Optimal. Leaf size=33 \[ -x+\frac {x}{-5+\frac {-x+\log \left (e^5+\frac {x}{4}\right )}{-8-\log (x)}} \]
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Rubi [F] time = 4.29, 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 {-1912 x+81 x^2-x^3+e^5 \left (-7680+324 x-4 x^2\right )+\left (-479 x+10 x^2+e^5 (-1920+40 x)\right ) \log (x)+\left (-120 e^5-30 x\right ) \log ^2(x)+\left (-89 x+2 x^2+e^5 (-356+8 x)+\left (-44 e^5-11 x\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^5+x\right )\right )+\left (-4 e^5-x\right ) \log ^2\left (\frac {1}{4} \left (4 e^5+x\right )\right )}{1600 x-80 x^2+x^3+e^5 \left (6400-320 x+4 x^2\right )+\left (e^5 (1600-40 x)+400 x-10 x^2\right ) \log (x)+\left (100 e^5+25 x\right ) \log ^2(x)+\left (e^5 (320-8 x)+80 x-2 x^2+\left (40 e^5+10 x\right ) \log (x)\right ) \log \left (\frac {1}{4} \left (4 e^5+x\right )\right )+\left (4 e^5+x\right ) \log ^2\left (\frac {1}{4} \left (4 e^5+x\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 {-7680 e^5-1912 \left (1-\frac {81 e^5}{478}\right ) x+81 \left (1-\frac {4 e^5}{81}\right ) x^2-x^3-\left (4 e^5+x\right ) \log ^2\left (e^5+\frac {x}{4}\right )+\left (4 e^5+x\right ) \log \left (e^5+\frac {x}{4}\right ) (-89+2 x-11 \log (x))-\left (-40 e^5 (-48+x)+(479-10 x) x\right ) \log (x)-30 \left (4 e^5+x\right ) \log ^2(x)}{\left (4 e^5+x\right ) \left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2} \, dx\\ &=\int \left (-\frac {6}{5}+\frac {\left (20 e^5+2 \left (3-2 e^5\right ) x-x^2\right ) \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{5 \left (4 e^5+x\right ) \left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2}+\frac {\left (1-8 e^5\right ) x-2 x^2+4 e^5 \log \left (e^5+\frac {x}{4}\right )+x \log \left (e^5+\frac {x}{4}\right )}{5 \left (4 e^5+x\right ) \left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )}\right ) \, dx\\ &=-\frac {6 x}{5}+\frac {1}{5} \int \frac {\left (20 e^5+2 \left (3-2 e^5\right ) x-x^2\right ) \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{\left (4 e^5+x\right ) \left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {\left (1-8 e^5\right ) x-2 x^2+4 e^5 \log \left (e^5+\frac {x}{4}\right )+x \log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )} \, dx\\ &=-\frac {6 x}{5}+\frac {1}{5} \int \left (\frac {6 \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {x \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {4 e^5 \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}\right ) \, dx+\frac {1}{5} \int \left (\frac {\left (-1+8 e^5\right ) x}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}+\frac {2 x^2}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}-\frac {4 e^5 \log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}-\frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}\right ) \, dx\\ &=-\frac {6 x}{5}-\frac {1}{5} \int \frac {x \left (x-\log \left (e^5+\frac {x}{4}\right )\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \int \frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {2}{5} \int \frac {x^2}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {6}{5} \int \frac {x-\log \left (e^5+\frac {x}{4}\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (4 e^5\right ) \int \frac {x-\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (4 e^5\right ) \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (-1+8 e^5\right ) \int \frac {x}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx\\ &=-\frac {6 x}{5}-\frac {1}{5} \int \left (\frac {x^2}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}\right ) \, dx-\frac {1}{5} \int \left (-\frac {4 e^5 \log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}-\frac {\log \left (e^5+\frac {x}{4}\right )}{40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)}\right ) \, dx+\frac {2}{5} \int \left (-\frac {4 e^5}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)}+\frac {x}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)}+\frac {16 e^{10}}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}\right ) \, dx+\frac {6}{5} \int \left (\frac {x}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {\log \left (e^5+\frac {x}{4}\right )}{\left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2}\right ) \, dx-\frac {1}{5} \left (4 e^5\right ) \int \left (\frac {x}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}\right ) \, dx-\frac {1}{5} \left (4 e^5\right ) \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (-1+8 e^5\right ) \int \left (\frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)}-\frac {4 e^5}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )}\right ) \, dx\\ &=-\frac {6 x}{5}-\frac {1}{5} \int \frac {x^2}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)} \, dx+\frac {2}{5} \int \frac {x}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {6}{5} \int \frac {x}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {6}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (4 e^5\right ) \int \frac {x}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \left (4 e^5\right ) \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {1}{5} \left (32 e^{10}\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (4 e^5 \left (1-8 e^5\right )\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (-1+8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx\\ &=-\frac {6 x}{5}-\frac {1}{5} \int \frac {x^2}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)} \, dx+\frac {2}{5} \int \frac {x}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {6}{5} \int \frac {x}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {6}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (4 e^5\right ) \int \left (\frac {1}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}-\frac {4 e^5}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2}\right ) \, dx+\frac {1}{5} \left (4 e^5\right ) \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {1}{5} \left (32 e^{10}\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (4 e^5 \left (1-8 e^5\right )\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (-1+8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx\\ &=-\frac {6 x}{5}-\frac {1}{5} \int \frac {x^2}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {x \log \left (e^5+\frac {x}{4}\right )}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)} \, dx+\frac {2}{5} \int \frac {x}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {6}{5} \int \frac {x}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {6}{5} \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (4 e^5\right ) \int \frac {1}{\left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \left (4 e^5\right ) \int \frac {\log \left (e^5+\frac {x}{4}\right )}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx-\frac {1}{5} \left (8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx+\frac {1}{5} \left (16 e^{10}\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )^2} \, dx+\frac {1}{5} \left (32 e^{10}\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (4 e^5 \left (1-8 e^5\right )\right ) \int \frac {1}{\left (4 e^5+x\right ) \left (-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)\right )} \, dx+\frac {1}{5} \left (-1+8 e^5\right ) \int \frac {1}{-40+x-\log \left (e^5+\frac {x}{4}\right )-5 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.16, size = 43, normalized size = 1.30 \begin {gather*} -\frac {1}{5} x \left (6+\frac {x-\log \left (e^5+\frac {x}{4}\right )}{40-x+\log \left (e^5+\frac {x}{4}\right )+5 \log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.71, size = 42, normalized size = 1.27 \begin {gather*} -\frac {x^{2} - 6 \, x \log \relax (x) - x \log \left (\frac {1}{4} \, x + e^{5}\right ) - 48 \, x}{x - 5 \, \log \relax (x) - \log \left (\frac {1}{4} \, x + e^{5}\right ) - 40} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 51, normalized size = 1.55 \begin {gather*} -\frac {x^{2} + 2 \, x \log \relax (2) - x \log \left (x + 4 \, e^{5}\right ) - 6 \, x \log \relax (x) - 48 \, x}{x + 2 \, \log \relax (2) - \log \left (x + 4 \, e^{5}\right ) - 5 \, \log \relax (x) - 40} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 29, normalized size = 0.88
method | result | size |
risch | \(-x +\frac {\left (\ln \relax (x )+8\right ) x}{x -5 \ln \relax (x )-\ln \left ({\mathrm e}^{5}+\frac {x}{4}\right )-40}\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.57, size = 50, normalized size = 1.52 \begin {gather*} -\frac {x^{2} + 2 \, x {\left (\log \relax (2) - 24\right )} - x \log \left (x + 4 \, e^{5}\right ) - 6 \, x \log \relax (x)}{x + 2 \, \log \relax (2) - \log \left (x + 4 \, e^{5}\right ) - 5 \, \log \relax (x) - 40} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.44, size = 187, normalized size = 5.67 \begin {gather*} \ln \relax (x)-x+\frac {54\,x+180\,{\mathrm {e}}^5}{x^2+\left (4\,{\mathrm {e}}^5-6\right )\,x-20\,{\mathrm {e}}^5}+\frac {\frac {x\,\left (312\,x+1280\,{\mathrm {e}}^5+20\,{\mathrm {e}}^5\,{\ln \relax (x)}^2+5\,x\,{\ln \relax (x)}^2-4\,x\,{\mathrm {e}}^5+320\,{\mathrm {e}}^5\,\ln \relax (x)+79\,x\,\ln \relax (x)-x^2\right )}{6\,x+20\,{\mathrm {e}}^5-4\,x\,{\mathrm {e}}^5-x^2}+\frac {x\,\ln \left (\frac {x}{4}+{\mathrm {e}}^5\right )\,\left (\ln \relax (x)+9\right )\,\left (x+4\,{\mathrm {e}}^5\right )}{6\,x+20\,{\mathrm {e}}^5-4\,x\,{\mathrm {e}}^5-x^2}}{\ln \left (\frac {x}{4}+{\mathrm {e}}^5\right )-x+5\,\ln \relax (x)+40}+\frac {\ln \relax (x)\,\left (6\,x+20\,{\mathrm {e}}^5\right )}{x^2+\left (4\,{\mathrm {e}}^5-6\right )\,x-20\,{\mathrm {e}}^5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.41, size = 27, normalized size = 0.82 \begin {gather*} - x + \frac {- x \log {\relax (x )} - 8 x}{- x + 5 \log {\relax (x )} + \log {\left (\frac {x}{4} + e^{5} \right )} + 40} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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