Optimal. Leaf size=28 \[ -e^4+e^x+e^{\frac {e^{3-x^2}}{x+x^4}} \]
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Rubi [F] time = 7.91, 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^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2+2 x^5+x^8} \, 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^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2 \left (1+2 x^3+x^6\right )} \, dx\\ &=\int \frac {e^{-x^2} \left (e^{3+\frac {e^{3-x^2}}{x+x^4}} \left (-1-2 x^2-4 x^3-2 x^5\right )+e^{x+x^2} \left (x^2+2 x^5+x^8\right )\right )}{x^2 \left (1+x^3\right )^2} \, dx\\ &=\int \left (e^x-\frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}} \left (1+2 x^2+4 x^3+2 x^5\right )}{x^2 \left (1+x^3\right )^2}\right ) \, dx\\ &=\int e^x \, dx-\int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}} \left (1+2 x^2+4 x^3+2 x^5\right )}{x^2 \left (1+x^3\right )^2} \, dx\\ &=e^x-\int \left (\frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{x^2}-\frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 (1+x)^2}+\frac {2 e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 (1+x)}+\frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{\left (1-x+x^2\right )^2}-\frac {2 e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}} (-1+x)}{3 \left (1-x+x^2\right )}\right ) \, dx\\ &=e^x+\frac {1}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{1+x} \, dx+\frac {2}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}} (-1+x)}{1-x+x^2} \, dx-\int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{x^2} \, dx-\int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{\left (1-x+x^2\right )^2} \, dx\\ &=e^x+\frac {1}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{1+x} \, dx+\frac {2}{3} \int \left (\frac {\left (1+\frac {i}{\sqrt {3}}\right ) e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{-1-i \sqrt {3}+2 x}+\frac {\left (1-\frac {i}{\sqrt {3}}\right ) e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{-1+i \sqrt {3}+2 x}\right ) \, dx-\int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{x^2} \, dx-\int \left (-\frac {4 e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 \left (1+i \sqrt {3}-2 x\right )^2}+\frac {4 i e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 \sqrt {3} \left (1+i \sqrt {3}-2 x\right )}-\frac {4 e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 \left (-1+i \sqrt {3}+2 x\right )^2}+\frac {4 i e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{3 \sqrt {3} \left (-1+i \sqrt {3}+2 x\right )}\right ) \, dx\\ &=e^x+\frac {1}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{1+x} \, dx+\frac {4}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{\left (1+i \sqrt {3}-2 x\right )^2} \, dx+\frac {4}{3} \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{\left (-1+i \sqrt {3}+2 x\right )^2} \, dx-\frac {(4 i) \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{1+i \sqrt {3}-2 x} \, dx}{3 \sqrt {3}}-\frac {(4 i) \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{-1+i \sqrt {3}+2 x} \, dx}{3 \sqrt {3}}+\frac {1}{9} \left (2 \left (3-i \sqrt {3}\right )\right ) \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{-1+i \sqrt {3}+2 x} \, dx+\frac {1}{9} \left (2 \left (3+i \sqrt {3}\right )\right ) \int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{-1-i \sqrt {3}+2 x} \, dx-\int \frac {e^{3-x^2+\frac {e^{3-x^2}}{x+x^4}}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.10, size = 26, normalized size = 0.93 \begin {gather*} e^x+e^{\frac {e^{3-x^2}}{x \left (1+x^3\right )}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.44, size = 34, normalized size = 1.21 \begin {gather*} {\left (e^{\left (x + 3\right )} + e^{\left (\frac {3 \, x^{4} + 3 \, x + e^{\left (-x^{2} + 3\right )}}{x^{4} + x}\right )}\right )} e^{\left (-3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x^{8} + 2 \, x^{5} + x^{2}\right )} e^{\left (x^{2} + x\right )} - {\left (2 \, x^{5} + 4 \, x^{3} + 2 \, x^{2} + 1\right )} e^{\left (\frac {e^{\left (-x^{2} + 3\right )}}{x^{4} + x} + 3\right )}\right )} e^{\left (-x^{2}\right )}}{x^{8} + 2 \, x^{5} + x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 32, normalized size = 1.14
method | result | size |
risch | \({\mathrm e}^{x}+{\mathrm e}^{\frac {{\mathrm e}^{-x^{2}+3}}{x \left (x +1\right ) \left (x^{2}-x +1\right )}}\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 91, normalized size = 3.25 \begin {gather*} {\left (e^{\left (x + \frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x + 1\right )}}\right )} + e^{\left (-\frac {2 \, x e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x^{2} - x + 1\right )}} + \frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x^{2} - x + 1\right )}} + \frac {e^{\left (-x^{2} + 3\right )}}{x}\right )}\right )} e^{\left (-\frac {e^{\left (-x^{2} + 3\right )}}{3 \, {\left (x + 1\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.82, size = 20, normalized size = 0.71 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{-x^2}}{x^4+x}}+{\mathrm {e}}^x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.62, size = 17, normalized size = 0.61 \begin {gather*} e^{x} + e^{\frac {e^{3} e^{- x^{2}}}{x^{4} + x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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