Integrand size = 193, antiderivative size = 32 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^x \left (x+\log \left (1+x-x^2\right )\right )+\log \left (\frac {1}{2}-e^3+x-\log (x)\right ) \]
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Time = 1.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 21, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6873, 6860, 6816, 6820, 2225, 2207, 2209, 2634, 2302} \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^x \log \left (-x^2+x+1\right )-e^x x+\log \left (2 x-2 \log (x)-2 e^3+1\right ) \]
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Rule 2207
Rule 2209
Rule 2225
Rule 2302
Rule 2634
Rule 6816
Rule 6820
Rule 6860
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x \left (1+x-x^2\right ) \left (1-2 e^3+2 x-2 \log (x)\right )} \, dx \\ & = \int \left (\frac {2 (-1+x)}{x \left (1-2 e^3+2 x-2 \log (x)\right )}-\frac {e^x \left (-2+x^3-\log \left (1+x-x^2\right )-x \log \left (1+x-x^2\right )+x^2 \log \left (1+x-x^2\right )\right )}{-1-x+x^2}\right ) \, dx \\ & = 2 \int \frac {-1+x}{x \left (1-2 e^3+2 x-2 \log (x)\right )} \, dx-\int \frac {e^x \left (-2+x^3-\log \left (1+x-x^2\right )-x \log \left (1+x-x^2\right )+x^2 \log \left (1+x-x^2\right )\right )}{-1-x+x^2} \, dx \\ & = \log \left (1-2 e^3+2 x-2 \log (x)\right )-\int \frac {e^x \left (2-x^3-\left (-1-x+x^2\right ) \log \left (1+x-x^2\right )\right )}{1+x-x^2} \, dx \\ & = \log \left (1-2 e^3+2 x-2 \log (x)\right )-\int \left (\frac {e^x \left (-2+x^3\right )}{-1-x+x^2}+e^x \log \left (1+x-x^2\right )\right ) \, dx \\ & = \log \left (1-2 e^3+2 x-2 \log (x)\right )-\int \frac {e^x \left (-2+x^3\right )}{-1-x+x^2} \, dx-\int e^x \log \left (1+x-x^2\right ) \, dx \\ & = -e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right )+\int \frac {e^x (1-2 x)}{1+x-x^2} \, dx-\int \left (e^x+e^x x-\frac {e^x (1-2 x)}{-1-x+x^2}\right ) \, dx \\ & = -e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right )-\int e^x \, dx+\int \left (-\frac {2 e^x}{1-\sqrt {5}-2 x}-\frac {2 e^x}{1+\sqrt {5}-2 x}\right ) \, dx-\int e^x x \, dx+\int \frac {e^x (1-2 x)}{-1-x+x^2} \, dx \\ & = -e^x-e^x x-e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right )-2 \int \frac {e^x}{1-\sqrt {5}-2 x} \, dx-2 \int \frac {e^x}{1+\sqrt {5}-2 x} \, dx+\int e^x \, dx+\int \left (-\frac {2 e^x}{-1-\sqrt {5}+2 x}-\frac {2 e^x}{-1+\sqrt {5}+2 x}\right ) \, dx \\ & = -e^x x+e^{\frac {1}{2} \left (1+\sqrt {5}\right )} \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (-1-\sqrt {5}+2 x\right )\right )+e^{\frac {1}{2}-\frac {\sqrt {5}}{2}} \operatorname {ExpIntegralEi}\left (\frac {1}{2} \left (-1+\sqrt {5}+2 x\right )\right )-e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right )-2 \int \frac {e^x}{-1-\sqrt {5}+2 x} \, dx-2 \int \frac {e^x}{-1+\sqrt {5}+2 x} \, dx \\ & = -e^x x-e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right ) \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-e^x x-e^x \log \left (1+x-x^2\right )+\log \left (1-2 e^3+2 x-2 \log (x)\right ) \]
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Time = 45.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-{\mathrm e}^{x} \ln \left (-x^{2}+x +1\right )-{\mathrm e}^{x} x +\ln \left ({\mathrm e}^{3}-x +\ln \left (x \right )-\frac {1}{2}\right )\) | \(30\) |
parallelrisch | \(-{\mathrm e}^{x} x -{\mathrm e}^{x} \ln \left (-x^{2}+x +1\right )+\ln \left (\frac {1}{2}-{\mathrm e}^{3}+x -\ln \left (x \right )\right )\) | \(32\) |
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-2 \, x + 2 \, e^{3} + 2 \, \log \left (x\right ) - 1\right ) \]
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Time = 0.53 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=\left (- x - \log {\left (- x^{2} + x + 1 \right )}\right ) e^{x} + \log {\left (- x + \log {\left (x \right )} - \frac {1}{2} + e^{3} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-x + e^{3} + \log \left (x\right ) - \frac {1}{2}\right ) \]
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Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=-x e^{x} - e^{x} \log \left (-x^{2} + x + 1\right ) + \log \left (-2 \, x + 2 \, e^{3} + 2 \, \log \left (x\right ) - 1\right ) \]
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Time = 12.80 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+4 x^2-2 x^3+e^x \left (-2 x-4 x^2+x^4+2 x^5+e^3 \left (4 x-2 x^4\right )\right )+e^x \left (-x-3 x^2-x^3+2 x^4+e^3 \left (2 x+2 x^2-2 x^3\right )\right ) \log \left (1+x-x^2\right )+\log (x) \left (e^x \left (4 x-2 x^4\right )+e^x \left (2 x+2 x^2-2 x^3\right ) \log \left (1+x-x^2\right )\right )}{x+3 x^2+x^3-2 x^4+e^3 \left (-2 x-2 x^2+2 x^3\right )+\left (-2 x-2 x^2+2 x^3\right ) \log (x)} \, dx=\ln \left ({\mathrm {e}}^3-x+\ln \left (x\right )-\frac {1}{2}\right )-x\,{\mathrm {e}}^x-{\mathrm {e}}^x\,\ln \left (-x^2+x+1\right ) \]
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