Integrand size = 83, antiderivative size = 29 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\log \left (-2+x-\left (-\frac {e^{-x}}{5}+x\right )^2-\log (4)-\log (2 x)\right ) \]
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\[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1-x+2 x^2}{x \left (2-x+x^2+\log (8 x)\right )}+\frac {-1-2 x^3+10 e^x x^4+20 e^x x^2 (1+\log (2))-10 e^x x (1+\log (4))-3 x \left (1+\frac {\log (16)}{3}\right )-2 x \log (2 x)-10 e^x x \log (2 x)+10 e^x x^2 \log (2 x)}{x \left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}\right ) \, dx \\ & = \int \frac {1-x+2 x^2}{x \left (2-x+x^2+\log (8 x)\right )} \, dx+\int \frac {-1-2 x^3+10 e^x x^4+20 e^x x^2 (1+\log (2))-10 e^x x (1+\log (4))-3 x \left (1+\frac {\log (16)}{3}\right )-2 x \log (2 x)-10 e^x x \log (2 x)+10 e^x x^2 \log (2 x)}{x \left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx \\ & = \log \left (2-x+x^2+\log (8 x)\right )+\int \left (\frac {1}{x \left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {2 x^2}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {2 \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x x^3}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {20 e^x x (1+\log (2))}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x (-1-\log (4))}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {-3-\log (16)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}+\frac {10 e^x x \log (2 x)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )}\right ) \, dx \\ & = \log \left (2-x+x^2+\log (8 x)\right )+2 \int \frac {x^2}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+2 \int \frac {\log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x \log (2 x)}{\left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x x^3}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+10 \int \frac {e^x x \log (2 x)}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+(20 (1+\log (2))) \int \frac {e^x x}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx-(10 (1+\log (4))) \int \frac {e^x}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+(-3-\log (16)) \int \frac {1}{\left (1-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+50 e^{2 x} (1+\log (2))+25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx+\int \frac {1}{x \left (-1+10 e^x x+25 e^{2 x} x-25 e^{2 x} x^2-50 e^{2 x} (1+\log (2))-25 e^{2 x} \log (2 x)\right ) \left (2-x+x^2+\log (8 x)\right )} \, dx \\ \end{align*}
Time = 5.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 x+\log \left (1+50 e^{2 x}-10 e^x x-25 e^{2 x} x+25 e^{2 x} x^2+25 e^{2 x} \log (4)+25 e^{2 x} \log (2 x)\right ) \]
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Time = 0.71 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\ln \left (\ln \left (2 x \right )+x^{2}+2 \ln \left (2\right )-\frac {2 x \,{\mathrm e}^{-x}}{5}-x +2+\frac {{\mathrm e}^{-2 x}}{25}\right )\) | \(31\) |
parallelrisch | \(\ln \left ({\mathrm e}^{2 x} x^{2}+2 \ln \left (2\right ) {\mathrm e}^{2 x}+{\mathrm e}^{2 x} \ln \left (2 x \right )-x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{2 x}-\frac {2 \,{\mathrm e}^{x} x}{5}+\frac {1}{25}\right )-2 x\) | \(51\) |
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Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\log \left ({\left (25 \, {\left (x^{2} - x + 2 \, \log \left (2\right ) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 25 \, e^{\left (2 \, x\right )} \log \left (2 \, x\right ) + 1\right )} e^{\left (-2 \, x\right )}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (24) = 48\).
Time = 9.37 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.76 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=- 2 x + \log {\left (- \frac {2 x e^{x}}{5 x^{2} - 5 x + 5 \log {\left (2 x \right )} + 10 \log {\left (2 \right )} + 10} + e^{2 x} + \frac {1}{25 x^{2} - 25 x + 25 \log {\left (2 x \right )} + 50 \log {\left (2 \right )} + 50} \right )} + \log {\left (x^{2} - x + \log {\left (2 x \right )} + 2 \log {\left (2 \right )} + 2 \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (28) = 56\).
Time = 0.31 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 \, x + \log \left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right ) + \log \left (\frac {25 \, {\left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right )} e^{\left (2 \, x\right )} - 10 \, x e^{x} + 1}{25 \, {\left (x^{2} - x + 3 \, \log \left (2\right ) + \log \left (x\right ) + 2\right )}}\right ) \]
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Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=-2 \, x + \log \left (25 \, x^{2} e^{\left (2 \, x\right )} - 25 \, x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 75 \, e^{\left (2 \, x\right )} \log \left (2\right ) + 25 \, e^{\left (2 \, x\right )} \log \left (x\right ) + 50 \, e^{\left (2 \, x\right )} + 1\right ) \]
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Timed out. \[ \int \frac {-2 x+e^x \left (-10 x+10 x^2\right )+e^{2 x} \left (25-25 x+50 x^2\right )}{x-10 e^x x^2+e^{2 x} \left (50 x-25 x^2+25 x^3+25 x \log (4)\right )+25 e^{2 x} x \log (2 x)} \, dx=\int -\frac {2\,x-{\mathrm {e}}^{2\,x}\,\left (50\,x^2-25\,x+25\right )+{\mathrm {e}}^x\,\left (10\,x-10\,x^2\right )}{x-10\,x^2\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (50\,x+50\,x\,\ln \left (2\right )-25\,x^2+25\,x^3\right )+25\,x\,\ln \left (2\,x\right )\,{\mathrm {e}}^{2\,x}} \,d x \]
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