}-16 a^{2} \ln \left (a x +i\right ) x^{2}+8 i a x}{2 \left (a^{2} x^{2}+1\right ) x^{2}}\) | \(91\) |
meijerg | \(\frac {a^{2} \left (\frac {3 a^{2} x^{2}}{3 a^{2} x^{2}+3}+2 \ln \left (a^{2} x^{2}+1\right )-1-4 \ln \left (x \right )-2 \ln \left (a^{2}\right )-\frac {1}{a^{2} x^{2}}\right )}{2}+\frac {2 i a^{3} \left (-\frac {2 \left (3 a^{2} x^{2}+2\right )}{x \sqrt {a^{2}}\, \left (2 a^{2} x^{2}+2\right )}-\frac {3 a \arctan \left (a x \right )}{\sqrt {a^{2}}}\right )}{\sqrt {a^{2}}}-3 a^{2} \left (-\frac {2 a^{2} x^{2}}{2 a^{2} x^{2}+2}-\ln \left (a^{2} x^{2}+1\right )+1+2 \ln \left (x \right )+\ln \left (a^{2}\right )\right )-\frac {2 i a^{3} \left (\frac {2 x \sqrt {a^{2}}}{2 a^{2} x^{2}+2}+\frac {\sqrt {a^{2}}\, \arctan \left (a x \right )}{a}\right )}{\sqrt {a^{2}}}+\frac {a^{4} x^{2}}{2 a^{2} x^{2}+2}\) | \(226\) |
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Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.48 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=\frac {-16 i \, a^{2} x^{2} + 7 \, a x - 16 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (x\right ) + 16 \, {\left (a^{3} x^{3} + i \, a^{2} x^{2}\right )} \log \left (\frac {a x + i}{a}\right ) - i}{2 \, {\left (a x^{3} + i \, x^{2}\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=8 a^{2} \left (- \log {\left (16 a^{3} x \right )} + \log {\left (16 a^{3} x + 16 i a^{2} \right )}\right ) + \frac {- 16 i a^{2} x^{2} + 7 a x - i}{2 a x^{3} + 2 i x^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.33 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=-8 i \, a^{2} \arctan \left (a x\right ) + 4 \, a^{2} \log \left (a^{2} x^{2} + 1\right ) - 8 \, a^{2} \log \left (x\right ) + \frac {-16 i \, a^{3} x^{3} - 9 \, a^{2} x^{2} - 8 i \, a x - 1}{2 \, {\left (a^{2} x^{4} + x^{2}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=8 \, a^{2} \log \left (a x + i\right ) - 8 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {16 i \, a^{2} x^{2} - 7 \, a x + i}{2 \, {\left (a x + i\right )} x^{2}} \]
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Time = 0.57 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \frac {e^{4 i \arctan (a x)}}{x^3} \, dx=-a^2\,\mathrm {atan}\left (2\,a\,x+1{}\mathrm {i}\right )\,16{}\mathrm {i}+\frac {8\,a^2\,x^2+\frac {a\,x\,7{}\mathrm {i}}{2}+\frac {1}{2}}{x^2\,\left (-1+a\,x\,1{}\mathrm {i}\right )} \]
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