Optimal. Leaf size=27 \[ \log \left (-x+e^x \left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\right ) \]
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Rubi [F] time = 2.46, 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 {-1+6 e^x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\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 {6+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}+\frac {-5+6 x-\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )+x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}\right ) \, dx\\ &=\int \frac {6+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )} \, dx+\int \frac {-5+6 x-\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )+x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\\ &=\int \frac {-5+6 x+(-1+x) \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx+\operatorname {Subst}\left (\int \frac {6+\log \left (\frac {5 x}{-e^5+\log (5)}\right )}{x \left (5+\log \left (\frac {5 x}{-e^5+\log (5)}\right )\right )} \, dx,x,e^x\right )\\ &=\int \left (-\frac {6 x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {5}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}-\frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )}\right ) \, dx+\operatorname {Subst}\left (\int \frac {6+x}{5+x} \, dx,x,\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\\ &=-\left (5 \int \frac {1}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\right )-6 \int \frac {x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx+\operatorname {Subst}\left (\int \left (1+\frac {1}{5+x}\right ) \, dx,x,\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )\\ &=x+\log \left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )-5 \int \frac {1}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-6 \int \frac {x}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (-5 e^x+x-e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx-\int \frac {\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )}{\left (5+\log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 30, normalized size = 1.11 \begin {gather*} \log \left (5 e^x-x+e^x \log \left (\frac {5 e^x}{-e^5+\log (5)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.81, size = 172, normalized size = 6.37 \begin {gather*} \frac {1}{2} \, \log \left (\pi ^{2} + x^{2} + 2 \, {\left (x + 5\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right ) + \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + 10 \, x + 25\right ) + \frac {1}{2} \, \log \left (\frac {e^{\left (2 \, x\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + x^{2} + {\left (\pi ^{2} + x^{2} + 10 \, x + 25\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{2} + 5 \, x\right )} e^{x} + 2 \, {\left ({\left (x + 5\right )} e^{\left (2 \, x\right )} - x e^{x}\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )}{\pi ^{2} + x^{2} + 2 \, {\left (x + 5\right )} \log \left (\frac {5}{e^{5} - \log \relax (5)}\right ) + \log \left (\frac {5}{e^{5} - \log \relax (5)}\right )^{2} + 10 \, x + 25}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.58, size = 154, normalized size = 5.70 \begin {gather*} \frac {1}{2} \, \log \left (\pi ^{2} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} e^{x} + 2 \, x e^{\left (2 \, x\right )} \log \relax (5) - 2 \, x e^{x} \log \relax (5) + e^{\left (2 \, x\right )} \log \relax (5)^{2} - 2 \, x e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right ) + 2 \, x e^{x} \log \left (e^{5} - \log \relax (5)\right ) - 2 \, e^{\left (2 \, x\right )} \log \relax (5) \log \left (e^{5} - \log \relax (5)\right ) + e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right )^{2} + x^{2} + 10 \, x e^{\left (2 \, x\right )} - 10 \, x e^{x} + 10 \, e^{\left (2 \, x\right )} \log \relax (5) - 10 \, e^{\left (2 \, x\right )} \log \left (e^{5} - \log \relax (5)\right ) + 25 \, e^{\left (2 \, x\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 26, normalized size = 0.96
method | result | size |
norman | \(\ln \left (-{\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+x -5 \,{\mathrm e}^{x}\right )\) | \(26\) |
derivativedivides | \(\ln \left ({\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+5 \,{\mathrm e}^{x}-x \right )\) | \(27\) |
default | \(\ln \left ({\mathrm e}^{x} \ln \left (\frac {5 \,{\mathrm e}^{x}}{\ln \relax (5)-{\mathrm e}^{5}}\right )+5 \,{\mathrm e}^{x}-x \right )\) | \(27\) |
risch | \(x +\ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (2 \,{\mathrm e}^{x} \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 \,{\mathrm e}^{x} \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3}+2 i {\mathrm e}^{x} \ln \relax (5)-2 i {\mathrm e}^{x} \ln \left (-\ln \relax (5)+{\mathrm e}^{5}\right )-2 \,{\mathrm e}^{x} \pi -2 i x +10 i {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{2}\right )\) | \(76\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 26, normalized size = 0.96 \begin {gather*} \log \left (e^{x} \log \left (-\frac {5 \, e^{x}}{e^{5} - \log \relax (5)}\right ) - x + 5 \, e^{x}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.90, size = 33, normalized size = 1.22 \begin {gather*} \ln \left (5\,{\mathrm {e}}^x-x+{\mathrm {e}}^x\,\left (\ln \relax (5)-\ln \left ({\mathrm {e}}^5-\ln \relax (5)\right )+\pi \,1{}\mathrm {i}\right )+x\,{\mathrm {e}}^x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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