Optimal. Leaf size=28 \[ e^{e^{e^{\frac {e^{5 \left (-3-x+5 x^2\right )}}{x}+x}}}+x \]
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Rubi [F] time = 4.41, 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 {x^2+\exp \left (e^{e^{\frac {e^{-15-5 x+25 x^2}+x^2}{x}}}+e^{\frac {e^{-15-5 x+25 x^2}+x^2}{x}}+\frac {e^{-15-5 x+25 x^2}+x^2}{x}\right ) \left (x^2+e^{-15-5 x+25 x^2} \left (-1-5 x+50 x^2\right )\right )}{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+\exp \left (e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}+x\right )+\frac {\exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right ) (-1+5 x) (1+10 x)}{x^2}\right ) \, dx\\ &=x+\int \exp \left (e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}+x\right ) \, dx+\int \frac {\exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right ) (-1+5 x) (1+10 x)}{x^2} \, dx\\ &=x+\int \exp \left (e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}+x\right ) \, dx+\int \left (50 \exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right )-\frac {\exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right )}{x^2}-\frac {5 \exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right )}{x}\right ) \, dx\\ &=x-5 \int \frac {\exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right )}{x} \, dx+50 \int \exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right ) \, dx+\int \exp \left (e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}+x\right ) \, dx-\int \frac {\exp \left (-15+e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}+e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}+\frac {e^{-15-5 x+25 x^2}}{x}-4 x+25 x^2\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.21, size = 26, normalized size = 0.93 \begin {gather*} e^{e^{e^{\frac {e^{-15-5 x+25 x^2}}{x}+x}}}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.70, size = 154, normalized size = 5.50 \begin {gather*} {\left (x e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x} + e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\right )} + e^{\left (\frac {x^{2} + x e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )} + x e^{\left (e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\right )} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\right )} e^{\left (-\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x} - e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\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 {x^{2} + {\left (x^{2} + {\left (50 \, x^{2} - 5 \, x - 1\right )} e^{\left (25 \, x^{2} - 5 \, x - 15\right )}\right )} e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x} + e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )} + e^{\left (e^{\left (\frac {x^{2} + e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\right )}\right )}}{x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 25, normalized size = 0.89
method | result | size |
risch | \(x +{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{25 x^{2}-5 x -15}+x^{2}}{x}}}}\) | \(25\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.82, size = 22, normalized size = 0.79 \begin {gather*} x + e^{\left (e^{\left (e^{\left (x + \frac {e^{\left (25 \, x^{2} - 5 \, x - 15\right )}}{x}\right )}\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.24, size = 24, normalized size = 0.86 \begin {gather*} x+{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^{-15}\,{\mathrm {e}}^{25\,x^2}}{x}}\,{\mathrm {e}}^x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 14.72, size = 22, normalized size = 0.79 \begin {gather*} x + e^{e^{e^{\frac {x^{2} + e^{25 x^{2} - 5 x - 15}}{x}}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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