Integrand size = 119, antiderivative size = 25 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {e^{4/3} x^2}{5 \left (-1+x+\frac {1}{x-\log (x)}\right )} \]
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\[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4/3} x \left (-1+3 x-2 x^2+x^3-2 (-1+x)^2 \log (x)+(-2+x) \log ^2(x)\right )}{5 \left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx \\ & = \frac {1}{5} e^{4/3} \int \frac {x \left (-1+3 x-2 x^2+x^3-2 (-1+x)^2 \log (x)+(-2+x) \log ^2(x)\right )}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx \\ & = \frac {1}{5} e^{4/3} \int \left (\frac {(-2+x) x}{(-1+x)^2}+\frac {x \left (-1+2 x-3 x^2+x^3\right )}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )^2}+\frac {2 x}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )}\right ) \, dx \\ & = \frac {1}{5} e^{4/3} \int \frac {(-2+x) x}{(-1+x)^2} \, dx+\frac {1}{5} e^{4/3} \int \frac {x \left (-1+2 x-3 x^2+x^3\right )}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx+\frac {1}{5} \left (2 e^{4/3}\right ) \int \frac {x}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )} \, dx \\ & = -\frac {e^{4/3} x^2}{5 (1-x)}+\frac {1}{5} e^{4/3} \int \left (-\frac {1}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2}-\frac {1}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )^2}-\frac {2}{(-1+x) \left (1-x+x^2+\log (x)-x \log (x)\right )^2}-\frac {x}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2}+\frac {x^2}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2}\right ) \, dx+\frac {1}{5} \left (2 e^{4/3}\right ) \int \left (\frac {1}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )}+\frac {1}{(-1+x) \left (1-x+x^2+\log (x)-x \log (x)\right )}\right ) \, dx \\ & = -\frac {e^{4/3} x^2}{5 (1-x)}-\frac {1}{5} e^{4/3} \int \frac {1}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx-\frac {1}{5} e^{4/3} \int \frac {1}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx-\frac {1}{5} e^{4/3} \int \frac {x}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx+\frac {1}{5} e^{4/3} \int \frac {x^2}{\left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx-\frac {1}{5} \left (2 e^{4/3}\right ) \int \frac {1}{(-1+x) \left (1-x+x^2+\log (x)-x \log (x)\right )^2} \, dx+\frac {1}{5} \left (2 e^{4/3}\right ) \int \frac {1}{(-1+x)^2 \left (1-x+x^2+\log (x)-x \log (x)\right )} \, dx+\frac {1}{5} \left (2 e^{4/3}\right ) \int \frac {1}{(-1+x) \left (1-x+x^2+\log (x)-x \log (x)\right )} \, dx \\ \end{align*}
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {1}{5} e^{4/3} \left (\frac {1}{-1+x}+x+\frac {x^2}{(-1+x) \left (-1+x-x^2-\log (x)+x \log (x)\right )}\right ) \]
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Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48
method | result | size |
default | \(\frac {{\mathrm e}^{\frac {4}{3}} \left (x^{2} \ln \left (x \right )-x^{3}\right )}{5 x \ln \left (x \right )-5 x^{2}-5 \ln \left (x \right )+5 x -5}\) | \(37\) |
norman | \(\frac {\frac {{\mathrm e}^{\frac {4}{3}} x^{3}}{5}-\frac {{\mathrm e}^{\frac {4}{3}} x^{2} \ln \left (x \right )}{5}}{x^{2}-x \ln \left (x \right )-x +\ln \left (x \right )+1}\) | \(40\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {4}{3}} x^{3}-{\mathrm e}^{\frac {4}{3}} x^{2} \ln \left (x \right )}{5 x^{2}-5 x \ln \left (x \right )-5 x +5 \ln \left (x \right )+5}\) | \(40\) |
risch | \(\frac {{\mathrm e}^{\frac {4}{3}} \left (x^{2}-x +1\right )}{5 x -5}-\frac {{\mathrm e}^{\frac {4}{3}} x^{2}}{5 \left (-1+x \right ) \left (x^{2}-x \ln \left (x \right )-x +\ln \left (x \right )+1\right )}\) | \(48\) |
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=-\frac {{\left (x^{2} - x + 1\right )} e^{\frac {4}{3}} \log \left (x\right ) - {\left (x^{3} - x^{2} + x - 1\right )} e^{\frac {4}{3}}}{5 \, {\left (x^{2} - {\left (x - 1\right )} \log \left (x\right ) - x + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (19) = 38\).
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\frac {4}{3}}}{- 5 x^{3} + 10 x^{2} - 10 x + \left (5 x^{2} - 10 x + 5\right ) \log {\left (x \right )} + 5} + \frac {x e^{\frac {4}{3}}}{5} + \frac {e^{\frac {4}{3}}}{5 x - 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (20) = 40\).
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.36 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{3} e^{\frac {4}{3}} - x^{2} e^{\frac {4}{3}} + x e^{\frac {4}{3}} - {\left (x^{2} e^{\frac {4}{3}} - x e^{\frac {4}{3}} + e^{\frac {4}{3}}\right )} \log \left (x\right ) - e^{\frac {4}{3}}}{5 \, {\left (x^{2} - {\left (x - 1\right )} \log \left (x\right ) - x + 1\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.48 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=\frac {x^{3} e^{\frac {4}{3}} - x^{2} e^{\frac {4}{3}} \log \left (x\right ) - x^{2} e^{\frac {4}{3}} + x e^{\frac {4}{3}} \log \left (x\right ) + x e^{\frac {4}{3}} - e^{\frac {4}{3}} \log \left (x\right ) - e^{\frac {4}{3}}}{5 \, {\left (x^{2} - x \log \left (x\right ) - x + \log \left (x\right ) + 1\right )}} \]
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Time = 8.63 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {e^{4/3} \left (-x+3 x^2-2 x^3+x^4\right )+e^{4/3} \left (-2 x+4 x^2-2 x^3\right ) \log (x)+e^{4/3} \left (-2 x+x^2\right ) \log ^2(x)}{5-10 x+15 x^2-10 x^3+5 x^4+\left (10-20 x+20 x^2-10 x^3\right ) \log (x)+\left (5-10 x+5 x^2\right ) \log ^2(x)} \, dx=-\frac {{\mathrm {e}}^{4/3}\,\left (\ln \left (x\right )-x+x^2\,\ln \left (x\right )-x\,\ln \left (x\right )+x^2-x^3+1\right )}{5\,\left (\ln \left (x\right )-x-x\,\ln \left (x\right )+x^2+1\right )} \]
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