Integrand size = 13, antiderivative size = 43 \[ \int x^2 \tan (a+i \log (x)) \, dx=-2 i e^{2 i a} x+\frac {i x^3}{3}+2 i e^{3 i a} \arctan \left (e^{-i a} x\right ) \]
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Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4591, 456, 470, 327, 209} \[ \int x^2 \tan (a+i \log (x)) \, dx=2 i e^{3 i a} \arctan \left (e^{-i a} x\right )-2 i e^{2 i a} x+\frac {i x^3}{3} \]
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Rule 209
Rule 327
Rule 456
Rule 470
Rule 4591
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (i-\frac {i e^{2 i a}}{x^2}\right ) x^2}{1+\frac {e^{2 i a}}{x^2}} \, dx \\ & = \int \frac {x^2 \left (-i e^{2 i a}+i x^2\right )}{e^{2 i a}+x^2} \, dx \\ & = \frac {i x^3}{3}-\left (2 i e^{2 i a}\right ) \int \frac {x^2}{e^{2 i a}+x^2} \, dx \\ & = -2 i e^{2 i a} x+\frac {i x^3}{3}+\left (2 i e^{4 i a}\right ) \int \frac {1}{e^{2 i a}+x^2} \, dx \\ & = -2 i e^{2 i a} x+\frac {i x^3}{3}+2 i e^{3 i a} \arctan \left (e^{-i a} x\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.53 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {i x^3}{3}-2 i x \cos (2 a)+2 i \arctan (x \cos (a)-i x \sin (a)) \cos (3 a)+2 x \sin (2 a)-2 \arctan (x \cos (a)-i x \sin (a)) \sin (3 a) \]
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Time = 3.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {i x^{3}}{3}-2 i {\mathrm e}^{2 i a} x +2 i \arctan \left (x \,{\mathrm e}^{-i a}\right ) {\mathrm e}^{3 i a}\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} - 2 i \, x e^{\left (2 i \, a\right )} - e^{\left (3 i \, a\right )} \log \left (x + i \, e^{\left (i \, a\right )}\right ) + e^{\left (3 i \, a\right )} \log \left (x - i \, e^{\left (i \, a\right )}\right ) \]
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Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.42 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {i x^{3}}{3} - 2 i x e^{2 i a} + \left (\log {\left (x e^{2 i a} - i e^{3 i a} \right )} - \log {\left (x e^{2 i a} + i e^{3 i a} \right )}\right ) e^{3 i a} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (26) = 52\).
Time = 0.30 (sec) , antiderivative size = 149, normalized size of antiderivative = 3.47 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} - 2 \, x {\left (i \, \cos \left (2 \, a\right ) - \sin \left (2 \, a\right )\right )} - {\left (i \, \cos \left (3 \, a\right ) - \sin \left (3 \, a\right )\right )} \arctan \left (\frac {2 \, x \cos \left (a\right )}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}, \frac {x^{2} - \cos \left (a\right )^{2} - \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + \frac {1}{2} \, {\left (\cos \left (3 \, a\right ) + i \, \sin \left (3 \, a\right )\right )} \log \left (\frac {x^{2} + \cos \left (a\right )^{2} + 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) \]
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Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.60 \[ \int x^2 \tan (a+i \log (x)) \, dx=\frac {1}{3} i \, x^{3} + 2 i \, \arctan \left (x e^{\left (-i \, a\right )}\right ) e^{\left (3 i \, a\right )} - 2 i \, x e^{\left (2 i \, a\right )} \]
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Time = 26.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int x^2 \tan (a+i \log (x)) \, dx={\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\right )}^{3/2}\,\mathrm {atan}\left (\frac {x}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}}}\right )\,2{}\mathrm {i}+\frac {x^3\,1{}\mathrm {i}}{3}-x\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,2{}\mathrm {i} \]
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