Optimal. Leaf size=34 \[ \frac {x}{4}-\frac {x}{\log \left (\log \left (\log \left (\frac {1}{4} e^2 \left (-3 e^{x/5}+x\right ) \log (2)\right )\right )\right )} \]
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Rubi [F]
time = 7.35, 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 {-20 x+12 e^{x/5} x+\left (-60 e^{x/5}+20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log \left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )+\left (15 e^{x/5}-5 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right )\right )}{\left (60 e^{x/5}-20 x\right ) \log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right ) \log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\right ) \log ^2\left (\log \left (\log \left (\frac {1}{4} \left (-3 e^{2+\frac {x}{5}} \log (2)+e^2 x \log (2)\right )\right )\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 \left (\frac {1}{4}+\frac {\left (-5+3 e^{x/5}\right ) x}{5 \left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}-\frac {1}{\log \left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {\left (-5+3 e^{x/5}\right ) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-\int \frac {1}{\log \left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx\\ &=\frac {x}{4}+\frac {1}{5} \int \left (\frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}+\frac {(-5+x) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx-5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {(-5+x) x}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \left (\frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}+\frac {5 x}{\left (-3 e^{x/5}+x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )}\right ) \, dx+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )\\ &=\frac {x}{4}+\frac {1}{5} \int \frac {x}{\left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx+\frac {1}{5} \int \frac {x^2}{\left (3 e^{x/5}-x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx-5 \text {Subst}\left (\int \frac {1}{\log \left (\log \left (\log \left (-3 e^x+5 x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx,x,\frac {x}{5}\right )+\int \frac {x}{\left (-3 e^{x/5}+x\right ) \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right ) \log ^2\left (\log \left (\log \left (-3 e^{x/5}+x\right )+2 \left (1+\frac {1}{2} \log \left (\frac {\log (2)}{4}\right )\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.21, size = 34, normalized size = 1.00 \begin {gather*} \frac {x}{4}-\frac {x}{\log \left (\log \left (2+\log \left (-3 e^{x/5}+x\right )+\log \left (\frac {\log (2)}{4}\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 31, normalized size = 0.91
method | result | size |
risch | \(\frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (-\frac {3 \ln \left (2\right ) {\mathrm e}^{2+\frac {x}{5}}}{4}+\frac {x \,{\mathrm e}^{2} \ln \left (2\right )}{4}\right )\right )\right )}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.61, size = 50, normalized size = 1.47 \begin {gather*} \frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 52, normalized size = 1.53 \begin {gather*} \frac {x \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (\log \left (\frac {1}{4} \, x e^{2} \log \left (2\right ) - \frac {3}{4} \, e^{\left (\frac {1}{5} \, x + 2\right )} \log \left (2\right )\right )\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 7.15, size = 34, normalized size = 1.00 \begin {gather*} \frac {x}{4} - \frac {x}{\log {\left (\log {\left (\log {\left (\frac {x e^{2} \log {\left (2 \right )}}{4} - \frac {3 e^{2} e^{\frac {x}{5}} \log {\left (2 \right )}}{4} \right )} \right )} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.73, size = 50, normalized size = 1.47 \begin {gather*} \frac {x \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right ) - 4 \, x}{4 \, \log \left (\log \left (-2 \, \log \left (2\right ) + \log \left (x - 3 \, e^{\left (\frac {1}{5} \, x\right )}\right ) + \log \left (\log \left (2\right )\right ) + 2\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 14.09, size = 30, normalized size = 0.88 \begin {gather*} \frac {x}{4}-\frac {x}{\ln \left (\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^2\,\ln \left (2\right )}{4}-\frac {3\,{\mathrm {e}}^2\,\ln \left (2\right )\,{\left ({\mathrm {e}}^x\right )}^{1/5}}{4}\right )\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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