Optimal. Leaf size=24 \[ 3+e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))} \]
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Rubi [A] time = 0.36, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 7, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6688, 2282, 2194, 2390, 12, 2302, 30} \begin {gather*} e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 30
Rule 2194
Rule 2282
Rule 2302
Rule 2390
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^x \left (2+e^{e^x}\right )-\frac {25}{(-4+x) \log ^2(20-5 x)}\right ) \, dx\\ &=-\left (25 \int \frac {1}{(-4+x) \log ^2(20-5 x)} \, dx\right )+\int e^x \left (2+e^{e^x}\right ) \, dx\\ &=5 \operatorname {Subst}\left (\int -\frac {5}{x \log ^2(x)} \, dx,x,20-5 x\right )+\operatorname {Subst}\left (\int \left (2+e^x\right ) \, dx,x,e^x\right )\\ &=2 e^x-25 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,20-5 x\right )+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+2 e^x-25 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-5 (-4+x))\right )\\ &=e^{e^x}+2 e^x+\frac {25}{\log (5 (4-x))}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.07, size = 21, normalized size = 0.88 \begin {gather*} e^{e^x}+2 e^x+\frac {25}{\log (-5 (-4+x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.87, size = 42, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, e^{\left (2 \, x\right )} \log \left (-5 \, x + 20\right ) + e^{\left (x + e^{x}\right )} \log \left (-5 \, x + 20\right ) + 25 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (-5 \, x + 20\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.31, size = 42, normalized size = 1.75 \begin {gather*} \frac {{\left (2 \, e^{\left (2 \, x\right )} \log \left (-5 \, x + 20\right ) + e^{\left (x + e^{x}\right )} \log \left (-5 \, x + 20\right ) + 25 \, e^{x}\right )} e^{\left (-x\right )}}{\log \left (-5 \, x + 20\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.28, size = 19, normalized size = 0.79
method | result | size |
default | \(2 \,{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{\ln \left (-5 x +20\right )}\) | \(19\) |
risch | \(2 \,{\mathrm e}^{x}+{\mathrm e}^{{\mathrm e}^{x}}+\frac {25}{\ln \left (-5 x +20\right )}\) | \(19\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 8 \, e^{4} E_{1}\left (-x + 4\right ) + \frac {2 \, x e^{x} + {\left (x - 4\right )} e^{\left (e^{x}\right )}}{x - 4} + \frac {25}{i \, \pi + \log \relax (5) + \log \left (x - 4\right )} + 8 \, \int \frac {e^{x}}{x^{2} - 8 \, x + 16}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.14, size = 18, normalized size = 0.75 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+2\,{\mathrm {e}}^x+\frac {25}{\ln \left (20-5\,x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.44, size = 17, normalized size = 0.71 \begin {gather*} 2 e^{x} + e^{e^{x}} + \frac {25}{\log {\left (20 - 5 x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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