Integrand size = 12, antiderivative size = 23 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {2 i \log (i-a-b x)}{b} \]
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Time = 0.01 (sec) , antiderivative size = 23, 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 = {5201, 45} \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {2 i \log (-a-b x+i)}{b} \]
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Rule 45
Rule 5201
Rubi steps \begin{align*} \text {integral}& = \int \frac {1-i a-i b x}{1+i a+i b x} \, dx \\ & = \int \left (-1-\frac {2 i}{-i+a+b x}\right ) \, dx \\ & = -x-\frac {2 i \log (i-a-b x)}{b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x+\frac {2 \arctan (a+b x)}{b}-\frac {i \log \left (1+(a+b x)^2\right )}{b} \]
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Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
default | \(-x -\frac {2 i \ln \left (-b x -a +i\right )}{b}\) | \(22\) |
risch | \(-x -\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b}+\frac {2 \arctan \left (b x +a \right )}{b}\) | \(40\) |
parallelrisch | \(\frac {2 i \ln \left (b x +a -i\right ) x b +b^{2} x^{2}+2 i \ln \left (b x +a -i\right ) a +1+2 i a -a^{2}+2 \ln \left (b x +a -i\right )}{b \left (-b x -a +i\right )}\) | \(70\) |
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none
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-\frac {b x + 2 i \, \log \left (\frac {b x + a - i}{b}\right )}{b} \]
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Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int e^{-2 i \arctan (a+b x)} \, dx=- x - \frac {2 i \log {\left (a + b x - i \right )}}{b} \]
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Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x - \frac {2 i \, \log \left (i \, b x + i \, a + 1\right )}{b} \]
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Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int e^{-2 i \arctan (a+b x)} \, dx=\frac {i \, {\left (i \, b x + i \, a + 1\right )}}{b} + \frac {2 i \, \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b} \]
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Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int e^{-2 i \arctan (a+b x)} \, dx=-x-\frac {\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,2{}\mathrm {i}}{b} \]
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