Optimal. Leaf size=24 \[ 2+x-5 \left (x^2-\frac {x}{1+e^{-1+x}+x}\right )^2 \]
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Rubi [F] time = 1.69, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1-7 x+33 x^2+31 x^3-40 x^4-60 x^5-20 x^6+e^{-3+3 x} \left (1-20 x^3\right )+e^{-2+2 x} \left (3+3 x+30 x^2-70 x^3-60 x^4\right )+e^{-1+x} \left (3-4 x+73 x^2-20 x^3-130 x^4-60 x^5\right )}{1+e^{-3+3 x}+3 x+3 x^2+x^3+e^{-2+2 x} (3+3 x)+e^{-1+x} \left (3+6 x+3 x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (1-7 x+33 x^2+31 x^3-40 x^4-60 x^5-20 x^6+e^{-3+3 x} \left (1-20 x^3\right )+e^{-2+2 x} \left (3+3 x+30 x^2-70 x^3-60 x^4\right )+e^{-1+x} \left (3-4 x+73 x^2-20 x^3-130 x^4-60 x^5\right )\right )}{\left (e+e^x+e x\right )^3} \, dx\\ &=e^3 \int \frac {1-7 x+33 x^2+31 x^3-40 x^4-60 x^5-20 x^6+e^{-3+3 x} \left (1-20 x^3\right )+e^{-2+2 x} \left (3+3 x+30 x^2-70 x^3-60 x^4\right )+e^{-1+x} \left (3-4 x+73 x^2-20 x^3-130 x^4-60 x^5\right )}{\left (e+e^x+e x\right )^3} \, dx\\ &=e^3 \int \left (-\frac {10 x^3}{\left (e+e^x+e x\right )^3}-\frac {10 (-3+x) x^2}{e^2 \left (e+e^x+e x\right )}+\frac {1-20 x^3}{e^3}+\frac {10 x \left (-1+x+x^3\right )}{e \left (e+e^x+e x\right )^2}\right ) \, dx\\ &=-\left ((10 e) \int \frac {(-3+x) x^2}{e+e^x+e x} \, dx\right )+\left (10 e^2\right ) \int \frac {x \left (-1+x+x^3\right )}{\left (e+e^x+e x\right )^2} \, dx-\left (10 e^3\right ) \int \frac {x^3}{\left (e+e^x+e x\right )^3} \, dx+\int \left (1-20 x^3\right ) \, dx\\ &=x-5 x^4-(10 e) \int \left (-\frac {3 x^2}{e+e^x+e x}+\frac {x^3}{e+e^x+e x}\right ) \, dx+\left (10 e^2\right ) \int \left (-\frac {x}{\left (e+e^x+e x\right )^2}+\frac {x^2}{\left (e+e^x+e x\right )^2}+\frac {x^4}{\left (e+e^x+e x\right )^2}\right ) \, dx-\left (10 e^3\right ) \int \frac {x^3}{\left (e+e^x+e x\right )^3} \, dx\\ &=x-5 x^4-(10 e) \int \frac {x^3}{e+e^x+e x} \, dx+(30 e) \int \frac {x^2}{e+e^x+e x} \, dx-\left (10 e^2\right ) \int \frac {x}{\left (e+e^x+e x\right )^2} \, dx+\left (10 e^2\right ) \int \frac {x^2}{\left (e+e^x+e x\right )^2} \, dx+\left (10 e^2\right ) \int \frac {x^4}{\left (e+e^x+e x\right )^2} \, dx-\left (10 e^3\right ) \int \frac {x^3}{\left (e+e^x+e x\right )^3} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 41, normalized size = 1.71 \begin {gather*} x-5 x^4-\frac {5 e^2 x^2}{\left (e+e^x+e x\right )^2}+\frac {10 e x^3}{e+e^x+e x} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 1.08, size = 102, normalized size = 4.25 \begin {gather*} -\frac {5 \, x^{6} + 10 \, x^{5} - 5 \, x^{4} - 11 \, x^{3} + 3 \, x^{2} + {\left (5 \, x^{4} - x\right )} e^{\left (2 \, x - 2\right )} + 2 \, {\left (5 \, x^{5} + 5 \, x^{4} - 5 \, x^{3} - x^{2} - x\right )} e^{\left (x - 1\right )} - x}{x^{2} + 2 \, {\left (x + 1\right )} e^{\left (x - 1\right )} + 2 \, x + e^{\left (2 \, x - 2\right )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.16, size = 135, normalized size = 5.62 \begin {gather*} -\frac {5 \, x^{6} e^{2} + 10 \, x^{5} e^{2} + 10 \, x^{5} e^{\left (x + 1\right )} - 5 \, x^{4} e^{2} + 5 \, x^{4} e^{\left (2 \, x\right )} + 10 \, x^{4} e^{\left (x + 1\right )} - 11 \, x^{3} e^{2} - 10 \, x^{3} e^{\left (x + 1\right )} + 3 \, x^{2} e^{2} - 2 \, x^{2} e^{\left (x + 1\right )} - x e^{2} - x e^{\left (2 \, x\right )} - 2 \, x e^{\left (x + 1\right )}}{x^{2} e^{2} + 2 \, x e^{2} + 2 \, x e^{\left (x + 1\right )} + e^{2} + e^{\left (2 \, x\right )} + 2 \, e^{\left (x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 39, normalized size = 1.62
method | result | size |
risch | \(-5 x^{4}+x +\frac {5 x^{2} \left (2 x^{2}+2 x \,{\mathrm e}^{x -1}+2 x -1\right )}{\left (1+x +{\mathrm e}^{x -1}\right )^{2}}\) | \(39\) |
norman | \(\frac {x \,{\mathrm e}^{2 x -2}-4 x^{2}-2 \,{\mathrm e}^{x -1}-x -{\mathrm e}^{2 x -2}+11 x^{3}+5 x^{4}-10 x^{5}-5 x^{6}+2 x^{2} {\mathrm e}^{x -1}+10 x^{3} {\mathrm e}^{x -1}-10 x^{4} {\mathrm e}^{x -1}-10 \,{\mathrm e}^{x -1} x^{5}-5 \,{\mathrm e}^{2 x -2} x^{4}-1}{\left (1+x +{\mathrm e}^{x -1}\right )^{2}}\) | \(110\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.42, size = 126, normalized size = 5.25 \begin {gather*} -\frac {5 \, x^{6} e^{2} + 10 \, x^{5} e^{2} - 5 \, x^{4} e^{2} - 11 \, x^{3} e^{2} + 3 \, x^{2} e^{2} - x e^{2} + {\left (5 \, x^{4} - x\right )} e^{\left (2 \, x\right )} + 2 \, {\left (5 \, x^{5} e + 5 \, x^{4} e - 5 \, x^{3} e - x^{2} e - x e\right )} e^{x}}{x^{2} e^{2} + 2 \, x e^{2} + 2 \, {\left (x e + e\right )} e^{x} + e^{2} + e^{\left (2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.14, size = 85, normalized size = 3.54 \begin {gather*} -\frac {x\,\left (3\,x-{\mathrm {e}}^{2\,x-2}+5\,x^3\,{\mathrm {e}}^{2\,x-2}-11\,x^2-5\,x^3+10\,x^4+5\,x^5-1\right )-x\,{\mathrm {e}}^{x-1}\,\left (-10\,x^4-10\,x^3+10\,x^2+2\,x+2\right )}{{\left (x+{\mathrm {e}}^{x-1}+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.25, size = 56, normalized size = 2.33 \begin {gather*} - 5 x^{4} + x + \frac {10 x^{4} + 10 x^{3} e^{x - 1} + 10 x^{3} - 5 x^{2}}{x^{2} + 2 x + \left (2 x + 2\right ) e^{x - 1} + e^{2 x - 2} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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