Optimal. Leaf size=22 \[ -1+\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \]
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Rubi [F]
time = 1.69, 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^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, 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^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 x^2 \left (e^4+x+\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{x^2 \left (e^4+x+\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \left (\frac {1}{x^2}+\frac {5 \left (e^4+x+\log (x)\right )^{-1+x} \left (-e^4+x^2-\log (x)+e^4 x \log \left (e^4+x+\log (x)\right )+x^2 \log \left (e^4+x+\log (x)\right )+x \log (x) \log \left (e^4+x+\log (x)\right )\right )}{x^2}\right ) \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^{-1+x} \left (-e^4+x^2-\log (x)+e^4 x \log \left (e^4+x+\log (x)\right )+x^2 \log \left (e^4+x+\log (x)\right )+x \log (x) \log \left (e^4+x+\log (x)\right )\right )}{x^2} \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \left (\frac {\left (-e^4+x^2-\log (x)\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}+\frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x}\right ) \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (-e^4+x^2-\log (x)\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \left (\frac {\left (-e^4+x^2\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}-\frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}\right ) \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \frac {\left (-e^4+x^2\right ) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx\\ &=-\frac {1}{2 x}-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \left (\left (e^4+x+\log (x)\right )^{-1+x}-\frac {e^4 \left (e^4+x+\log (x)\right )^{-1+x}}{x^2}\right ) \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx\\ &=-\frac {1}{2 x}+\frac {5}{2} \int \left (e^4+x+\log (x)\right )^{-1+x} \, dx-\frac {5}{2} \int \frac {\log (x) \left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx+\frac {5}{2} \int \frac {\left (e^4+x+\log (x)\right )^x \log \left (e^4+x+\log (x)\right )}{x} \, dx-\frac {1}{2} \left (5 e^4\right ) \int \frac {\left (e^4+x+\log (x)\right )^{-1+x}}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.46, size = 20, normalized size = 0.91 \begin {gather*} \frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.70, size = 20, normalized size = 0.91
method | result | size |
risch | \(-\frac {1}{2 x}+\frac {5 \left (\ln \left (x \right )+x +{\mathrm e}^{4}\right )^{x}}{2 x}\) | \(20\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.31, size = 17, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.49, size = 17, normalized size = 0.77 \begin {gather*} \frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.75, size = 22, normalized size = 1.00 \begin {gather*} \frac {5 e^{x \log {\left (x + \log {\left (x \right )} + e^{4} \right )}}}{2 x} - \frac {1}{2 x} \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*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.81, size = 17, normalized size = 0.77 \begin {gather*} \frac {5\,{\left (x+{\mathrm {e}}^4+\ln \left (x\right )\right )}^x-1}{2\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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