Optimal. Leaf size=28 \[ e^{-\frac {5}{5+e^x}-x+x^2+x (-1+5 \log (x))^2} \]
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Rubi [F] time = 12.23, 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 {\exp \left (\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}\right ) \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{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^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \left (50 (-5+x)+2 e^{2 x} (-5+x)+5 e^x (-19+4 x)+40 \left (5+e^x\right )^2 \log (x)+25 \left (5+e^x\right )^2 \log ^2(x)\right )}{\left (5+e^x\right )^2} \, dx\\ &=\int \left (2 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x}-10 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}-\frac {25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2}+\frac {5 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x}+40 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x)+25 e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x)\right ) \, dx\\ &=2 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{1-10 x} \, dx+5 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{5+e^x} \, dx-10 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \, dx-25 \int \frac {e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x}}{\left (5+e^x\right )^2} \, dx+25 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log ^2(x) \, dx+40 \int e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \log (x) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.31, size = 28, normalized size = 1.00 \begin {gather*} e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 47, normalized size = 1.68 \begin {gather*} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x)^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x) - 5}{e^{x} + 5}\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 (25 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \relax (x)^{2} + 2 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} + 5 \, {\left (4 \, x - 19\right )} e^{x} + 40 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \relax (x) + 50 \, x - 250\right )} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x)^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \relax (x) - 5}{e^{x} + 5}\right )}}{e^{\left (2 \, x\right )} + 10 \, e^{x} + 25}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 50, normalized size = 1.79
method | result | size |
risch | \({\mathrm e}^{\frac {25 x \,{\mathrm e}^{x} \ln \relax (x )^{2}+125 x \ln \relax (x )^{2}-10 x \,{\mathrm e}^{x} \ln \relax (x )+{\mathrm e}^{x} x^{2}-50 x \ln \relax (x )+5 x^{2}-5}{{\mathrm e}^{x}+5}}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.62, size = 85, normalized size = 3.04 \begin {gather*} e^{\left (\frac {25 \, x e^{x} \log \relax (x)^{2}}{e^{x} + 5} + \frac {x^{2} e^{x}}{e^{x} + 5} - \frac {10 \, x e^{x} \log \relax (x)}{e^{x} + 5} + \frac {125 \, x \log \relax (x)^{2}}{e^{x} + 5} + \frac {5 \, x^{2}}{e^{x} + 5} - \frac {50 \, x \log \relax (x)}{e^{x} + 5} - \frac {5}{e^{x} + 5}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.41, size = 72, normalized size = 2.57 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {125\,x\,{\ln \relax (x)}^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {5\,x^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2}{{\mathrm {e}}^x+5}}}{x^{10\,x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.87, size = 49, normalized size = 1.75 \begin {gather*} e^{\frac {x^{2} e^{x} + 5 x^{2} + \left (- 10 x e^{x} - 50 x\right ) \log {\relax (x )} + \left (25 x e^{x} + 125 x\right ) \log {\relax (x )}^{2} - 5}{e^{x} + 5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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