Optimal. Leaf size=27 \[ 1+5 e^{-x \left (1+16 x+\frac {1}{5} \log ^2\left (\frac {5}{x}\right )\right )}+x \]
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Rubi [F]
time = 3.20, 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 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \left (-5+e^{\frac {1}{5} \left (5 x+80 x^2+x \log ^2\left (\frac {5}{x}\right )\right )}-160 x+2 \log \left (\frac {5}{x}\right )-\log ^2\left (\frac {5}{x}\right )\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-5 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )}+\exp \left (\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )+\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )\right )-160 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x+2 e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right )-e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right )\right ) \, dx\\ &=2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx+\int \exp \left (\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )+\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )\right ) \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ &=2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx+\int 1 \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ &=x+2 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log \left (\frac {5}{x}\right ) \, dx-5 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \, dx-160 \int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} x \, dx-\int e^{\frac {1}{5} \left (-5 x-80 x^2-x \log ^2\left (\frac {5}{x}\right )\right )} \log ^2\left (\frac {5}{x}\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 1.65, size = 24, normalized size = 0.89 \begin {gather*} 5 e^{-\frac {1}{5} x \left (5+80 x+\log ^2\left (\frac {5}{x}\right )\right )}+x \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 26, normalized size = 0.96
method | result | size |
risch | \(x +5 \,{\mathrm e}^{-\frac {x \left (\ln \left (\frac {5}{x}\right )^{2}+80 x +5\right )}{5}}\) | \(22\) |
default | \(x +5 \,{\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\) | \(26\) |
norman | \(\left (5+x \,{\mathrm e}^{\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}+16 x^{2}+x}\right ) {\mathrm e}^{-\frac {x \ln \left (\frac {5}{x}\right )^{2}}{5}-16 x^{2}-x}\) | \(46\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.56, size = 35, normalized size = 1.30 \begin {gather*} x + 5 \, e^{\left (-\frac {1}{5} \, x \log \left (5\right )^{2} + \frac {2}{5} \, x \log \left (5\right ) \log \left (x\right ) - \frac {1}{5} \, x \log \left (x\right )^{2} - 16 \, x^{2} - x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs.
\(2 (22) = 44\).
time = 0.38, size = 45, normalized size = 1.67 \begin {gather*} {\left (x e^{\left (\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} + 16 \, x^{2} + x\right )} + 5\right )} e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.20, size = 22, normalized size = 0.81 \begin {gather*} x + 5 e^{- 16 x^{2} - \frac {x \log {\left (\frac {5}{x} \right )}^{2}}{5} - x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.47, size = 25, normalized size = 0.93 \begin {gather*} x + 5 \, e^{\left (-\frac {1}{5} \, x \log \left (\frac {5}{x}\right )^{2} - 16 \, x^{2} - x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.41, size = 43, normalized size = 1.59 \begin {gather*} x+\frac {5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-16\,x^2}\,{\mathrm {e}}^{-\frac {x\,{\ln \left (\frac {1}{x}\right )}^2}{5}}\,{\mathrm {e}}^{-\frac {2\,x\,\ln \left (\frac {1}{x}\right )\,\ln \left (5\right )}{5}}}{{\left ({\mathrm {e}}^{x\,{\ln \left (5\right )}^2}\right )}^{1/5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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