Integrand size = 105, antiderivative size = 31 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\log \left (\frac {3}{x+x^2-\left (e^{x^2+\frac {5+x}{x}}+5 x\right )^2}\right ) \]
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\[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \left (-5+2 x^3\right )}{x^2}+\frac {10-50 e^{1+\frac {5}{x}+x^2}-239 x-10 e^{1+\frac {5}{x}+x^2} x-48 x^2-4 x^3+20 e^{1+\frac {5}{x}+x^2} x^3+96 x^4}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )}\right ) \, dx \\ & = -\left (2 \int \frac {-5+2 x^3}{x^2} \, dx\right )+\int \frac {10-50 e^{1+\frac {5}{x}+x^2}-239 x-10 e^{1+\frac {5}{x}+x^2} x-48 x^2-4 x^3+20 e^{1+\frac {5}{x}+x^2} x^3+96 x^4}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )} \, dx \\ & = -\left (2 \int \left (-\frac {5}{x^2}+2 x\right ) \, dx\right )+\int \left (-\frac {239}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}-\frac {10 e^{1+\frac {5}{x}+x^2}}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}+\frac {10}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )}-\frac {50 e^{1+\frac {5}{x}+x^2}}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )}-\frac {48 x}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}-\frac {4 x^2}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}+\frac {20 e^{1+\frac {5}{x}+x^2} x^2}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}+\frac {96 x^3}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2}\right ) \, dx \\ & = -\frac {10}{x}-2 x^2-4 \int \frac {x^2}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx-10 \int \frac {e^{1+\frac {5}{x}+x^2}}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx+10 \int \frac {1}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )} \, dx+20 \int \frac {e^{1+\frac {5}{x}+x^2} x^2}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx-48 \int \frac {x}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx-50 \int \frac {e^{1+\frac {5}{x}+x^2}}{x \left (e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2\right )} \, dx+96 \int \frac {x^3}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx-239 \int \frac {1}{e^{2+\frac {10}{x}+2 x^2}-x+10 e^{1+\frac {5}{x}+x^2} x+24 x^2} \, dx \\ \end{align*}
\[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx \]
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Time = 0.29 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29
method | result | size |
norman | \(-\ln \left (10 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}+24 x^{2}+{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}-x \right )\) | \(40\) |
parallelrisch | \(-\ln \left (x^{2}+\frac {5 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}}{12}+\frac {{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}}{24}-\frac {x}{24}\right )\) | \(40\) |
risch | \(-2 x^{2}-\frac {10}{x}+\frac {2 x^{3}+2 x +10}{x}-\ln \left (10 x \,{\mathrm e}^{\frac {x^{3}+x +5}{x}}+24 x^{2}+{\mathrm e}^{\frac {2 x^{3}+2 x +10}{x}}-x \right )\) | \(61\) |
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Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\log \left (24 \, x^{2} + 10 \, x e^{\left (\frac {x^{3} + x + 5}{x}\right )} - x + e^{\left (\frac {2 \, {\left (x^{3} + x + 5\right )}}{x}\right )}\right ) \]
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Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=- \log {\left (24 x^{2} + 10 x e^{\frac {x^{3} + x + 5}{x}} - x + e^{\frac {2 \left (x^{3} + x + 5\right )}{x}} \right )} \]
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Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\frac {10}{x} - \log \left ({\left (24 \, x^{2} + 10 \, x e^{\left (x^{2} + \frac {5}{x} + 1\right )} - x + e^{\left (2 \, x^{2} + \frac {10}{x} + 2\right )}\right )} e^{\left (-\frac {10}{x} - 2\right )}\right ) \]
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Time = 0.35 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=-\log \left (24 \, x^{2} + 10 \, x e^{\left (\frac {x^{3} + x + 5}{x}\right )} - x + e^{\left (\frac {2 \, {\left (x^{3} + x + 5\right )}}{x}\right )}\right ) \]
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Timed out. \[ \int \frac {x^2-48 x^3+e^{\frac {2 \left (5+x+x^3\right )}{x}} \left (10-4 x^3\right )+e^{\frac {5+x+x^3}{x}} \left (50 x-10 x^2-20 x^4\right )}{e^{\frac {2 \left (5+x+x^3\right )}{x}} x^2-x^3+10 e^{\frac {5+x+x^3}{x}} x^3+24 x^4} \, dx=\int -\frac {{\mathrm {e}}^{\frac {2\,\left (x^3+x+5\right )}{x}}\,\left (4\,x^3-10\right )+{\mathrm {e}}^{\frac {x^3+x+5}{x}}\,\left (20\,x^4+10\,x^2-50\,x\right )-x^2+48\,x^3}{10\,x^3\,{\mathrm {e}}^{\frac {x^3+x+5}{x}}+x^2\,{\mathrm {e}}^{\frac {2\,\left (x^3+x+5\right )}{x}}-x^3+24\,x^4} \,d x \]
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