Optimal. Leaf size=23 \[ \frac {x}{\log \left (-1+\frac {e^{-2 x}}{x}-(5+x)^2\right )} \]
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Rubi [F] time = 6.54, 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 {10 x+2 x^2+\frac {e^{-2 x} (1+2 x)}{x}+\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {26+72 x+23 x^2+2 x^3}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {-10 x-2 x^2+26 \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )+10 x \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )+x^2 \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx\\ &=-\int \frac {26+72 x+23 x^2+2 x^3}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {-10 x-2 x^2+26 \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )+10 x \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )+x^2 \log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=-\int \left (\frac {3}{\left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {2 x}{\left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}-\frac {2 (26+5 x)}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx+\int \frac {-\frac {2 x (5+x)}{26+10 x+x^2}+\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=-\left (2 \int \frac {x}{\left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\right )+2 \int \frac {26+5 x}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \left (-\frac {2 x (5+x)}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x (5+x)}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\right )+2 \int \frac {-26-5 x}{\left (26+10 x+x^2\right ) \left (1-e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=-\left (2 \int \left (\frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {-26-5 x}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx\right )+2 \int \left (\frac {26}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {5 x}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=-\left (2 \int \frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\right )-2 \int \frac {-26-5 x}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+10 \int \frac {x}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 \int \frac {1}{\left (26+10 x+x^2\right ) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=-\left (2 \int \left (-\frac {26}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}-\frac {5 x}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx\right )-2 \int \frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+10 \int \left (\frac {1+5 i}{((10-2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {1-5 i}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx+52 \int \left (\frac {i}{((-10+2 i)-2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {i}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=52 i \int \frac {1}{((-10+2 i)-2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 i \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+10 \int \frac {x}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10-50 i) \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10+50 i) \int \frac {1}{((10-2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 \int \frac {1}{\left (26+10 x+x^2\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=52 i \int \frac {1}{((-10+2 i)-2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 i \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+10 \int \left (\frac {1+5 i}{((10-2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {1-5 i}{((10+2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx+(10-50 i) \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10+50 i) \int \frac {1}{((10-2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 \int \left (\frac {i}{((-10+2 i)-2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}+\frac {i}{((10+2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )}\right ) \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ &=52 i \int \frac {1}{((-10+2 i)-2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 i \int \frac {1}{((10+2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 i \int \frac {1}{((-10+2 i)-2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+52 i \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {1}{\log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-2 \int \frac {x}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx-3 \int \frac {1}{\left (-1+e^{2 x} x \left (26+10 x+x^2\right )\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10-50 i) \int \frac {1}{((10+2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10-50 i) \int \frac {1}{((10+2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10+50 i) \int \frac {1}{((10-2 i)+2 x) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+(10+50 i) \int \frac {1}{((10-2 i)+2 x) \left (-1+26 e^{2 x} x+10 e^{2 x} x^2+e^{2 x} x^3\right ) \log ^2\left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx+\int \frac {1}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 24, normalized size = 1.04 \begin {gather*} \frac {x}{\log \left (-26+\frac {e^{-2 x}}{x}-10 x-x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 28, normalized size = 1.22 \begin {gather*} \frac {x}{\log \left (-x^{2} - 10 \, x + e^{\left (-2 \, x + \log \left (\frac {e^{\left (-1\right )}}{x}\right ) + 1\right )} - 26\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.74, size = 28, normalized size = 1.22 \begin {gather*} \frac {x}{\log \left (-x^{3} - 10 \, x^{2} - 26 \, x + e^{\left (-2 \, x\right )}\right ) - \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right ) \ln \left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right )+\left (2 x +1\right ) {\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}+2 x^{2}+10 x}{\left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right ) \ln \left ({\mathrm e}^{\ln \left (\frac {{\mathrm e}^{-1}}{x}\right )-2 x +1}-x^{2}-10 x -26\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 34, normalized size = 1.48 \begin {gather*} -\frac {x}{2 \, x - \log \left (-{\left (x^{3} + 10 \, x^{2} + 26 \, x\right )} e^{\left (2 \, x\right )} + 1\right ) + \log \relax (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 = -0.04 \begin {gather*} \int -\frac {10\,x+{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}\,\left (2\,x+1\right )+2\,x^2-\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )}{{\ln \left ({\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}-10\,x-x^2-26\right )}^2\,\left (10\,x-{\mathrm {e}}^{\ln \left (\frac {{\mathrm {e}}^{-1}}{x}\right )-2\,x+1}+x^2+26\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 22, normalized size = 0.96 \begin {gather*} \frac {x}{\log {\left (- x^{2} - 10 x - 26 + \frac {e^{1 - 2 x}}{e x} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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