Integrand size = 81, antiderivative size = 26 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\frac {i \pi -x-\log (2)}{(-2-2 x)^2 x^3}} \]
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Time = 1.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6873, 12, 6838} \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=2^{-\frac {1}{4 x^3 (x+1)^2}} e^{-\frac {x+i \pi }{4 x^3 (x+1)^2}} \]
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Rule 12
Rule 6838
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3 (1+x)^2}} \left (3 i \pi +4 x^2+\log (8)+x (2+5 i \pi +\log (32))\right )}{4 x^4 (1+x)^3} \, dx \\ & = \frac {1}{4} \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3 (1+x)^2}} \left (3 i \pi +4 x^2+\log (8)+x (2+5 i \pi +\log (32))\right )}{x^4 (1+x)^3} \, dx \\ & = 2^{-\frac {1}{4 x^3 (1+x)^2}} e^{-\frac {i \pi +x}{4 x^3 (1+x)^2}} \\ \end{align*}
\[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx \]
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Time = 0.58 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77
method | result | size |
risch | \({\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (1+x \right )^{2}}}\) | \(20\) |
parallelrisch | \({\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (x^{2}+2 x +1\right )}}\) | \(25\) |
gosper | \(-\frac {i {\mathrm e}^{-\frac {i \pi +\ln \left (2\right )+x}{4 x^{3} \left (x^{2}+2 x +1\right )}} \left (5 i \pi x +3 i \pi +5 x \ln \left (2\right )+4 x^{2}+3 \ln \left (2\right )+2 x \right )}{-5 i x \ln \left (2\right )-4 i x^{2}+5 \pi x -3 i \ln \left (2\right )-2 i x +3 \pi }\) | \(86\) |
norman | \(\frac {x^{3} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}+x^{5} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}+2 x^{4} {\mathrm e}^{\frac {-\ln \left (2\right )-i \pi -x}{4 x^{5}+8 x^{4}+4 x^{3}}}}{x^{3} \left (1+x \right )^{2}}\) | \(120\) |
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Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (-\frac {i \, \pi }{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {x}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {\log \left (2\right )}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\text {Timed out} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (50) = 100\).
Time = 0.65 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.38 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (\frac {i \, \pi }{4 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 i \, \pi }{4 \, {\left (x + 1\right )}} - \frac {3 i \, \pi }{4 \, x} + \frac {\log \left (2\right )}{4 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {3 \, \log \left (2\right )}{4 \, {\left (x + 1\right )}} - \frac {3 \, \log \left (2\right )}{4 \, x} - \frac {1}{4 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {1}{2 \, {\left (x + 1\right )}} + \frac {i \, \pi }{2 \, x^{2}} + \frac {1}{2 \, x} + \frac {\log \left (2\right )}{2 \, x^{2}} - \frac {i \, \pi }{4 \, x^{3}} - \frac {1}{4 \, x^{2}} - \frac {\log \left (2\right )}{4 \, x^{3}}\right )} \]
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Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=e^{\left (-\frac {i \, \pi }{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {x}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}} - \frac {\log \left (2\right )}{4 \, {\left (x^{5} + 2 \, x^{4} + x^{3}\right )}}\right )} \]
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Time = 8.61 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \frac {e^{\frac {-i \pi -x-\log (2)}{4 x^3+8 x^4+4 x^5}} \left (2 x+4 x^2+(3+5 x) (i \pi +\log (2))\right )}{4 x^4+12 x^5+12 x^6+4 x^7} \, dx=\frac {{\mathrm {e}}^{-\frac {\Pi \,1{}\mathrm {i}}{4\,x^5+8\,x^4+4\,x^3}}\,{\mathrm {e}}^{-\frac {x}{4\,x^5+8\,x^4+4\,x^3}}}{2^{\frac {1}{4\,x^5+8\,x^4+4\,x^3}}} \]
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