Integrand size = 93, antiderivative size = 27 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=-e^{-5+x}-\frac {x}{-2+x}+x \left (e^x+x\right ) \log ^2(x) \]
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\[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=\int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{(-2+x)^2} \, dx \\ & = \int \left (-e^{-5+x}+\frac {2}{(-2+x)^2}+2 \left (e^x+x\right ) \log (x)+\left (2 x+e^x (1+x)\right ) \log ^2(x)\right ) \, dx \\ & = \frac {2}{2-x}+2 \int \left (e^x+x\right ) \log (x) \, dx-\int e^{-5+x} \, dx+\int \left (2 x+e^x (1+x)\right ) \log ^2(x) \, dx \\ & = -e^{-5+x}+\frac {2}{2-x}+2 e^x \log (x)+x^2 \log (x)-2 \int \left (\frac {e^x}{x}+\frac {x}{2}\right ) \, dx+\int \left (2 x \log ^2(x)+e^x (1+x) \log ^2(x)\right ) \, dx \\ & = -e^{-5+x}+\frac {2}{2-x}-\frac {x^2}{2}+2 e^x \log (x)+x^2 \log (x)-2 \int \frac {e^x}{x} \, dx+2 \int x \log ^2(x) \, dx+\int e^x (1+x) \log ^2(x) \, dx \\ & = -e^{-5+x}+\frac {2}{2-x}-\frac {x^2}{2}-2 \operatorname {ExpIntegralEi}(x)+2 e^x \log (x)+x^2 \log (x)+x^2 \log ^2(x)-2 \int x \log (x) \, dx+\int \left (e^x \log ^2(x)+e^x x \log ^2(x)\right ) \, dx \\ & = -e^{-5+x}+\frac {2}{2-x}-2 \operatorname {ExpIntegralEi}(x)+2 e^x \log (x)+x^2 \log ^2(x)+\int e^x \log ^2(x) \, dx+\int e^x x \log ^2(x) \, dx \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=-e^{-5+x}-\frac {2}{-2+x}+x \left (e^x+x\right ) \log ^2(x) \]
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Time = 0.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
default | \(x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-\frac {2}{-2+x}+x^{2} \ln \left (x \right )^{2}-{\mathrm e}^{-5+x}\) | \(31\) |
parts | \(x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-\frac {2}{-2+x}+x^{2} \ln \left (x \right )^{2}-{\mathrm e}^{-5+x}\) | \(31\) |
risch | \(\left ({\mathrm e}^{x} x +x^{2}\right ) \ln \left (x \right )^{2}-\frac {x \,{\mathrm e}^{-5+x}-2 \,{\mathrm e}^{-5+x}+2}{-2+x}\) | \(36\) |
parallelrisch | \(\frac {2 x^{3} \ln \left (x \right )^{2}+2 x^{2} {\mathrm e}^{x} \ln \left (x \right )^{2}-4 x^{2} \ln \left (x \right )^{2}-4 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-2 x \,{\mathrm e}^{-5+x}-4+4 \,{\mathrm e}^{-5+x}}{2 x -4}\) | \(61\) |
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Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=\frac {{\left ({\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{5} + {\left (x^{2} - 2 \, x\right )} e^{\left (x + 5\right )}\right )} \log \left (x\right )^{2} - {\left (x - 2\right )} e^{x} - 2 \, e^{5}\right )} e^{\left (-5\right )}}{x - 2} \]
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Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=x^{2} \log {\left (x \right )}^{2} + \frac {\left (x e^{5} \log {\left (x \right )}^{2} - 1\right ) e^{x}}{e^{5}} - \frac {2}{x - 2} \]
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\[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=\int { \frac {{\left (2 \, x^{3} - 8 \, x^{2} + {\left (x^{3} - 3 \, x^{2} + 4\right )} e^{x} + 8 \, x\right )} \log \left (x\right )^{2} - {\left (x^{2} - 4 \, x + 4\right )} e^{\left (x - 5\right )} + 2 \, {\left (x^{3} - 4 \, x^{2} + {\left (x^{2} - 4 \, x + 4\right )} e^{x} + 4 \, x\right )} \log \left (x\right ) + 2}{x^{2} - 4 \, x + 4} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.59 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx=\frac {x^{3} e^{5} \log \left (x\right )^{2} - 2 \, x^{2} e^{5} \log \left (x\right )^{2} + x^{2} e^{\left (x + 5\right )} \log \left (x\right )^{2} - 2 \, x e^{\left (x + 5\right )} \log \left (x\right )^{2} - x e^{x} - 2 \, e^{5} + 2 \, e^{x}}{x e^{5} - 2 \, e^{5}} \]
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Time = 9.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {2+e^{-5+x} \left (-4+4 x-x^2\right )+\left (8 x-8 x^2+2 x^3+e^x \left (8-8 x+2 x^2\right )\right ) \log (x)+\left (8 x-8 x^2+2 x^3+e^x \left (4-3 x^2+x^3\right )\right ) \log ^2(x)}{4-4 x+x^2} \, dx={\ln \left (x\right )}^2\,\left (x\,{\mathrm {e}}^x+x^2\right )-\frac {2}{x-2}-{\mathrm {e}}^{x-5} \]
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