Optimal. Leaf size=30 \[ \log (2)+\frac {1}{25 \left (-x+4 x \left (-x+e^{e^{x^2}} \log (x)\right )\right )} \]
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Rubi [F] time = 2.38, 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 {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^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 {1-4 e^{e^{x^2}}+8 x-4 e^{e^{x^2}} \left (1+2 e^{x^2} x^2\right ) \log (x)}{25 x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ &=\frac {1}{25} \int \frac {1-4 e^{e^{x^2}}+8 x-4 e^{e^{x^2}} \left (1+2 e^{x^2} x^2\right ) \log (x)}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ &=\frac {1}{25} \int \left (\frac {1}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2}+\frac {8}{x \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2}-\frac {4 e^{e^{x^2}}}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}-\frac {8 e^{e^{x^2}+x^2} \log (x)}{\left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}-\frac {4 e^{e^{x^2}} \log (x)}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{25} \int \frac {1}{x^2 \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {4}{25} \int \frac {e^{e^{x^2}}}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {4}{25} \int \frac {e^{e^{x^2}} \log (x)}{x^2 \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx+\frac {8}{25} \int \frac {1}{x \left (1+4 x-4 e^{e^{x^2}} \log (x)\right )^2} \, dx-\frac {8}{25} \int \frac {e^{e^{x^2}+x^2} \log (x)}{\left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.61, size = 25, normalized size = 0.83 \begin {gather*} \frac {1}{25 x \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.69, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 22, normalized size = 0.73
method | result | size |
risch | \(-\frac {1}{25 x \left (-4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}} \ln \relax (x )+4 x +1\right )}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \relax (x) - 4 \, x^{2} - x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.76, size = 86, normalized size = 2.87 \begin {gather*} -\frac {\frac {4\,x}{25}+\ln \relax (x)\,\left (\frac {2\,x^2\,{\mathrm {e}}^{x^2}}{25}-\frac {4\,x}{25}+\frac {8\,x^3\,{\mathrm {e}}^{x^2}}{25}\right )+\frac {1}{25}}{\left (4\,x-4\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \relax (x)+1\right )\,\left (x-4\,x^2\,\ln \relax (x)+4\,x^2+2\,x^3\,{\mathrm {e}}^{x^2}\,\ln \relax (x)+8\,x^4\,{\mathrm {e}}^{x^2}\,\ln \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.35, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{- 100 x^{2} + 100 x e^{e^{x^{2}}} \log {\relax (x )} - 25 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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