Integrand size = 131, antiderivative size = 36 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=1-2 x-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right ) \]
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\[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=\int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-e^{\frac {(1+x)^2}{4 x^2}} \left (1+x+4 x^2\right )+x^2 \log (16)}{2 x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}-\log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )\right ) \, dx \\ & = \frac {1}{2} \int \frac {-e^{\frac {(1+x)^2}{4 x^2}} \left (1+x+4 x^2\right )+x^2 \log (16)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\int \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right ) \, dx \\ & = -x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \left (\frac {-1-x-4 x^2}{x^2}-\frac {(1+x) \log (2)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx+\int \frac {e^{\frac {(1+x)^2}{4 x^2}} (1+x)}{2 x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx \\ & = -x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \frac {-1-x-4 x^2}{x^2} \, dx+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}} (1+x)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1+x}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx \\ & = -x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \left (-4-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx+\frac {1}{2} \int \left (\frac {e^{\frac {(1+x)^2}{4 x^2}}}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}+\frac {e^{\frac {(1+x)^2}{4 x^2}}}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx-\frac {1}{2} \log (2) \int \left (\frac {1}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}+\frac {1}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx \\ & = \frac {1}{2 x}-2 x-\frac {\log (x)}{2}-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}}}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}}}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx \\ \end{align*}
Time = 1.62 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=-x \left (2+\log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )\right ) \]
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Time = 0.99 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-x \ln \left (-\frac {{\mathrm e}^{{\mathrm e}^{2}}}{-{\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}+\ln \left (2\right )}\right )-2 x\) | \(35\) |
norman | \(\frac {-2 x^{2}-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}}}{{\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \left (2\right )}\right )}{x}\) | \(42\) |
default | \(\frac {\left (-2 \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}}}{{\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \left (2\right )}\right )-2 \ln \left ({\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \left (2\right )\right )-4\right ) x^{2}+2 x^{2} \ln \left ({\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \left (2\right )\right )}{2 x}\) | \(88\) |
risch | \(x \ln \left (-{\mathrm e}^{\frac {\left (1+x \right )^{2}}{4 x^{2}}}+\ln \left (2\right )\right )-\frac {x \left (-2 i \pi \operatorname {csgn}\left (\frac {i}{-{\mathrm e}^{\frac {\left (1+x \right )^{2}}{4 x^{2}}}+\ln \left (2\right )}\right )^{2}+2 i \pi \operatorname {csgn}\left (\frac {i}{-{\mathrm e}^{\frac {\left (1+x \right )^{2}}{4 x^{2}}}+\ln \left (2\right )}\right )^{3}+2 i \pi +2 \,{\mathrm e}^{2}+4\right )}{2}\) | \(90\) |
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Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=-x \log \left (\frac {e^{\left (e^{2}\right )}}{e^{\left (\frac {x^{2} + 2 \, x + 1}{4 \, x^{2}}\right )} - \log \left (2\right )}\right ) - 2 \, x \]
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Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=- x \log {\left (\frac {e^{e^{2}}}{e^{\frac {\frac {x^{2}}{4} + \frac {x}{2} + \frac {1}{4}}{x^{2}}} - \log {\left (2 \right )}} \right )} - 2 x \]
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Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=-x {\left (e^{2} + 2\right )} + x \log \left (e^{\left (\frac {1}{2 \, x} + \frac {1}{4 \, x^{2}} + \frac {1}{4}\right )} - \log \left (2\right )\right ) \]
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Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=-x e^{2} + x \log \left (e^{\left (\frac {x^{2} + 2 \, x + 1}{4 \, x^{2}}\right )} - \log \left (2\right )\right ) - 2 \, x \]
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Time = 9.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx=-2\,x-x\,{\mathrm {e}}^2-x\,\ln \left (-\frac {1}{\ln \left (2\right )-\sqrt {{\mathrm {e}}^{1/x}}\,{\left ({\mathrm {e}}^{\frac {1}{x^2}}\right )}^{1/4}\,{\mathrm {e}}^{1/4}}\right ) \]
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