Optimal. Leaf size=30 \[ -x+e^{x-\frac {e^4 (-5+x)}{x-(-1+x) x^2}} x \]
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Rubi [F] time = 10.87, 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 {e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \left (-1-2 x+x^3+e^{-x+\frac {e^4 (-5+x)}{x+x^2-x^3}} \left (1+x-x^2\right )+\frac {e^4 \left (5+10 x-16 x^2+2 x^3\right )}{x+x^2-x^3}\right )}{-1-x+x^2} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \left (-5 e^4+\left (1-10 e^4\right ) x+3 \left (1+\frac {16 e^4}{3}\right ) x^2+\left (1-2 e^4\right ) x^3-3 x^4-x^5+x^6\right )}{x \left (1+x-x^2\right )^2}\right ) \, dx\\ &=-x+\int \frac {e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \left (-5 e^4+\left (1-10 e^4\right ) x+3 \left (1+\frac {16 e^4}{3}\right ) x^2+\left (1-2 e^4\right ) x^3-3 x^4-x^5+x^6\right )}{x \left (1+x-x^2\right )^2} \, dx\\ &=-x+\int \frac {e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \left (x (1+x) \left (1+x-x^2\right )^2-e^4 \left (5+10 x-16 x^2+2 x^3\right )\right )}{x \left (1+x-x^2\right )^2} \, dx\\ &=-x+\int \left (e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}}-\frac {5 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x}+e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x+\frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} (-7+9 x)}{\left (-1-x+x^2\right )^2}+\frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} (-7+5 x)}{-1-x+x^2}\right ) \, dx\\ &=-x-5 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x \, dx+\int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} (-7+9 x)}{\left (-1-x+x^2\right )^2} \, dx+\int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} (-7+5 x)}{-1-x+x^2} \, dx\\ &=-x-5 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x \, dx+\int \left (\frac {\left (5-\frac {9}{\sqrt {5}}\right ) e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1-\sqrt {5}+2 x}+\frac {\left (5+\frac {9}{\sqrt {5}}\right ) e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x}\right ) \, dx+\int \left (-\frac {7 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (-1-x+x^2\right )^2}+\frac {9 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x}{\left (-1-x+x^2\right )^2}\right ) \, dx\\ &=-x-5 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x} \, dx-7 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (-1-x+x^2\right )^2} \, dx+9 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x}{\left (-1-x+x^2\right )^2} \, dx+\frac {1}{5} \left (25-9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1-\sqrt {5}+2 x} \, dx+\frac {1}{5} \left (25+9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x \, dx\\ &=-x-5 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x} \, dx-7 \int \left (\frac {4 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \left (1+\sqrt {5}-2 x\right )^2}+\frac {4 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \sqrt {5} \left (1+\sqrt {5}-2 x\right )}+\frac {4 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \left (-1+\sqrt {5}+2 x\right )^2}+\frac {4 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \sqrt {5} \left (-1+\sqrt {5}+2 x\right )}\right ) \, dx+9 \int \left (\frac {2 \left (1+\sqrt {5}\right ) e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \left (1+\sqrt {5}-2 x\right )^2}+\frac {2 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \sqrt {5} \left (1+\sqrt {5}-2 x\right )}+\frac {2 \left (1-\sqrt {5}\right ) e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \left (-1+\sqrt {5}+2 x\right )^2}+\frac {2 e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{5 \sqrt {5} \left (-1+\sqrt {5}+2 x\right )}\right ) \, dx+\frac {1}{5} \left (25-9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1-\sqrt {5}+2 x} \, dx+\frac {1}{5} \left (25+9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x \, dx\\ &=-x-5 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{x} \, dx-\frac {28}{5} \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (1+\sqrt {5}-2 x\right )^2} \, dx-\frac {28}{5} \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (-1+\sqrt {5}+2 x\right )^2} \, dx+\frac {18 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{1+\sqrt {5}-2 x} \, dx}{5 \sqrt {5}}+\frac {18 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x} \, dx}{5 \sqrt {5}}-\frac {28 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{1+\sqrt {5}-2 x} \, dx}{5 \sqrt {5}}-\frac {28 \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x} \, dx}{5 \sqrt {5}}+\frac {1}{5} \left (25-9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1-\sqrt {5}+2 x} \, dx+\frac {1}{5} \left (18 \left (1-\sqrt {5}\right )\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (-1+\sqrt {5}+2 x\right )^2} \, dx+\frac {1}{5} \left (18 \left (1+\sqrt {5}\right )\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{\left (1+\sqrt {5}-2 x\right )^2} \, dx+\frac {1}{5} \left (25+9 \sqrt {5}\right ) \int \frac {e^{4+x-\frac {e^4 (-5+x)}{x+x^2-x^3}}}{-1+\sqrt {5}+2 x} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} \, dx+\int e^{x-\frac {e^4 (-5+x)}{x+x^2-x^3}} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 28, normalized size = 0.93 \begin {gather*} \left (-1+e^{x+\frac {e^4 (-5+x)}{x \left (-1-x+x^2\right )}}\right ) x \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.67, size = 42, normalized size = 1.40 \begin {gather*} x e^{\left (\frac {x^{4} - x^{3} - x^{2} + {\left (x - 5\right )} e^{4}}{x^{3} - x^{2} - x}\right )} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.49, size = 44, normalized size = 1.47 \begin {gather*} x e^{\left (\frac {x^{4} - x^{3} - x^{2} + x e^{4} - 5 \, e^{4}}{x^{3} - x^{2} - x}\right )} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.51, size = 44, normalized size = 1.47
method | result | size |
risch | \(-x +x \,{\mathrm e}^{\frac {x^{4}-x^{3}+x \,{\mathrm e}^{4}-x^{2}-5 \,{\mathrm e}^{4}}{\left (x^{2}-x -1\right ) x}}\) | \(44\) |
default | \(-x +\frac {\left (x^{3}-x^{2}-x \right ) {\mathrm e}^{-\frac {\left (x -5\right ) {\mathrm e}^{4}}{-x^{3}+x^{2}+x}+x}}{x^{2}-x -1}\) | \(53\) |
norman | \(\frac {\left (x^{3}+{\mathrm e}^{\frac {\left (x -5\right ) {\mathrm e}^{4}}{-x^{3}+x^{2}+x}-x}+2 x \,{\mathrm e}^{\frac {\left (x -5\right ) {\mathrm e}^{4}}{-x^{3}+x^{2}+x}-x}-x -x^{2}-x^{3} {\mathrm e}^{\frac {\left (x -5\right ) {\mathrm e}^{4}}{-x^{3}+x^{2}+x}-x}\right ) {\mathrm e}^{-\frac {\left (x -5\right ) {\mathrm e}^{4}}{-x^{3}+x^{2}+x}+x}}{x^{2}-x -1}\) | \(126\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.06, size = 45, normalized size = 1.50 \begin {gather*} x e^{\left (x - \frac {5 \, x e^{4}}{x^{2} - x - 1} + \frac {6 \, e^{4}}{x^{2} - x - 1} + \frac {5 \, e^{4}}{x}\right )} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.22, size = 43, normalized size = 1.43 \begin {gather*} x\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^4}{-x^3+x^2+x}}\,{\mathrm {e}}^x\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^4}{-x^3+x^2+x}}-x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.43, size = 20, normalized size = 0.67 \begin {gather*} x e^{x - \frac {\left (x - 5\right ) e^{4}}{- x^{3} + x^{2} + x}} - x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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