Integrand size = 14, antiderivative size = 159 \[ \int e^{3 i \arctan (a x)} x^m \, dx=-\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}+\frac {4 i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m} \]
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Time = 0.65 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5168, 6874, 371, 864, 822} \[ \int e^{3 i \arctan (a x)} x^m \, dx=-\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}+\frac {4 x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{2},\frac {m+3}{2},-a^2 x^2\right )}{m+1}-\frac {i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2}+\frac {4 i a x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-a^2 x^2\right )}{m+2} \]
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Rule 371
Rule 822
Rule 864
Rule 5168
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^m (1+i a x)^2}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx \\ & = \int \left (-\frac {3 x^m}{\sqrt {1+a^2 x^2}}-\frac {i a x^{1+m}}{\sqrt {1+a^2 x^2}}+\frac {4 x^m}{(1-i a x) \sqrt {1+a^2 x^2}}\right ) \, dx \\ & = -\left (3 \int \frac {x^m}{\sqrt {1+a^2 x^2}} \, dx\right )+4 \int \frac {x^m}{(1-i a x) \sqrt {1+a^2 x^2}} \, dx-(i a) \int \frac {x^{1+m}}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m}+4 \int \frac {x^m (1+i a x)}{\left (1+a^2 x^2\right )^{3/2}} \, dx \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m}+4 \int \frac {x^m}{\left (1+a^2 x^2\right )^{3/2}} \, dx+(4 i a) \int \frac {x^{1+m}}{\left (1+a^2 x^2\right )^{3/2}} \, dx \\ & = -\frac {3 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}-\frac {i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m}+\frac {4 x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-a^2 x^2\right )}{1+m}+\frac {4 i a x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-a^2 x^2\right )}{2+m} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71 \[ \int e^{3 i \arctan (a x)} x^m \, dx=-\frac {i x^{1+m} \sqrt {1-i a x} \sqrt {-i+a x} \left (\operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {1}{2},2+m,-i a x,i a x\right )-2 \operatorname {AppellF1}\left (1+m,-\frac {1}{2},\frac {3}{2},2+m,-i a x,i a x\right )\right )}{(1+m) \sqrt {1+i a x} \sqrt {i+a x}} \]
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Time = 0.23 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.92
method | result | size |
meijerg | \(\frac {x^{1+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {1}{2}+\frac {m}{2}\right ], \left [\frac {3}{2}+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{1+m}+\frac {3 i a \,x^{2+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 1+\frac {m}{2}\right ], \left [2+\frac {m}{2}\right ], -a^{2} x^{2}\right )}{2+m}-\frac {3 a^{2} x^{3+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {3}{2}+\frac {m}{2}\right ], \left [\frac {m}{2}+\frac {5}{2}\right ], -a^{2} x^{2}\right )}{3+m}-\frac {i a^{3} x^{4+m} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 2+\frac {m}{2}\right ], \left [\frac {m}{2}+3\right ], -a^{2} x^{2}\right )}{4+m}\) | \(146\) |
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\[ \int e^{3 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int e^{3 i \arctan (a x)} x^m \, dx=- i \left (\int \frac {i x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 a x x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx + \int \frac {a^{3} x^{3} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\, dx + \int \left (- \frac {3 i a^{2} x^{2} x^{m}}{a^{2} x^{2} \sqrt {a^{2} x^{2} + 1} + \sqrt {a^{2} x^{2} + 1}}\right )\, dx\right ) \]
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\[ \int e^{3 i \arctan (a x)} x^m \, dx=\int { \frac {{\left (i \, a x + 1\right )}^{3} x^{m}}{{\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
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Exception generated. \[ \int e^{3 i \arctan (a x)} x^m \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int e^{3 i \arctan (a x)} x^m \, dx=\int \frac {x^m\,{\left (1+a\,x\,1{}\mathrm {i}\right )}^3}{{\left (a^2\,x^2+1\right )}^{3/2}} \,d x \]
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