Optimal. Leaf size=25 \[ 5-e^x-2 x^2-\frac {5 x}{\log (2)}+2 \log ^4(x) \]
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Rubi [A] time = 0.06, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 2194, 2302, 30} \begin {gather*} -2 x^2-e^x+2 \log ^4(x)-\frac {5 x}{\log (2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 30
Rule 2194
Rule 2302
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-5 x-e^x x \log (2)-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x} \, dx}{\log (2)}\\ &=\frac {\int \left (-e^x \log (2)+\frac {-5 x-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x}\right ) \, dx}{\log (2)}\\ &=\frac {\int \frac {-5 x-4 x^2 \log (2)+8 \log (2) \log ^3(x)}{x} \, dx}{\log (2)}-\int e^x \, dx\\ &=-e^x+\frac {\int \left (-5-4 x \log (2)+\frac {8 \log (2) \log ^3(x)}{x}\right ) \, dx}{\log (2)}\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+8 \int \frac {\log ^3(x)}{x} \, dx\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+8 \operatorname {Subst}\left (\int x^3 \, dx,x,\log (x)\right )\\ &=-e^x-2 x^2-\frac {5 x}{\log (2)}+2 \log ^4(x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 32, normalized size = 1.28 \begin {gather*} -e^x-\frac {5 x}{\log (2)}-\frac {x^2 \log (16)}{2 \log (2)}+2 \log ^4(x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 24, normalized size = 0.96
method | result | size |
risch | \(-\frac {5 x}{\ln \relax (2)}-2 x^{2}-{\mathrm e}^{x}+2 \ln \relax (x )^{4}\) | \(24\) |
default | \(\frac {-5 x -2 x^{2} \ln \relax (2)-{\mathrm e}^{x} \ln \relax (2)+2 \ln \relax (2) \ln \relax (x )^{4}}{\ln \relax (2)}\) | \(31\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 30, normalized size = 1.20 \begin {gather*} \frac {2 \, \log \relax (2) \log \relax (x)^{4} - 2 \, x^{2} \log \relax (2) - e^{x} \log \relax (2) - 5 \, x}{\log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.28, size = 23, normalized size = 0.92 \begin {gather*} 2\,{\ln \relax (x)}^4-{\mathrm {e}}^x-\frac {5\,x}{\ln \relax (2)}-2\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.28, size = 20, normalized size = 0.80 \begin {gather*} - 2 x^{2} - \frac {5 x}{\log {\relax (2 )}} - e^{x} + 2 \log {\relax (x )}^{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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