Optimal. Leaf size=30 \[ -e^x x+\frac {5 \left (5+x^2+\frac {\left (2+x^2\right ) (1+\log (x))}{x}\right )}{x} \]
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Rubi [A] time = 0.05, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {14, 2176, 2194, 2304} \begin {gather*} \frac {10}{x^2}+\frac {10 \log (x)}{x^2}+5 x+e^x-e^x (x+1)+\frac {25}{x}+5 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 2176
Rule 2194
Rule 2304
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-e^x (1+x)+\frac {5 \left (-2-5 x+x^2+x^3-4 \log (x)\right )}{x^3}\right ) \, dx\\ &=5 \int \frac {-2-5 x+x^2+x^3-4 \log (x)}{x^3} \, dx-\int e^x (1+x) \, dx\\ &=-e^x (1+x)+5 \int \left (\frac {-2-5 x+x^2+x^3}{x^3}-\frac {4 \log (x)}{x^3}\right ) \, dx+\int e^x \, dx\\ &=e^x-e^x (1+x)+5 \int \frac {-2-5 x+x^2+x^3}{x^3} \, dx-20 \int \frac {\log (x)}{x^3} \, dx\\ &=e^x+\frac {5}{x^2}-e^x (1+x)+\frac {10 \log (x)}{x^2}+5 \int \left (1-\frac {2}{x^3}-\frac {5}{x^2}+\frac {1}{x}\right ) \, dx\\ &=e^x+\frac {10}{x^2}+\frac {25}{x}+5 x-e^x (1+x)+5 \log (x)+\frac {10 \log (x)}{x^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 28, normalized size = 0.93 \begin {gather*} -\frac {-10-25 x+\left (-5+e^x\right ) x^3-5 \left (2+x^2\right ) \log (x)}{x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 30, normalized size = 1.00 \begin {gather*} -\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, {\left (x^{2} + 2\right )} \log \relax (x) - 25 \, x - 10}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 32, normalized size = 1.07 \begin {gather*} -\frac {x^{3} e^{x} - 5 \, x^{3} - 5 \, x^{2} \log \relax (x) - 25 \, x - 10 \, \log \relax (x) - 10}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 31, normalized size = 1.03
method | result | size |
default | \(-{\mathrm e}^{x} x +5 x +5 \ln \relax (x )+\frac {25}{x}+\frac {10}{x^{2}}+\frac {10 \ln \relax (x )}{x^{2}}\) | \(31\) |
norman | \(\frac {10+25 x +5 x^{3}-{\mathrm e}^{x} x^{3}+10 \ln \relax (x )}{x^{2}}+5 \ln \relax (x )\) | \(31\) |
risch | \(\frac {10 \ln \relax (x )}{x^{2}}+\frac {-{\mathrm e}^{x} x^{3}+5 x^{2} \ln \relax (x )+5 x^{3}+25 x +10}{x^{2}}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.38, size = 36, normalized size = 1.20 \begin {gather*} -{\left (x - 1\right )} e^{x} + 5 \, x + \frac {25}{x} + \frac {10 \, \log \relax (x)}{x^{2}} + \frac {10}{x^{2}} - e^{x} + 5 \, \log \relax (x) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.24, size = 25, normalized size = 0.83 \begin {gather*} 5\,\ln \relax (x)-x\,\left ({\mathrm {e}}^x-5\right )+\frac {25\,x+10\,\ln \relax (x)+10}{x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.30, size = 29, normalized size = 0.97 \begin {gather*} - x e^{x} + 5 x + 5 \log {\relax (x )} + \frac {25 x + 10}{x^{2}} + \frac {10 \log {\relax (x )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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