Optimal. Leaf size=29 \[ \frac {2 \log \left (\frac {3 e^x}{x}\right )}{\frac {1+x}{e^5}-\log ^2(5+x)} \]
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Rubi [F]
time = 1.96, 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^5 \left (-10-2 x+10 x^2+2 x^3\right )+e^{10} \left (10-8 x-2 x^2\right ) \log ^2(5+x)+\log \left (\frac {3 e^x}{x}\right ) \left (e^5 \left (-10 x-2 x^2\right )+4 e^{10} x \log (5+x)\right )}{5 x+11 x^2+7 x^3+x^4+e^5 \left (-10 x-12 x^2-2 x^3\right ) \log ^2(5+x)+e^{10} \left (5 x+x^2\right ) \log ^4(5+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 {2 e^5 \left (-x \log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )+\left (-5+4 x+x^2\right ) \left (1+x-e^5 \log ^2(5+x)\right )\right )}{x (5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\\ &=\left (2 e^5\right ) \int \frac {-x \log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )+\left (-5+4 x+x^2\right ) \left (1+x-e^5 \log ^2(5+x)\right )}{x (5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\\ &=\left (2 e^5\right ) \int \left (-\frac {\log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}+\frac {-1+x}{x \left (1+x-e^5 \log ^2(5+x)\right )}\right ) \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \left (5+x-2 e^5 \log (5+x)\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {-1+x}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx\\ &=-\left (\left (2 e^5\right ) \int \left (\frac {5 \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}+\frac {x \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}-\frac {2 e^5 \log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}\right ) \, dx\right )+\left (2 e^5\right ) \int \left (\frac {1}{1+x-e^5 \log ^2(5+x)}-\frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )}\right ) \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {x \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx-\left (10 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\\ &=\left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx-\left (2 e^5\right ) \int \left (\frac {\log \left (\frac {3 e^x}{x}\right )}{\left (1+x-e^5 \log ^2(5+x)\right )^2}-\frac {5 \log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2}\right ) \, dx-\left (10 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\\ &=-\left (\left (2 e^5\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right )}{\left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\right )+\left (2 e^5\right ) \int \frac {1}{1+x-e^5 \log ^2(5+x)} \, dx-\left (2 e^5\right ) \int \frac {1}{x \left (1+x-e^5 \log ^2(5+x)\right )} \, dx+\left (4 e^{10}\right ) \int \frac {\log \left (\frac {3 e^x}{x}\right ) \log (5+x)}{(5+x) \left (1+x-e^5 \log ^2(5+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.23, size = 31, normalized size = 1.07 \begin {gather*} -\frac {2 e^5 \log \left (\frac {3 e^x}{x}\right )}{-1-x+e^5 \log ^2(5+x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 172.84, size = 135, normalized size = 4.66
method | result | size |
risch | \(-\frac {2 \,{\mathrm e}^{5} \ln \left (3 \,{\mathrm e}^{x}\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}+\frac {{\mathrm e}^{5} \left (-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )+i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+2 \ln \left (x \right )\right )}{{\mathrm e}^{5} \ln \left (5+x \right )^{2}-x -1}\) | \(135\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.53, size = 34, normalized size = 1.17 \begin {gather*} -\frac {2 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 30, normalized size = 1.03 \begin {gather*} -\frac {2 \, e^{5} \log \left (\frac {e^{\left (x + \log \left (3\right )\right )}}{x}\right )}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.62, size = 34, normalized size = 1.17 \begin {gather*} -\frac {4 \, {\left (x e^{5} + e^{5} \log \left (3\right ) - e^{5} \log \left (x\right )\right )}}{e^{5} \log \left (x + 5\right )^{2} - x - 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.62, size = 27, normalized size = 0.93 \begin {gather*} \frac {2\,{\mathrm {e}}^5\,\ln \left (\frac {3\,{\mathrm {e}}^x}{x}\right )}{-{\mathrm {e}}^5\,{\ln \left (x+5\right )}^2+x+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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