Integrand size = 85, antiderivative size = 22 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x^2 \left (9 e^4+x\right )^2}{5-x+\log (x)} \]
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\[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (9 e^4+x\right ) \left (-9 e^4 (-9+x)-x (-19+3 x)+2 \left (9 e^4+2 x\right ) \log (x)\right )}{(5-x+\log (x))^2} \, dx \\ & = \int \left (\frac {(-1+x) x \left (9 e^4+x\right )^2}{(-5+x-\log (x))^2}-\frac {2 x \left (81 e^8+27 e^4 x+2 x^2\right )}{-5+x-\log (x)}\right ) \, dx \\ & = -\left (2 \int \frac {x \left (81 e^8+27 e^4 x+2 x^2\right )}{-5+x-\log (x)} \, dx\right )+\int \frac {(-1+x) x \left (9 e^4+x\right )^2}{(-5+x-\log (x))^2} \, dx \\ & = -\left (2 \int \left (\frac {81 e^8 x}{-5+x-\log (x)}+\frac {27 e^4 x^2}{-5+x-\log (x)}+\frac {2 x^3}{-5+x-\log (x)}\right ) \, dx\right )+\int \left (-\frac {81 e^8 x}{(-5+x-\log (x))^2}+\frac {9 e^4 \left (-2+9 e^4\right ) x^2}{(-5+x-\log (x))^2}+\frac {\left (-1+18 e^4\right ) x^3}{(-5+x-\log (x))^2}+\frac {x^4}{(-5+x-\log (x))^2}\right ) \, dx \\ & = -\left (4 \int \frac {x^3}{-5+x-\log (x)} \, dx\right )-\left (54 e^4\right ) \int \frac {x^2}{-5+x-\log (x)} \, dx-\left (81 e^8\right ) \int \frac {x}{(-5+x-\log (x))^2} \, dx-\left (162 e^8\right ) \int \frac {x}{-5+x-\log (x)} \, dx-\left (9 e^4 \left (2-9 e^4\right )\right ) \int \frac {x^2}{(-5+x-\log (x))^2} \, dx+\left (-1+18 e^4\right ) \int \frac {x^3}{(-5+x-\log (x))^2} \, dx+\int \frac {x^4}{(-5+x-\log (x))^2} \, dx \\ \end{align*}
Time = 0.92 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x^2 \left (9 e^4+x\right )^2}{-5+x-\log (x)} \]
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Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
risch | \(-\frac {\left (81 \,{\mathrm e}^{8}+18 x \,{\mathrm e}^{4}+x^{2}\right ) x^{2}}{-\ln \left (x \right )+x -5}\) | \(28\) |
default | \(\frac {x^{4}+18 x^{3} {\mathrm e}^{4}+81 x^{2} {\mathrm e}^{8}}{\ln \left (x \right )-x +5}\) | \(31\) |
parallelrisch | \(-\frac {x^{4}+18 x^{3} {\mathrm e}^{4}+81 x^{2} {\mathrm e}^{8}}{-\ln \left (x \right )+x -5}\) | \(32\) |
norman | \(\frac {-x^{4}-81 x^{2} {\mathrm e}^{8}-18 x^{3} {\mathrm e}^{4}}{-\ln \left (x \right )+x -5}\) | \(33\) |
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Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x^{4} + 18 \, x^{3} e^{4} + 81 \, x^{2} e^{8}}{x - \log \left (x\right ) - 5} \]
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Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x^{4} + 18 x^{3} e^{4} + 81 x^{2} e^{8}}{- x + \log {\left (x \right )} + 5} \]
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x^{4} + 18 \, x^{3} e^{4} + 81 \, x^{2} e^{8}}{x - \log \left (x\right ) - 5} \]
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Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=-\frac {x^{4} + 18 \, x^{3} e^{4} + 81 \, x^{2} e^{8}}{x - \log \left (x\right ) - 5} \]
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Time = 7.66 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {19 x^3-3 x^4+e^8 \left (729 x-81 x^2\right )+e^4 \left (252 x^2-36 x^3\right )+\left (162 e^8 x+54 e^4 x^2+4 x^3\right ) \log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx=\frac {x^2\,{\left (x+9\,{\mathrm {e}}^4\right )}^2}{\ln \left (x\right )-x+5} \]
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