Optimal. Leaf size=30 \[ -5+e^{10-\frac {2 x^2}{e^{e^{x^2}}+x}} \log ^2(x+\log (2)) \]
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Rubi [B] time = 4.23, antiderivative size = 182, normalized size of antiderivative = 6.07, number of steps used = 3, number of rules used = 3, integrand size = 180, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 2288} \begin {gather*} -\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (-e^{x^2+e^{x^2}} x^2 (2 x+\log (4))+e^{e^{x^2}} (2 x+\log (4))+x (x+\log (2))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {10 e^{x^2+e^{x^2}} x-2 x+5}{e^{e^{x^2}}+x}-\frac {\left (2 e^{x^2+e^{x^2}} x+1\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2288
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx\\ &=2 \int \frac {e^{\frac {10 e^{e^{x^2}}-2 (-5+x) x}{e^{e^{x^2}}+x}} \log (x+\log (2)) \left (\left (e^{e^{x^2}}+x\right )^2+x \left (-x (x+\log (2))-e^{e^{x^2}} (2 x+\log (4))+e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log (x+\log (2))\right )}{\left (e^{e^{x^2}}+x\right )^2 (x+\log (2))} \, dx\\ &=-\frac {e^{\frac {2 \left (5 e^{e^{x^2}}+(5-x) x\right )}{e^{e^{x^2}}+x}} x \left (x (x+\log (2))+e^{e^{x^2}} (2 x+\log (4))-e^{e^{x^2}+x^2} x^2 (2 x+\log (4))\right ) \log ^2(x+\log (2))}{\left (e^{e^{x^2}}+x\right )^2 \left (\frac {5-2 x+10 e^{e^{x^2}+x^2} x}{e^{e^{x^2}}+x}-\frac {\left (1+2 e^{e^{x^2}+x^2} x\right ) \left (5 e^{e^{x^2}}+(5-x) x\right )}{\left (e^{e^{x^2}}+x\right )^2}\right ) (x+\log (2))}\\ \end {aligned} \end {gather*}
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Mathematica [F] time = 13.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{-\frac {2 \left (-5 e^{e^{x^2}}-5 x+x^2\right )}{e^{e^{x^2}}+x}} \left (2 e^{2 e^{x^2}} \log (x+\log (2))+2 x^2 \log (x+\log (2))+\left (-2 x^3-2 x^2 \log (2)\right ) \log ^2(x+\log (2))+e^{e^{x^2}} \left (4 x \log (x+\log (2))+\left (-4 x^2-4 x \log (2)+e^{x^2} \left (4 x^4+4 x^3 \log (2)\right )\right ) \log ^2(x+\log (2))\right )\right )}{x^3+x^2 \log (2)+e^{2 e^{x^2}} (x+\log (2))+e^{e^{x^2}} \left (2 x^2+2 x \log (2)\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.65, size = 34, normalized size = 1.13 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{2} - 5 \, x - 5 \, e^{\left (e^{\left (x^{2}\right )}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}}\right )} \log \left (x + \log \relax (2)\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 2.26, size = 74, normalized size = 2.47 \begin {gather*} e^{\left (-x^{2} + \frac {x^{3} + x^{2} e^{\left (e^{\left (x^{2}\right )}\right )} - 2 \, x^{2} + x e^{\left (x^{2}\right )} + 10 \, x + e^{\left (x^{2} + e^{\left (x^{2}\right )}\right )} + 10 \, e^{\left (e^{\left (x^{2}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}} - e^{\left (x^{2}\right )}\right )} \log \left (x + \log \relax (2)\right )^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 35, normalized size = 1.17
method | result | size |
risch | \(\ln \left (\ln \relax (2)+x \right )^{2} {\mathrm e}^{-\frac {2 \left (-5 \,{\mathrm e}^{{\mathrm e}^{x^{2}}}+x^{2}-5 x \right )}{{\mathrm e}^{{\mathrm e}^{x^{2}}}+x}}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, \int \frac {{\left (x^{2} \log \left (x + \log \relax (2)\right ) - {\left (x^{3} + x^{2} \log \relax (2)\right )} \log \left (x + \log \relax (2)\right )^{2} - 2 \, {\left ({\left (x^{2} - {\left (x^{4} + x^{3} \log \relax (2)\right )} e^{\left (x^{2}\right )} + x \log \relax (2)\right )} \log \left (x + \log \relax (2)\right )^{2} - x \log \left (x + \log \relax (2)\right )\right )} e^{\left (e^{\left (x^{2}\right )}\right )} + e^{\left (2 \, e^{\left (x^{2}\right )}\right )} \log \left (x + \log \relax (2)\right )\right )} e^{\left (-\frac {2 \, {\left (x^{2} - 5 \, x - 5 \, e^{\left (e^{\left (x^{2}\right )}\right )}\right )}}{x + e^{\left (e^{\left (x^{2}\right )}\right )}}\right )}}{x^{3} + x^{2} \log \relax (2) + {\left (x + \log \relax (2)\right )} e^{\left (2 \, e^{\left (x^{2}\right )}\right )} + 2 \, {\left (x^{2} + x \log \relax (2)\right )} e^{\left (e^{\left (x^{2}\right )}\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {2\,\left (5\,x+5\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}-x^2\right )}{x+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left ({\ln \left (x+\ln \relax (2)\right )}^2\,\left (4\,x\,\ln \relax (2)-{\mathrm {e}}^{x^2}\,\left (4\,x^4+4\,\ln \relax (2)\,x^3\right )+4\,x^2\right )-4\,x\,\ln \left (x+\ln \relax (2)\right )\right )-2\,{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\ln \left (x+\ln \relax (2)\right )+{\ln \left (x+\ln \relax (2)\right )}^2\,\left (2\,x^3+2\,\ln \relax (2)\,x^2\right )-2\,x^2\,\ln \left (x+\ln \relax (2)\right )\right )}{x^2\,\ln \relax (2)+{\mathrm {e}}^{2\,{\mathrm {e}}^{x^2}}\,\left (x+\ln \relax (2)\right )+x^3+{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\left (2\,x^2+2\,\ln \relax (2)\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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