Integrand size = 16, antiderivative size = 59 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=\frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {2 i (i-a)^2 \log (i-a-b x)}{b^3} \]
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Time = 0.04 (sec) , antiderivative size = 59, 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} \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=-\frac {2 i (-a+i)^2 \log (-a-b x+i)}{b^3}+\frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3} \]
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Rule 78
Rule 5203
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (1-i a-i b x)}{1+i a+i b x} \, dx \\ & = \int \left (\frac {2 i (-i+a)}{b^2}-\frac {2 i x}{b}-x^2-\frac {2 i (-i+a)^2}{b^2 (-i+a+b x)}\right ) \, dx \\ & = \frac {2 (1+i a) x}{b^2}-\frac {i x^2}{b}-\frac {x^3}{3}-\frac {2 i (i-a)^2 \log (i-a-b x)}{b^3} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=\frac {b x \left (6+6 i a-3 i b x-b^2 x^2\right )-6 i (-i+a)^2 \log (i-a-b x)}{3 b^3} \]
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Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {i \left (\frac {1}{3} i b^{2} x^{3}-x^{2} b -2 i x +2 a x \right )}{b^{2}}+\frac {\left (-2 i a^{2}-4 a +2 i\right ) \ln \left (-b x -a +i\right )}{b^{3}}\) | \(59\) |
risch | \(-\frac {x^{3}}{3}-\frac {i x^{2}}{b}+\frac {2 x}{b^{2}}+\frac {2 i a x}{b^{2}}-\frac {2 \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a}{b^{3}}-\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right ) a^{2}}{b^{3}}+\frac {i \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{b^{3}}-\frac {4 i \arctan \left (b x +a \right ) a}{b^{3}}+\frac {2 \arctan \left (b x +a \right ) a^{2}}{b^{3}}-\frac {2 \arctan \left (b x +a \right )}{b^{3}}\) | \(143\) |
parallelrisch | \(\frac {x^{4} b^{4}-3 i a \,b^{2} x^{2}+a \,b^{3} x^{3}+6 i \ln \left (b x +a -i\right ) a^{3}+6 i \ln \left (b x +a -i\right ) x \,a^{2} b -18 i a -6+6 i a^{3}+2 i b^{3} x^{3}+12 \ln \left (b x +a -i\right ) x a b -3 b^{2} x^{2}-18 i \ln \left (b x +a -i\right ) a +18 \ln \left (b x +a -i\right ) a^{2}-6 i \ln \left (b x +a -i\right ) x b +18 a^{2}-6 \ln \left (b x +a -i\right )}{3 b^{3} \left (-b x -a +i\right )}\) | \(168\) |
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Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=-\frac {b^{3} x^{3} + 3 i \, b^{2} x^{2} + 6 \, {\left (-i \, a - 1\right )} b x + 6 \, {\left (i \, a^{2} + 2 \, a - i\right )} \log \left (\frac {b x + a - i}{b}\right )}{3 \, b^{3}} \]
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Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.83 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=- \frac {x^{3}}{3} - x \left (- \frac {2 i a}{b^{2}} - \frac {2}{b^{2}}\right ) - \frac {i x^{2}}{b} - \frac {2 i \left (a - i\right )^{2} \log {\left (a + b x - i \right )}}{b^{3}} \]
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Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=-\frac {b^{2} x^{3} + 3 i \, b x^{2} + 6 \, {\left (-i \, a - 1\right )} x}{3 \, b^{2}} - \frac {2 \, {\left (i \, a^{2} + 2 \, a - i\right )} \log \left (i \, b x + i \, a + 1\right )}{b^{3}} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (45) = 90\).
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.85 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=-\frac {i \, {\left (i \, b x + i \, a + 1\right )}^{3} {\left (-\frac {3 i \, {\left (a b - 2 i \, b\right )}}{{\left (i \, b x + i \, a + 1\right )} b} - \frac {3 \, {\left (a^{2} b^{2} - 6 i \, a b^{2} - 5 \, b^{2}\right )}}{{\left (i \, b x + i \, a + 1\right )}^{2} b^{2}} + 1\right )}}{3 \, b^{3}} - \frac {2 \, {\left (-i \, a^{2} - 2 \, a + i\right )} \log \left (\frac {1}{\sqrt {{\left (b x + a\right )}^{2} + 1} {\left | b \right |}}\right )}{b^{3}} \]
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Time = 0.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53 \[ \int e^{-2 i \arctan (a+b x)} x^2 \, dx=-\ln \left (x+\frac {a-\mathrm {i}}{b}\right )\,\left (\frac {4\,a}{b^3}+\frac {\left (2\,a^2-2\right )\,1{}\mathrm {i}}{b^3}\right )+x^2\,\left (\frac {a-\mathrm {i}}{2\,b}-\frac {a+1{}\mathrm {i}}{2\,b}\right )-\frac {x^3}{3}-\frac {x\,\left (\frac {a-\mathrm {i}}{b}-\frac {a+1{}\mathrm {i}}{b}\right )\,\left (a-\mathrm {i}\right )}{b} \]
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