Integrand size = 14, antiderivative size = 39 \[ \int e^{2 i \arctan (a x)} x^m \, dx=-\frac {x^{1+m}}{1+m}+\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x)}{1+m} \]
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Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5170, 81, 66} \[ \int e^{2 i \arctan (a x)} x^m \, dx=-\frac {x^{m+1}}{m+1}+\frac {2 x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,i a x)}{m+1} \]
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Rule 66
Rule 81
Rule 5170
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m (1+i a x)}{1-i a x} \, dx \\ & = -\frac {x^{1+m}}{1+m}+2 \int \frac {x^m}{1-i a x} \, dx \\ & = -\frac {x^{1+m}}{1+m}+\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x)}{1+m} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74 \[ \int e^{2 i \arctan (a x)} x^m \, dx=\frac {x^{1+m} (-1+2 \operatorname {Hypergeometric2F1}(1,1+m,2+m,i a x))}{1+m} \]
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Result contains higher order function than in optimal. Order 9 vs. order 5.
Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 4.49
method | result | size |
meijerg | \(\frac {x^{1+m} \left (\frac {1}{2}+\frac {m}{2}\right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}+\frac {i \left (a^{2}\right )^{-\frac {m}{2}} \left (\frac {2 x^{m} \left (a^{2}\right )^{\frac {m}{2}}}{m}+\frac {x^{m} \left (a^{2}\right )^{\frac {m}{2}} \left (-2-m \right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {m}{2}\right )}{2+m}\right )}{a}-\frac {\left (a^{2}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (a^{2}\right )^{\frac {3}{2}+\frac {m}{2}}}{\left (1+m \right ) a^{2}}+\frac {x^{1+m} \left (a^{2}\right )^{\frac {3}{2}+\frac {m}{2}} \left (-3-m \right ) \operatorname {LerchPhi}\left (-a^{2} x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{\left (3+m \right ) a^{2}}\right )}{2}\) | \(175\) |
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\[ \int e^{2 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} + 1} \,d x } \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (27) = 54\).
Time = 1.63 (sec) , antiderivative size = 126, normalized size of antiderivative = 3.23 \[ \int e^{2 i \arctan (a x)} x^m \, dx=\frac {i a m x^{m + 2} \Phi \left (a x e^{\frac {i \pi }{2}}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac {2 i a x^{m + 2} \Phi \left (a x e^{\frac {i \pi }{2}}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} + \frac {m x^{m + 1} \Phi \left (a x e^{\frac {i \pi }{2}}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} + \frac {x^{m + 1} \Phi \left (a x e^{\frac {i \pi }{2}}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} \]
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\[ \int e^{2 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} + 1} \,d x } \]
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\[ \int e^{2 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{2} x^{m}}{a^{2} x^{2} + 1} \,d x } \]
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Timed out. \[ \int e^{2 i \arctan (a x)} x^m \, dx=\int \frac {x^m\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^2}{a^2\,x^2+1} \,d x \]
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