Optimal. Leaf size=32 \[ \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (x \left (x+e^{-x (4+2 x)} x\right )\right )} \]
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Rubi [F] time = 7.39, 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 {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\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 \log \left (\frac {25 e^{-x}}{4}\right ) \left (-\frac {\left (1+e^{2 x (2+x)}-2 x-2 x^2\right ) \log \left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) x}-\log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{\log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\frac {\left (1+e^{2 x (2+x)}-2 x-2 x^2\right ) \log \left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) x}-\log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{\log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \left (\frac {2 (1+x) \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}+\frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\log \left (\frac {25 e^{-x}}{4}\right )-x \log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right ) \left (-\log \left (\frac {25 e^{-x}}{4}\right )-x \log \left (\left (1+e^{-2 x (2+x)}\right ) x^2\right )\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {(1+x) \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ &=2 \int \left (-\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}-\frac {\log \left (\frac {25 e^{-x}}{4}\right )}{\log \left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx+4 \int \left (\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}+\frac {x \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{x \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\right )-2 \int \frac {\log \left (\frac {25 e^{-x}}{4}\right )}{\log \left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx+4 \int \frac {x \log ^2\left (\frac {25 e^{-x}}{4}\right )}{\left (1+e^{2 x (2+x)}\right ) \log ^2\left (x^2+e^{-2 x (2+x)} x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 33, normalized size = 1.03 \begin {gather*} \frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (\left (1+e^{-4 x-2 x^2}\right ) x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.55, size = 46, normalized size = 1.44 \begin {gather*} \frac {x^{2} - 2 \, x \log \left (\frac {25}{4}\right ) + \log \left (\frac {25}{4}\right )^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.87, size = 65, normalized size = 2.03 \begin {gather*} \frac {x^{2} - 4 \, x \log \relax (5) + 4 \, \log \relax (5)^{2} + 4 \, x \log \relax (2) - 8 \, \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.69, size = 426, normalized size = 13.31
method | result | size |
risch | \(\frac {i \left (16 \ln \relax (5)^{2}+16 \ln \relax (2)^{2}-32 \ln \relax (2) \ln \relax (5)-16 \ln \relax (5) \ln \left ({\mathrm e}^{x}\right )+16 \ln \relax (2) \ln \left ({\mathrm e}^{x}\right )+4 \ln \left ({\mathrm e}^{x}\right )^{2}\right )}{2 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-4 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )-2 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2}+2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )-2 \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}+2 \pi \mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{3}-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{3}+8 i \ln \relax (x )+4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )-4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}\right )}\) | \(426\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.59, size = 64, normalized size = 2.00 \begin {gather*} -\frac {x^{2} - 4 \, x {\left (\log \relax (5) - \log \relax (2)\right )} + 4 \, \log \relax (5)^{2} - 8 \, \log \relax (5) \log \relax (2) + 4 \, \log \relax (2)^{2}}{2 \, x^{2} + 4 \, x - 2 \, \log \relax (x) - \log \left (e^{\left (2 \, x^{2} + 4 \, x\right )} + 1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.34, size = 226, normalized size = 7.06 \begin {gather*} \frac {{\left (x-\ln \left (\frac {25}{4}\right )\right )}^2+\frac {x\,\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )\,\left ({\mathrm {e}}^{2\,x^2+4\,x}+1\right )\,\left (x-\ln \left (\frac {25}{4}\right )\right )}{2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1}}{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )}-x\,\ln \left (\frac {25}{4}\right )+x^2-\frac {2\,\left (3\,x^2\,\ln \left (\frac {25}{4}\right )+3\,x^3\,\ln \left (\frac {25}{4}\right )-8\,x^4\,\ln \left (\frac {25}{4}\right )-12\,x^5\,\ln \left (\frac {25}{4}\right )-4\,x^6\,\ln \left (\frac {25}{4}\right )-3\,x^3-3\,x^4+8\,x^5+12\,x^6+4\,x^7\right )}{\left (4\,x^3+8\,x^2-3\right )\,\left (2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.67, size = 68, normalized size = 2.12 \begin {gather*} \frac {x^{2} - 4 x \log {\relax (5 )} + 4 x \log {\relax (2 )} - 8 \log {\relax (2 )} \log {\relax (5 )} + 4 \log {\relax (2 )}^{2} + 4 \log {\relax (5 )}^{2}}{\log {\left (\left (x^{2} e^{2 x^{2} + 4 x} + x^{2}\right ) e^{- 2 x^{2} - 4 x} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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