Integrand size = 14, antiderivative size = 65 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4}+\frac {4 i}{a^4 (i+a x)}+\frac {16 \log (i+a x)}{a^4} \]
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Time = 0.03 (sec) , antiderivative size = 65, 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} \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {4 i}{a^4 (a x+i)}+\frac {16 \log (a x+i)}{a^4}+\frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4} \]
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Rule 90
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 (1+i a x)^2}{(1-i a x)^2} \, dx \\ & = \int \left (\frac {12 i}{a^3}-\frac {8 x}{a^2}-\frac {4 i x^2}{a}+x^3-\frac {4 i}{a^3 (i+a x)^2}+\frac {16}{a^3 (i+a x)}\right ) \, dx \\ & = \frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4}+\frac {4 i}{a^4 (i+a x)}+\frac {16 \log (i+a x)}{a^4} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {12 i x}{a^3}-\frac {4 x^2}{a^2}-\frac {4 i x^3}{3 a}+\frac {x^4}{4}+\frac {4 i}{a^4 (i+a x)}+\frac {16 \log (i+a x)}{a^4} \]
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Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03
method | result | size |
default | \(-\frac {-\frac {1}{4} a^{3} x^{4}+\frac {4}{3} i a^{2} x^{3}+4 a \,x^{2}-12 i x}{a^{3}}-\frac {4 \left (-\frac {i}{a \left (a x +i\right )}-\frac {4 \ln \left (a x +i\right )}{a}\right )}{a^{3}}\) | \(67\) |
risch | \(\frac {x^{4}}{4}-\frac {4 i x^{3}}{3 a}-\frac {4 x^{2}}{a^{2}}+\frac {12 i x}{a^{3}}+\frac {4 i}{a^{4} \left (a x +i\right )}+\frac {8 \ln \left (a^{2} x^{2}+1\right )}{a^{4}}-\frac {16 i \arctan \left (a x \right )}{a^{4}}\) | \(70\) |
parallelrisch | \(-\frac {-3 x^{6} a^{6}+16 i x^{5} a^{5}+45 a^{4} x^{4}-128 i a^{3} x^{3}-192 a^{2} \ln \left (a x +i\right ) x^{2}+96 a^{2} x^{2}-192 i a x -192 \ln \left (a x +i\right )}{12 a^{4} \left (a^{2} x^{2}+1\right )}\) | \(89\) |
meijerg | \(\frac {-\frac {a^{2} x^{2}}{a^{2} x^{2}+1}+\ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}+\frac {2 i \left (\frac {x \left (a^{2}\right )^{\frac {5}{2}} \left (10 a^{2} x^{2}+15\right )}{5 a^{4} \left (a^{2} x^{2}+1\right )}-\frac {3 \left (a^{2}\right )^{\frac {5}{2}} \arctan \left (a x \right )}{a^{5}}\right )}{a^{3} \sqrt {a^{2}}}-\frac {3 \left (\frac {x^{2} a^{2} \left (3 a^{2} x^{2}+6\right )}{3 a^{2} x^{2}+3}-2 \ln \left (a^{2} x^{2}+1\right )\right )}{a^{4}}-\frac {2 i \left (-\frac {x \left (a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}+70 a^{2} x^{2}+105\right )}{21 a^{6} \left (a^{2} x^{2}+1\right )}+\frac {5 \left (a^{2}\right )^{\frac {7}{2}} \arctan \left (a x \right )}{a^{7}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {-\frac {x^{2} a^{2} \left (-2 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{4 \left (a^{2} x^{2}+1\right )}+3 \ln \left (a^{2} x^{2}+1\right )}{2 a^{4}}\) | \(263\) |
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Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {3 \, a^{5} x^{5} - 13 i \, a^{4} x^{4} - 32 \, a^{3} x^{3} + 96 i \, a^{2} x^{2} - 144 \, a x + 192 \, {\left (a x + i\right )} \log \left (\frac {a x + i}{a}\right ) + 48 i}{12 \, {\left (a^{5} x + i \, a^{4}\right )}} \]
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Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {x^{4}}{4} + \frac {4 i}{a^{5} x + i a^{4}} - \frac {4 i x^{3}}{3 a} - \frac {4 x^{2}}{a^{2}} + \frac {12 i x}{a^{3}} + \frac {16 \log {\left (a x + i \right )}}{a^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=-\frac {4 \, {\left (-i \, a x - 1\right )}}{a^{6} x^{2} + a^{4}} + \frac {3 \, a^{3} x^{4} - 16 i \, a^{2} x^{3} - 48 \, a x^{2} + 144 i \, x}{12 \, a^{3}} - \frac {16 i \, \arctan \left (a x\right )}{a^{4}} + \frac {8 \, \log \left (a^{2} x^{2} + 1\right )}{a^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {16 \, \log \left (a x + i\right )}{a^{4}} + \frac {4 i}{{\left (a x + i\right )} a^{4}} + \frac {3 \, a^{8} x^{4} - 16 i \, a^{7} x^{3} - 48 \, a^{6} x^{2} + 144 i \, a^{5} x}{12 \, a^{8}} \]
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Time = 0.51 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int e^{4 i \arctan (a x)} x^3 \, dx=\frac {16\,\ln \left (x+\frac {1{}\mathrm {i}}{a}\right )}{a^4}+\frac {x^4}{4}-\frac {4\,x^2}{a^2}+\frac {4{}\mathrm {i}}{a^5\,\left (x+\frac {1{}\mathrm {i}}{a}\right )}+\frac {x\,12{}\mathrm {i}}{a^3}-\frac {x^3\,4{}\mathrm {i}}{3\,a} \]
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