Integrand size = 16, antiderivative size = 38 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=\frac {(i-a) \log (x)}{i+a}-\frac {2 \log (i+a+b x)}{1-i a} \]
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Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5203, 78} \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=\frac {(-a+i) \log (x)}{a+i}-\frac {2 \log (a+b x+i)}{1-i a} \]
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Rule 78
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+i a+i b x}{x (1-i a-i b x)} \, dx \\ & = \int \left (\frac {i-a}{(i+a) x}-\frac {2 i b}{(i+a) (i+a+b x)}\right ) \, dx \\ & = \frac {(i-a) \log (x)}{i+a}-\frac {2 \log (i+a+b x)}{1-i a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=-\frac {(-i+a) \log (x)+2 i \log (i+a+b x)}{i+a} \]
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Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24
method | result | size |
parallelrisch | \(\frac {-2 \ln \left (b x +a +i\right )+2 i a \ln \left (x \right )-2 i \ln \left (b x +a +i\right ) a -a^{2} \ln \left (x \right )+\ln \left (x \right )}{a^{2}+1}\) | \(47\) |
risch | \(\frac {i \ln \left (-x \right )}{i+a}-\frac {\ln \left (-x \right ) a}{i+a}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{i+a}-\frac {2 \arctan \left (b x +a \right )}{i+a}\) | \(69\) |
default | \(\frac {\left (-a^{2}+2 i a +1\right ) \ln \left (x \right )}{a^{2}+1}-\frac {2 b \left (\frac {\left (i a b +b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (i a^{2}-i+2 a -\frac {\left (i a b +b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{a^{2}+1}\) | \(110\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=-\frac {{\left (a - i\right )} \log \left (x\right ) + 2 i \, \log \left (\frac {b x + a + i}{b}\right )}{a + i} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (24) = 48\).
Time = 0.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.63 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=- \frac {\left (a - i\right ) \log {\left (- \frac {a^{2} \left (a - i\right )}{a + i} + a^{2} - \frac {2 i a \left (a - i\right )}{a + i} + x \left (a b - 3 i b\right ) + \frac {a - i}{a + i} + 1 \right )}}{a + i} - \frac {2 i \log {\left (a^{2} - \frac {2 i a^{2}}{a + i} + \frac {4 a}{a + i} + x \left (a b - 3 i b\right ) + 1 + \frac {2 i}{a + i} \right )}}{a + i} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (29) = 58\).
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.05 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=-\frac {2 \, {\left (a - i\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{2} + 1} - \frac {{\left (i \, a + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{2} + 1} - \frac {{\left (a^{2} - 2 i \, a - 1\right )} \log \left (x\right )}{a^{2} + 1} \]
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=-\frac {2 i \, b \log \left (b x + a + i\right )}{a b + i \, b} - \frac {{\left (a - i\right )} \log \left ({\left | x \right |}\right )}{a + i} \]
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Time = 0.78 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {e^{2 i \arctan (a+b x)}}{x} \, dx=\ln \left (x\right )\,\left (-1+\frac {2{}\mathrm {i}}{a+1{}\mathrm {i}}\right )-\frac {\ln \left (a+b\,x+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{a+1{}\mathrm {i}} \]
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