Optimal. Leaf size=32 \[ x+\frac {2}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \]
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Rubi [F] time = 4.23, 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 {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (x^2+12 x^5 \log ^4(4)-8 x^9 \log ^8(4)+e^{25 x^2} \left (1-50 x^2-12 x^3 \log ^4(4)+8 x^7 \log ^8(4)\right )\right )+8 x^4 \left (e^{25 x^2}-x^2\right ) \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )-\left (-e^{25 x^2} x+x^3\right ) \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{\left (e^{25 x^2} x-x^3\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=\int \left (\frac {4 x \left (-1+25 x^2\right )}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)+8 x^4 \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )+x \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {x \left (-1+25 x^2\right )}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+\int \frac {2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)+8 x^4 \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )+x \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=4 \int \left (-\frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {25 x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx+\int \left (1-\frac {2 \left (-1+50 x^2+12 x^3 \log ^4(4)\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx\\ &=x-2 \int \frac {-1+50 x^2+12 x^3 \log ^4(4)}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=x-2 \int \left (-\frac {1}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {50 x}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {12 x^2 \log ^4(4)}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=x+2 \int \frac {1}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-100 \int \frac {x}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-\left (24 \log ^4(4)\right ) \int \frac {x^2}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.15, size = 58, normalized size = 1.81 \begin {gather*} \frac {2+4 x^4 \log ^4(4)+x \log \left (\frac {e^{25 x^2}}{x}-x\right )}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 62, normalized size = 1.94 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right ) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.95, size = 61, normalized size = 1.91 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \relax (x) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.18, size = 206, normalized size = 6.44
method | result | size |
risch | \(x +\frac {4 i}{128 i x^{3} \ln \relax (2)^{4}+\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{3}-2 \pi -2 i \ln \relax (x )+2 i \ln \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}\) | \(206\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.51, size = 61, normalized size = 1.91 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \relax (x) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.76, size = 33, normalized size = 1.03 \begin {gather*} x+\frac {2}{\ln \left (\frac {{\mathrm {e}}^{25\,x^2}-x^2}{x}\right )+64\,x^3\,{\ln \relax (2)}^4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.42, size = 26, normalized size = 0.81 \begin {gather*} x + \frac {2}{64 x^{3} \log {\relax (2 )}^{4} + \log {\left (\frac {- x^{2} + e^{25 x^{2}}}{x} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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