Integrand size = 17, antiderivative size = 184 \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=-\frac {i (1-m) m x (e x)^m}{2 (1+m)}+\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}-\frac {i \left (3+2 m+m^2\right ) x (e x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-m),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{1+m} \]
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Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {4591, 511, 479, 591, 470, 371} \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=-\frac {i \left (m^2+2 m+3\right ) x (e x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-m-1),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{m+1}+\frac {i e^{-2 i a} x \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}+\frac {i x \left (1-\frac {e^{2 i a}}{x^2}\right )^2 (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}-\frac {i (1-m) m x (e x)^m}{2 (m+1)} \]
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Rule 371
Rule 470
Rule 479
Rule 511
Rule 591
Rule 4591
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (i-\frac {i e^{2 i a}}{x^2}\right )^3 (e x)^m}{\left (1+\frac {e^{2 i a}}{x^2}\right )^3} \, dx \\ & = -\left (\left (\left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (i-i e^{2 i a} x^2\right )^3}{\left (1+e^{2 i a} x^2\right )^3} \, dx,x,\frac {1}{x}\right )\right ) \\ & = \frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {1}{4} \left (e^{-2 i a} \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (i-i e^{2 i a} x^2\right ) \left (2 e^{2 i a} (3+m)+2 e^{4 i a} (1-m) x^2\right )}{\left (1+e^{2 i a} x^2\right )^2} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}-\frac {1}{8} \left (e^{-4 i a} \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int \frac {x^{-2-m} \left (-4 i e^{4 i a} (2+m) (3+m)-4 i e^{6 i a} (1-m) m x^2\right )}{1+e^{2 i a} x^2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {i (1-m) m x (e x)^m}{2 (1+m)}+\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}+\left (i \left (3+2 m+m^2\right ) \left (\frac {1}{x}\right )^m (e x)^m\right ) \text {Subst}\left (\int \frac {x^{-2-m}}{1+e^{2 i a} x^2} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {i (1-m) m x (e x)^m}{2 (1+m)}+\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}-\frac {i \left (3+2 m+m^2\right ) x (e x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-m),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{1+m} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.68 \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\frac {i x (e x)^m \left (-1+6 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )-12 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )+8 \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )\right )}{1+m} \]
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\[\int \left (e x \right )^{m} \tan \left (a +i \ln \left (x \right )\right )^{3}d x\]
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\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
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\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int \left (e x\right )^{m} \tan ^{3}{\left (a + i \log {\left (x \right )} \right )}\, dx \]
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\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
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\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
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Timed out. \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int {\mathrm {tan}\left (a+\ln \left (x\right )\,1{}\mathrm {i}\right )}^3\,{\left (e\,x\right )}^m \,d x \]
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