Integrand size = 66, antiderivative size = 29 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=2+e^{\frac {1}{4} (5-x) x^2}+\frac {2}{1-\frac {1}{x}}+\log (x) \]
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Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1608, 27, 12, 6874, 6838, 907} \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=e^{\frac {1}{4} (5-x) x^2}-\frac {2}{1-x}+\log (x) \]
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Rule 12
Rule 27
Rule 907
Rule 1608
Rule 6838
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{x \left (4-8 x+4 x^2\right )} \, dx \\ & = \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 (-1+x)^2 x} \, dx \\ & = \frac {1}{4} \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{(-1+x)^2 x} \, dx \\ & = \frac {1}{4} \int \left (-e^{-\frac {1}{4} (-5+x) x^2} x (-10+3 x)+\frac {4 \left (1-4 x+x^2\right )}{(-1+x)^2 x}\right ) \, dx \\ & = -\left (\frac {1}{4} \int e^{-\frac {1}{4} (-5+x) x^2} x (-10+3 x) \, dx\right )+\int \frac {1-4 x+x^2}{(-1+x)^2 x} \, dx \\ & = e^{\frac {1}{4} (5-x) x^2}+\int \left (-\frac {2}{(-1+x)^2}+\frac {1}{x}\right ) \, dx \\ & = e^{\frac {1}{4} (5-x) x^2}-\frac {2}{1-x}+\log (x) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=e^{-\frac {1}{4} (-5+x) x^2}+\frac {2}{-1+x}+\log (x) \]
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Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {2}{-1+x}+\ln \left (x \right )+{\mathrm e}^{-\frac {x^{2} \left (-5+x \right )}{4}}\) | \(20\) |
parts | \(\frac {2}{-1+x}+\ln \left (x \right )+{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}}\) | \(23\) |
norman | \(\frac {{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}} x -{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}}+2}{-1+x}+\ln \left (x \right )\) | \(40\) |
parallelrisch | \(\frac {4 x \ln \left (x \right )+4 \,{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}} x +8-4 \ln \left (x \right )-4 \,{\mathrm e}^{-\frac {1}{4} x^{3}+\frac {5}{4} x^{2}}}{-4+4 x}\) | \(48\) |
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {{\left (x - 1\right )} e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + {\left (x - 1\right )} \log \left (x\right ) + 2}{x - 1} \]
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Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=e^{- \frac {x^{3}}{4} + \frac {5 x^{2}}{4}} + \log {\left (x \right )} + \frac {2}{x - 1} \]
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {2}{x - 1} + e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + \log \left (x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx=\frac {x e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} + x \log \left (x\right ) - e^{\left (-\frac {1}{4} \, x^{3} + \frac {5}{4} \, x^{2}\right )} - \log \left (x\right ) + 2}{x - 1} \]
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Time = 15.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {4-16 x+4 x^2+e^{\frac {1}{4} \left (5 x^2-x^3\right )} \left (10 x^2-23 x^3+16 x^4-3 x^5\right )}{4 x-8 x^2+4 x^3} \, dx={\mathrm {e}}^{\frac {5\,x^2}{4}-\frac {x^3}{4}}+\ln \left (x\right )+\frac {2}{x-1} \]
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