Integrand size = 145, antiderivative size = 25 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=\frac {10}{3}+x+\log (x)-\frac {4}{e^{2 x}-x+x^2+\log (x)} \]
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\[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=\int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+2 e^{2 x} x \left (3+x^2\right )+2 (1+x) \left (e^{2 x}+(-1+x) x\right ) \log (x)+(1+x) \log ^2(x)}{x \left (e^{2 x}+(-1+x) x+\log (x)\right )^2} \, dx \\ & = \int \left (\frac {1+x}{x}+\frac {8}{e^{2 x}-x+x^2+\log (x)}-\frac {4 \left (-1+x-4 x^2+2 x^3+2 x \log (x)\right )}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2}\right ) \, dx \\ & = -\left (4 \int \frac {-1+x-4 x^2+2 x^3+2 x \log (x)}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx\right )+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+\int \frac {1+x}{x} \, dx \\ & = -\left (4 \int \left (\frac {1}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}-\frac {1}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2}-\frac {4 x}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}+\frac {2 x^2}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}+\frac {2 \log (x)}{\left (e^{2 x}-x+x^2+\log (x)\right )^2}\right ) \, dx\right )+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+\int \left (1+\frac {1}{x}\right ) \, dx \\ & = x+\log (x)-4 \int \frac {1}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx+4 \int \frac {1}{x \left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx-8 \int \frac {x^2}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx-8 \int \frac {\log (x)}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx+8 \int \frac {1}{e^{2 x}-x+x^2+\log (x)} \, dx+16 \int \frac {x}{\left (e^{2 x}-x+x^2+\log (x)\right )^2} \, dx \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=x+\log (x)-\frac {4}{e^{2 x}-x+x^2+\log (x)} \]
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Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x +\ln \left (x \right )-\frac {4}{x^{2}+{\mathrm e}^{2 x}+\ln \left (x \right )-x}\) | \(22\) |
parallelrisch | \(\frac {x^{2} \ln \left (x \right )+x^{3}+\ln \left (x \right )^{2}+{\mathrm e}^{2 x} \ln \left (x \right )+x \,{\mathrm e}^{2 x}-x^{2}-4}{x^{2}+{\mathrm e}^{2 x}+\ln \left (x \right )-x}\) | \(50\) |
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Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x^{3} - x^{2} + x e^{\left (2 \, x\right )} + {\left (x^{2} + e^{\left (2 \, x\right )}\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \left (x\right )} \]
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Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=x + \log {\left (x \right )} - \frac {4}{x^{2} - x + e^{2 x} + \log {\left (x \right )}} \]
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x^{3} - x^{2} + x e^{\left (2 \, x\right )} + x \log \left (x\right ) - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \left (x\right )} + \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x^{3} + x^{2} \log \left (x\right ) - x^{2} + x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 4}{x^{2} - x + e^{\left (2 \, x\right )} + \log \left (x\right )} \]
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Time = 12.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {4-4 x+9 x^2-x^3-x^4+x^5+e^{4 x} (1+x)+e^{2 x} \left (6 x+2 x^3\right )+\left (-2 x+2 x^3+e^{2 x} (2+2 x)\right ) \log (x)+(1+x) \log ^2(x)}{e^{4 x} x+x^3-2 x^4+x^5+e^{2 x} \left (-2 x^2+2 x^3\right )+\left (2 e^{2 x} x-2 x^2+2 x^3\right ) \log (x)+x \log ^2(x)} \, dx=x+\ln \left (x\right )-\frac {4}{{\mathrm {e}}^{2\,x}-x+\ln \left (x\right )+x^2} \]
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