Optimal. Leaf size=27 \[ \frac {3 x}{5+e^{-8+50 x^2}+x-\log (5)-\log \left (x^2\right )} \]
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Rubi [F] time = 2.52, 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 {21+e^{-8+50 x^2} \left (3-300 x^2\right )-3 \log (5)-3 \log \left (x^2\right )}{25+e^{-16+100 x^2}+10 x+x^2+e^{-8+50 x^2} (10+2 x-2 \log (5))+(-10-2 x) \log (5)+\log ^2(5)+\left (-10-2 e^{-8+50 x^2}-2 x+2 \log (5)\right ) \log \left (x^2\right )+\log ^2\left (x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^8 \left (-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \frac {-e^{50 x^2} \left (-1+100 x^2\right )-e^8 (-7+\log (5))-e^8 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \left (\frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {e^8 \left (2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\left (3 e^8\right ) \int \frac {1-100 x^2}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (3 e^{16}\right ) \int \frac {2-x+100 x^3+500 x^2 \left (1-\frac {\log (5)}{5}\right )-100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx\\ &=\left (3 e^8\right ) \int \left (\frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )}+\frac {100 x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )}\right ) \, dx+\left (3 e^{16}\right ) \int \left (\frac {2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}+\frac {100 x^2 (5-\log (5))}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}-\frac {100 x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\left (3 e^8\right ) \int \frac {1}{e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )} \, dx+\left (300 e^8\right ) \int \frac {x^2}{-e^{50 x^2}-e^8 x-5 e^8 \left (1-\frac {\log (5)}{5}\right )+e^8 \log \left (x^2\right )} \, dx-\left (3 e^{16}\right ) \int \frac {x}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (6 e^{16}\right ) \int \frac {1}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16}\right ) \int \frac {x^3}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx-\left (300 e^{16}\right ) \int \frac {x^2 \log \left (x^2\right )}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx+\left (300 e^{16} (5-\log (5))\right ) \int \frac {x^2}{\left (e^{50 x^2}+e^8 x+5 e^8 \left (1-\frac {\log (5)}{5}\right )-e^8 \log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.69, size = 36, normalized size = 1.33 \begin {gather*} \frac {3 e^8 x}{e^{50 x^2}+e^8 (5+x-\log (5))-e^8 \log \left (x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.81, size = 26, normalized size = 0.96 \begin {gather*} \frac {3 \, x}{x + e^{\left (50 \, x^{2} - 8\right )} - \log \relax (5) - \log \left (x^{2}\right ) + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 36, normalized size = 1.33 \begin {gather*} \frac {3 \, x e^{8}}{x e^{8} - e^{8} \log \relax (5) - e^{8} \log \left (x^{2}\right ) + 5 \, e^{8} + e^{\left (50 \, x^{2}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.11, size = 83, normalized size = 3.07
method | result | size |
risch | \(\frac {6 x}{i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \,{\mathrm e}^{2 \left (5 x -2\right ) \left (5 x +2\right )}-2 \ln \relax (5)+2 x -4 \ln \relax (x )+10}\) | \(83\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.49, size = 32, normalized size = 1.19 \begin {gather*} \frac {3 \, x e^{8}}{x e^{8} - {\left (\log \relax (5) - 5\right )} e^{8} - 2 \, e^{8} \log \relax (x) + e^{\left (50 \, x^{2}\right )}} \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 = -0.04 \begin {gather*} \int -\frac {3\,\ln \left (x^2\right )+3\,\ln \relax (5)+{\mathrm {e}}^{50\,x^2-8}\,\left (300\,x^2-3\right )-21}{10\,x+{\mathrm {e}}^{100\,x^2-16}-\ln \left (x^2\right )\,\left (2\,x-2\,\ln \relax (5)+2\,{\mathrm {e}}^{50\,x^2-8}+10\right )-\ln \relax (5)\,\left (2\,x+10\right )+{\mathrm {e}}^{50\,x^2-8}\,\left (2\,x-2\,\ln \relax (5)+10\right )+{\ln \left (x^2\right )}^2+{\ln \relax (5)}^2+x^2+25} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.33, size = 22, normalized size = 0.81 \begin {gather*} \frac {3 x}{x + e^{50 x^{2} - 8} - \log {\left (x^{2} \right )} - \log {\relax (5 )} + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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