Integrand size = 86, antiderivative size = 28 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {1608, 28, 2326} \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log (x)}} \left (3 x^2+3 i \pi -\log \left (\frac {125}{8}\right )\right )}{3 \left (x^2+i \pi -\log \left (\frac {5}{2}\right )\right )^2} \]
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Rule 28
Rule 1608
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{x \left (2 x^4+4 x^2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx \\ & = 2 \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{x \left (2 x^2+2 \left (i \pi -\log \left (\frac {5}{2}\right )\right )\right )^2 \log ^2(x)} \, dx \\ & = \frac {e^{\frac {3}{2 \log (x)}} \left (3 i \pi +3 x^2-\log \left (\frac {125}{8}\right )\right )}{3 \left (i \pi +x^2-\log \left (\frac {5}{2}\right )\right )^2} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log (x)}}}{i \pi +x^2-\log \left (\frac {5}{2}\right )} \]
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Time = 5.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(-\frac {i {\mathrm e}^{\frac {3}{2 \ln \left (x \right )}}}{-i x^{2}-i \ln \left (\frac {2}{5}\right )+\pi }\) | \(26\) |
risch | \(-\frac {i {\mathrm e}^{\frac {3}{2 \ln \left (x \right )}}}{-i x^{2}+i \ln \left (5\right )-i \ln \left (2\right )+\pi }\) | \(31\) |
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{i \, \pi + x^{2} + \log \left (\frac {2}{5}\right )} \]
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Time = 0.56 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\frac {e^{\frac {3}{2 \log {\left (x \right )}}}}{x^{2} - \log {\left (5 \right )} + \log {\left (2 \right )} + i \pi } \]
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\[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\int { \frac {{\left (-3 i \, \pi - 4 \, x^{2} \log \left (x\right )^{2} - 3 \, x^{2} - 3 \, \log \left (\frac {2}{5}\right )\right )} e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{2 \, {\left (x^{5} + 2 \, {\left (i \, \pi + \log \left (\frac {2}{5}\right )\right )} x^{3} + {\left (i \, \pi + \log \left (\frac {2}{5}\right )\right )}^{2} x\right )} \log \left (x\right )^{2}} \,d x } \]
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Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=-\frac {i \, e^{\left (\frac {3}{2 \, \log \left (x\right )}\right )}}{\pi - i \, x^{2} + i \, \log \left (5\right ) - i \, \log \left (2\right )} \]
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Timed out. \[ \int \frac {e^{\frac {3}{2 \log (x)}} \left (-3 x^2-3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )-4 x^2 \log ^2(x)\right )}{\left (2 x^5+4 x^3 \left (i \pi -\log \left (\frac {5}{2}\right )\right )+2 x \left (i \pi -\log \left (\frac {5}{2}\right )\right )^2\right ) \log ^2(x)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {3}{2\,\ln \left (x\right )}}\,\left (4\,x^2\,{\ln \left (x\right )}^2+3\,x^2+\Pi \,3{}\mathrm {i}+3\,\ln \left (\frac {2}{5}\right )\right )}{{\ln \left (x\right )}^2\,\left (2\,x\,{\left (\ln \left (\frac {2}{5}\right )+\Pi \,1{}\mathrm {i}\right )}^2+4\,x^3\,\left (\ln \left (\frac {2}{5}\right )+\Pi \,1{}\mathrm {i}\right )+2\,x^5\right )} \,d x \]
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