Optimal. Leaf size=23 \[ e^{-x} \left (5+\frac {e}{2}+x\right ) \log \left (\log \left (5-e^x\right )\right ) \]
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Rubi [F] time = 1.44, 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 (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\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 {e^{-x} \left (e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (-10+2 e^x\right ) \log \left (5-e^x\right )} \, dx\\ &=\int \frac {1}{2} \left (\frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}-e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (\frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}-e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {e+2 (5+x)}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\frac {1}{2} \int e^{-x} (8+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {-10-e-2 x}{\left (5-e^x\right ) \log \left (5-e^x\right )} \, dx-\frac {1}{2} \int \left (8 \left (1+\frac {e}{8}\right ) e^{-x} \log \left (\log \left (5-e^x\right )\right )+2 e^{-x} x \log \left (\log \left (5-e^x\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (\frac {10 \left (1+\frac {e}{10}\right )}{\left (-5+e^x\right ) \log \left (5-e^x\right )}+\frac {2 x}{\left (-5+e^x\right ) \log \left (5-e^x\right )}\right ) \, dx-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \int \frac {1}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \operatorname {Subst}\left (\int \frac {1}{(-5+x) x \log (5-x)} \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=-\left (\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx\right )+\frac {1}{2} (10+e) \operatorname {Subst}\left (\int \left (\frac {1}{5 (-5+x) \log (5-x)}-\frac {1}{5 x \log (5-x)}\right ) \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{(-5+x) \log (5-x)} \, dx,x,e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,5-e^x\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\frac {1}{10} (10+e) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (5-e^x\right )\right )+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ &=\frac {1}{10} (10+e) \log \left (\log \left (5-e^x\right )\right )+\frac {1}{10} (-10-e) \operatorname {Subst}\left (\int \frac {1}{x \log (5-x)} \, dx,x,e^x\right )-\frac {1}{2} (8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right ) \, dx+\int \frac {x}{\left (-5+e^x\right ) \log \left (5-e^x\right )} \, dx-\int e^{-x} x \log \left (\log \left (5-e^x\right )\right ) \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.22, size = 24, normalized size = 1.04 \begin {gather*} \frac {1}{2} e^{-x} (10+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (2 \, x + e + 8\right )} e^{x} - 10 \, x - 5 \, e - 40\right )} \log \left (-e^{x} + 5\right ) \log \left (\log \left (-e^{x} + 5\right )\right ) - {\left (2 \, x + e + 10\right )} e^{x}}{2 \, {\left (e^{\left (2 \, x\right )} - 5 \, e^{x}\right )} \log \left (-e^{x} + 5\right )}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 22, normalized size = 0.96
method | result | size |
risch | \(\frac {\left ({\mathrm e}+2 x +10\right ) {\mathrm e}^{-x} \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )}{2}\) | \(22\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 21, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \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 {{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+10\right )+\ln \left (5-{\mathrm {e}}^x\right )\,\ln \left (\ln \left (5-{\mathrm {e}}^x\right )\right )\,\left (10\,x+5\,\mathrm {e}-{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+8\right )+40\right )}{\ln \left (5-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.55, size = 20, normalized size = 0.87 \begin {gather*} \frac {\left (2 x + e + 10\right ) e^{- x} \log {\left (\log {\left (5 - e^{x} \right )} \right )}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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