Optimal. Leaf size=27 \[ \frac {x (3+x)}{-5+e^{\left (-e^{-5+3 e^{10}}+x\right )^2}+x} \]
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Rubi [F] time = 14.75, 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 {-15-10 x+x^2+e^{e^{-10+6 e^{10}}-2 e^{-5+3 e^{10}} x+x^2} \left (3+2 x-6 x^2-2 x^3+e^{-5+3 e^{10}} \left (6 x+2 x^2\right )\right )}{25+\exp \left (2 e^{-10+6 e^{10}}-4 e^{-5+3 e^{10}} x+2 x^2\right )-10 x+x^2+e^{e^{-10+6 e^{10}}-2 e^{-5+3 e^{10}} x+x^2} (-10+2 x)} \, 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^{4 e^{-5+3 e^{10}} x} \left (-15-10 x+x^2+e^{e^{-10+6 e^{10}}-2 e^{-5+3 e^{10}} x+x^2} \left (3+2 x-6 x^2-2 x^3+e^{-5+3 e^{10}} \left (6 x+2 x^2\right )\right )\right )}{\left (5 e^{2 e^{-5+3 e^{10}} x}-e^{e^{-10+6 e^{10}}+x^2}-e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx\\ &=\int \left (\frac {e^{-5+4 e^{-5+3 e^{10}} x} x (3+x) \left (-e^5+10 e^{3 e^{10}}-2 \left (5 e^5+e^{3 e^{10}}\right ) x+2 e^5 x^2\right )}{\left (5 e^{2 e^{-5+3 e^{10}} x}-e^{e^{-10+6 e^{10}}+x^2}-e^{2 e^{-5+3 e^{10}} x} x\right )^2}+\frac {e^{-5+2 e^{-5+3 e^{10}} x} \left (-3 e^5-2 \left (e^5+3 e^{3 e^{10}}\right ) x+2 \left (3 e^5-e^{3 e^{10}}\right ) x^2+2 e^5 x^3\right )}{5 e^{2 e^{-5+3 e^{10}} x}-e^{e^{-10+6 e^{10}}+x^2}-e^{2 e^{-5+3 e^{10}} x} x}\right ) \, dx\\ &=\int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x (3+x) \left (-e^5+10 e^{3 e^{10}}-2 \left (5 e^5+e^{3 e^{10}}\right ) x+2 e^5 x^2\right )}{\left (5 e^{2 e^{-5+3 e^{10}} x}-e^{e^{-10+6 e^{10}}+x^2}-e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx+\int \frac {e^{-5+2 e^{-5+3 e^{10}} x} \left (-3 e^5-2 \left (e^5+3 e^{3 e^{10}}\right ) x+2 \left (3 e^5-e^{3 e^{10}}\right ) x^2+2 e^5 x^3\right )}{5 e^{2 e^{-5+3 e^{10}} x}-e^{e^{-10+6 e^{10}}+x^2}-e^{2 e^{-5+3 e^{10}} x} x} \, dx\\ &=\int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x (3+x) \left (-e^5+10 e^{3 e^{10}}-2 \left (5 e^5+e^{3 e^{10}}\right ) x+2 e^5 x^2\right )}{\left (e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)\right )^2} \, dx+\int \frac {-3 e^5-2 \left (e^5+3 e^{3 e^{10}}\right ) x+2 \left (3 e^5-e^{3 e^{10}}\right ) x^2+2 e^5 x^3}{e^5 \left (5-e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}-x\right )} \, dx\\ &=\frac {\int \frac {-3 e^5-2 \left (e^5+3 e^{3 e^{10}}\right ) x+2 \left (3 e^5-e^{3 e^{10}}\right ) x^2+2 e^5 x^3}{5-e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}-x} \, dx}{e^5}+\int \left (\frac {3 e^{-5+4 e^{-5+3 e^{10}} x} \left (-e^5+10 e^{3 e^{10}}\right ) x}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2}+\frac {e^{-5+4 e^{-5+3 e^{10}} x} \left (-31 e^5+4 e^{3 e^{10}}\right ) x^2}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2}-\frac {2 e^{-5+4 e^{-5+3 e^{10}} x} \left (2 e^5+e^{3 e^{10}}\right ) x^3}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2}+\frac {2 e^{4 e^{-5+3 e^{10}} x} x^4}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2}\right ) \, dx\\ &=2 \int \frac {e^{4 e^{-5+3 e^{10}} x} x^4}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx+\frac {\int \left (\frac {3 e^5}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x}+\frac {2 \left (e^5+3 e^{3 e^{10}}\right ) x}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x}+\frac {2 \left (-3 e^5+e^{3 e^{10}}\right ) x^2}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x}-\frac {2 e^5 x^3}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x}\right ) \, dx}{e^5}-\left (3 \left (e^5-10 e^{3 e^{10}}\right )\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx-\left (2 \left (2 e^5+e^{3 e^{10}}\right )\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x^3}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx+\left (-31 e^5+4 e^{3 e^{10}}\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x^2}{\left (-5 e^{2 e^{-5+3 e^{10}} x}+e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} x\right )^2} \, dx\\ &=2 \int \frac {e^{4 e^{-5+3 e^{10}} x} x^4}{\left (e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)\right )^2} \, dx-2 \int \frac {x^3}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x} \, dx+3 \int \frac {1}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x} \, dx-\left (3 \left (e^5-10 e^{3 e^{10}}\right )\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x}{\left (e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)\right )^2} \, dx-\left (2 \left (2 e^5+e^{3 e^{10}}\right )\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x^3}{\left (e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)\right )^2} \, dx+\left (-31 e^5+4 e^{3 e^{10}}\right ) \int \frac {e^{-5+4 e^{-5+3 e^{10}} x} x^2}{\left (e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)\right )^2} \, dx-\left (2 \left (3-e^{-5+3 e^{10}}\right )\right ) \int \frac {x^2}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x} \, dx+\left (2 \left (1+3 e^{-5+3 e^{10}}\right )\right ) \int \frac {x}{-5+e^{\frac {\left (e^{3 e^{10}}-e^5 x\right )^2}{e^{10}}}+x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.23, size = 55, normalized size = 2.04 \begin {gather*} \frac {e^{2 e^{-5+3 e^{10}} x} x (3+x)}{e^{e^{-10+6 e^{10}}+x^2}+e^{2 e^{-5+3 e^{10}} x} (-5+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.10, size = 35, normalized size = 1.30 \begin {gather*} \frac {x^{2} + 3 \, x}{x + e^{\left (x^{2} - 2 \, x e^{\left (3 \, e^{10} - 5\right )} + e^{\left (6 \, e^{10} - 10\right )}\right )} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.48, size = 35, normalized size = 1.30 \begin {gather*} \frac {x^{2} + 3 \, x}{x + e^{\left (x^{2} - 2 \, x e^{\left (3 \, e^{10} - 5\right )} + e^{\left (6 \, e^{10} - 10\right )}\right )} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.14, size = 33, normalized size = 1.22
method | result | size |
risch | \(\frac {\left (3+x \right ) x}{x +{\mathrm e}^{{\mathrm e}^{6 \,{\mathrm e}^{10}-10}-2 x \,{\mathrm e}^{3 \,{\mathrm e}^{10}-5}+x^{2}}-5}\) | \(33\) |
norman | \(\frac {x^{2}-3 \,{\mathrm e}^{{\mathrm e}^{6 \,{\mathrm e}^{10}-10}-2 x \,{\mathrm e}^{3 \,{\mathrm e}^{10}-5}+x^{2}}+15}{x +{\mathrm e}^{{\mathrm e}^{6 \,{\mathrm e}^{10}-10}-2 x \,{\mathrm e}^{3 \,{\mathrm e}^{10}-5}+x^{2}}-5}\) | \(70\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.49, size = 66, normalized size = 2.44 \begin {gather*} \frac {{\left (x^{2} - 5 \, x + 40\right )} e^{\left (2 \, x e^{\left (3 \, e^{10} - 5\right )}\right )} - 8 \, e^{\left (x^{2} + e^{\left (6 \, e^{10} - 10\right )}\right )}}{{\left (x - 5\right )} e^{\left (2 \, x e^{\left (3 \, e^{10} - 5\right )}\right )} + e^{\left (x^{2} + e^{\left (6 \, e^{10} - 10\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 72.59, size = 36, normalized size = 1.33 \begin {gather*} \frac {x^2+3\,x}{x+{\mathrm {e}}^{x^2-2\,{\mathrm {e}}^{3\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{-5}\,x+{\mathrm {e}}^{6\,{\mathrm {e}}^{10}}\,{\mathrm {e}}^{-10}}-5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.90, size = 34, normalized size = 1.26 \begin {gather*} \frac {x^{2} + 3 x}{x + e^{x^{2} - 2 x e^{-5 + 3 e^{10}} + e^{-10 + 6 e^{10}}} - 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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