Optimal. Leaf size=28 \[ x \log (5)+\frac {e^x (-1+\log (5-\log (1-x)))}{6 x} \]
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Rubi [F] time = 17.55, 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 {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-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 {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{6 (1-x) x^2 (5-\log (1-x))} \, dx\\ &=\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{(1-x) x^2 (5-\log (1-x))} \, dx\\ &=\frac {1}{6} \int \left (6 \log (5)+\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)-5 \log (5-\log (1-x))+10 x \log (5-\log (1-x))-5 x^2 \log (5-\log (1-x))+\log (1-x) \log (5-\log (1-x))-2 x \log (1-x) \log (5-\log (1-x))+x^2 \log (1-x) \log (5-\log (1-x))\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2+(-1+x)^2 \log (1-x) (-1+\log (5-\log (1-x)))-5 (-1+x)^2 \log (5-\log (1-x))\right )}{(1-x) x^2 (5-\log (1-x))} \, dx\\ &=x \log (5)+\frac {1}{6} \int \left (\frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))}+\frac {e^x (-1+x) \log (5-\log (1-x))}{x^2}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x \left (5-9 x+5 x^2-\log (1-x)+2 x \log (1-x)-x^2 \log (1-x)\right )}{(-1+x) x^2 (-5+\log (1-x))} \, dx+\frac {1}{6} \int \frac {e^x (-1+x) \log (5-\log (1-x))}{x^2} \, dx\\ &=x \log (5)+\frac {1}{6} \int \left (\frac {e^x (1-x)}{x^2}+\frac {e^x}{(-1+x) x (-5+\log (1-x))}\right ) \, dx+\frac {1}{6} \int \left (-\frac {e^x \log (5-\log (1-x))}{x^2}+\frac {e^x \log (5-\log (1-x))}{x}\right ) \, dx\\ &=x \log (5)+\frac {1}{6} \int \frac {e^x (1-x)}{x^2} \, dx+\frac {1}{6} \int \frac {e^x}{(-1+x) x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ &=-\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \left (\frac {e^x}{(-1+x) (-5+\log (1-x))}-\frac {e^x}{x (-5+\log (1-x))}\right ) \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ &=-\frac {e^x}{6 x}+x \log (5)+\frac {1}{6} \int \frac {e^x}{(-1+x) (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x}{x (-5+\log (1-x))} \, dx-\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x^2} \, dx+\frac {1}{6} \int \frac {e^x \log (5-\log (1-x))}{x} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.13, size = 36, normalized size = 1.29 \begin {gather*} \frac {1}{6} \left (-\frac {e^x}{x}+6 x \log (5)+\frac {e^x \log (5-\log (1-x))}{x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.78, size = 31, normalized size = 1.11 \begin {gather*} \frac {6 \, x^{2} \log \relax (5) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 31, normalized size = 1.11 \begin {gather*} \frac {6 \, x^{2} \log \relax (5) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 37, normalized size = 1.32
method | result | size |
risch | \(\frac {{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )}{6 x}+\frac {6 x^{2} \ln \relax (5)-{\mathrm e}^{x}}{6 x}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.48, size = 106, normalized size = 3.79 \begin {gather*} -\log \relax (5) \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + {\left ({\left (\log \left (-x + 1\right ) - 5\right )} \log \left (\log \left (-x + 1\right ) - 5\right ) - \log \left (-x + 1\right ) + 5\right )} \log \relax (5) + 5 \, \log \relax (5) \log \left (\log \left (-x + 1\right ) - 5\right ) + \frac {6 \, x^{2} \log \relax (5) + 6 \, x \log \relax (5) \log \left (-x + 1\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, 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 {\ln \left (5-\ln \left (1-x\right )\right )\,\left ({\mathrm {e}}^x\,\left (5\,x^2-10\,x+5\right )-{\mathrm {e}}^x\,\ln \left (1-x\right )\,\left (x^2-2\,x+1\right )\right )+\ln \left (1-x\right )\,\left ({\mathrm {e}}^x\,\left (x^2-2\,x+1\right )+\ln \relax (5)\,\left (6\,x^2-6\,x^3\right )\right )-\ln \relax (5)\,\left (30\,x^2-30\,x^3\right )-{\mathrm {e}}^x\,\left (5\,x^2-9\,x+5\right )}{\ln \left (1-x\right )\,\left (6\,x^2-6\,x^3\right )-30\,x^2+30\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.54, size = 20, normalized size = 0.71 \begin {gather*} x \log {\relax (5 )} + \frac {\left (\log {\left (5 - \log {\left (1 - x \right )} \right )} - 1\right ) e^{x}}{6 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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