Integrand size = 24, antiderivative size = 50 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {2 i}{a (1-i a x)^2}+\frac {4 i}{a (1-i a x)}+\frac {i \log (i+a x)}{a} \]
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Time = 0.04 (sec) , antiderivative size = 50, 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} \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {4 i}{a (1-i a x)}-\frac {2 i}{a (1-i a x)^2}+\frac {i \log (a x+i)}{a} \]
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Rule 45
Rule 5181
Rubi steps \begin{align*} \text {integral}& = \int \frac {(1+i a x)^2}{(1-i a x)^3} \, dx \\ & = \int \left (\frac {4}{(1-i a x)^3}-\frac {4}{(1-i a x)^2}+\frac {1}{1-i a x}\right ) \, dx \\ & = -\frac {2 i}{a (1-i a x)^2}+\frac {4 i}{a (1-i a x)}+\frac {i \log (i+a x)}{a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {i \left (-2+4 i a x+(i+a x)^2 \log (i+a x)\right )}{a (i+a x)^2} \]
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Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {-4 x -\frac {2 i}{a}}{\left (a x +i\right )^{2}}+\frac {i \ln \left (a x +i\right )}{a}\) | \(34\) |
risch | \(\frac {-4 x -\frac {2 i}{a}}{\left (a x +i\right )^{2}}+\frac {i \ln \left (a^{2} x^{2}+1\right )}{2 a}+\frac {\arctan \left (a x \right )}{a}\) | \(45\) |
parallelrisch | \(\frac {i \ln \left (a x +i\right ) x^{4} a^{4}-2 i x^{4} a^{4}+2 i \ln \left (a x +i\right ) x^{2} a^{2}-4 a^{3} x^{3}+2 i x^{2} a^{2}+i \ln \left (a x +i\right )}{\left (a^{2} x^{2}+1\right )^{2} a}\) | \(85\) |
meijerg | \(\frac {\frac {x \sqrt {a^{2}}\, \left (3 a^{2} x^{2}+5\right )}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \sqrt {a^{2}}\, \arctan \left (a x \right )}{2 a}}{4 \sqrt {a^{2}}}+\frac {5 i a \,x^{2} \left (a^{2} x^{2}+2\right )}{4 \left (a^{2} x^{2}+1\right )^{2}}-\frac {5 \left (-\frac {x \left (a^{2}\right )^{\frac {3}{2}} \left (-3 a^{2} x^{2}+3\right )}{6 a^{2} \left (a^{2} x^{2}+1\right )^{2}}+\frac {\left (a^{2}\right )^{\frac {3}{2}} \arctan \left (a x \right )}{2 a^{3}}\right )}{2 \sqrt {a^{2}}}-\frac {5 i a^{3} x^{4}}{2 \left (a^{2} x^{2}+1\right )^{2}}+\frac {-\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (25 a^{2} x^{2}+15\right )}{8 a^{4} \left (a^{2} x^{2}+1\right )^{2}}+\frac {15 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{8 a^{5}}}{\sqrt {a^{2}}}+\frac {i \left (-\frac {a^{2} x^{2} \left (9 a^{2} x^{2}+6\right )}{3 \left (a^{2} x^{2}+1\right )^{2}}+2 \ln \left (a^{2} x^{2}+1\right )\right )}{4 a}\) | \(256\) |
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Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {4 \, a x - {\left (i \, a^{2} x^{2} - 2 \, a x - i\right )} \log \left (\frac {a x + i}{a}\right ) + 2 i}{a^{3} x^{2} + 2 i \, a^{2} x - a} \]
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Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.72 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {- 4 a x - 2 i}{a^{3} x^{2} + 2 i a^{2} x - a} + \frac {i \log {\left (a x + i \right )}}{a} \]
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Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=-\frac {2 \, {\left (2 \, a^{3} x^{3} - 3 i \, a^{2} x^{2} - i\right )}}{a^{5} x^{4} + 2 \, a^{3} x^{2} + a} + \frac {\arctan \left (a x\right )}{a} + \frac {i \, \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.60 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {i \, \log \left (a x + i\right )}{a} - \frac {2 \, {\left (2 \, a x + i\right )}}{{\left (a x + i\right )}^{2} a} \]
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Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98 \[ \int \frac {e^{5 i \arctan (a x)}}{\sqrt {1+a^2 x^2}} \, dx=\frac {\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )\,1{}\mathrm {i}}{a}-\frac {\frac {4\,x}{a^2}+\frac {2{}\mathrm {i}}{a^3}}{x^2-\frac {1}{a^2}+\frac {x\,2{}\mathrm {i}}{a}} \]
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