Integrand size = 78, antiderivative size = 33 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {11}{3}-e^{2 x}-x+(1-x) x+\frac {x}{2+\frac {\log (x)}{x}} \]
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\[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{(2 x+\log (x))^2} \, dx \\ & = \int \left (-2 e^{2 x}-\frac {x}{(2 x+\log (x))^2}+\frac {2 x^2}{(2 x+\log (x))^2}-\frac {8 x^3}{(2 x+\log (x))^2}+\frac {2 x \log (x)}{(2 x+\log (x))^2}-\frac {8 x^2 \log (x)}{(2 x+\log (x))^2}-\frac {2 x \log ^2(x)}{(2 x+\log (x))^2}\right ) \, dx \\ & = -\left (2 \int e^{2 x} \, dx\right )+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \frac {x \log (x)}{(2 x+\log (x))^2} \, dx-2 \int \frac {x \log ^2(x)}{(2 x+\log (x))^2} \, dx-8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-8 \int \frac {x^2 \log (x)}{(2 x+\log (x))^2} \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ & = -e^{2 x}+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \left (-\frac {2 x^2}{(2 x+\log (x))^2}+\frac {x}{2 x+\log (x)}\right ) \, dx-2 \int \left (x+\frac {4 x^3}{(2 x+\log (x))^2}-\frac {4 x^2}{2 x+\log (x)}\right ) \, dx-8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-8 \int \left (-\frac {2 x^3}{(2 x+\log (x))^2}+\frac {x^2}{2 x+\log (x)}\right ) \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ & = -e^{2 x}-x^2+2 \int \frac {x^2}{(2 x+\log (x))^2} \, dx+2 \int \frac {x}{2 x+\log (x)} \, dx-4 \int \frac {x^2}{(2 x+\log (x))^2} \, dx-2 \left (8 \int \frac {x^3}{(2 x+\log (x))^2} \, dx\right )+16 \int \frac {x^3}{(2 x+\log (x))^2} \, dx-\int \frac {x}{(2 x+\log (x))^2} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-e^{2 x}-x^2+\frac {x^2}{2 x+\log (x)} \]
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-x^{2}-{\mathrm e}^{2 x}+\frac {x^{2}}{2 x +\ln \left (x \right )}\) | \(25\) |
parallelrisch | \(\frac {-2 x^{3}-x^{2} \ln \left (x \right )-2 x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (x \right )+x^{2}}{2 x +\ln \left (x \right )}\) | \(41\) |
default | \(-\frac {x^{2} \ln \left (x \right )-x^{2}+2 x^{3}}{2 x +\ln \left (x \right )}+\frac {-2 x \,{\mathrm e}^{2 x}-{\mathrm e}^{2 x} \ln \left (x \right )}{2 x +\ln \left (x \right )}\) | \(54\) |
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Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} - x^{2} + 2 \, x e^{\left (2 \, x\right )} + {\left (x^{2} + e^{\left (2 \, x\right )}\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]
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Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.52 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=- x^{2} + \frac {x^{2}}{2 x + \log {\left (x \right )}} - e^{2 x} \]
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Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} + x^{2} \log \left (x\right ) - x^{2} + {\left (2 \, x + \log \left (x\right )\right )} e^{\left (2 \, x\right )}}{2 \, x + \log \left (x\right )} \]
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Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=-\frac {2 \, x^{3} + x^{2} \log \left (x\right ) - x^{2} + 2 \, x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x\right )}{2 \, x + \log \left (x\right )} \]
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Time = 9.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {-x+2 x^2-8 e^{2 x} x^2-8 x^3+\left (2 x-8 e^{2 x} x-8 x^2\right ) \log (x)+\left (-2 e^{2 x}-2 x\right ) \log ^2(x)}{4 x^2+4 x \log (x)+\log ^2(x)} \, dx=\frac {x^3}{x\,\ln \left (x\right )+2\,x^2}-x^2-{\mathrm {e}}^{2\,x} \]
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