Optimal. Leaf size=29 \[ \log \left (\frac {3}{2}+\frac {5}{x}-\frac {\left (12-e^4\right ) (4+x)}{x \log (x)}\right ) \]
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Rubi [F] time = 1.47, 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 {96+e^4 (-8-2 x)+24 x+\left (96-8 e^4\right ) \log (x)-10 \log ^2(x)}{\left (-96 x-24 x^2+e^4 \left (8 x+2 x^2\right )\right ) \log (x)+\left (10 x+3 x^2\right ) \log ^2(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 {-96-e^4 (-8-2 x)-24 x-\left (96-8 e^4\right ) \log (x)+10 \log ^2(x)}{x \log (x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx\\ &=\int \frac {2 \left (-\left (\left (-12+e^4\right ) (4+x)\right )-4 \left (-12+e^4\right ) \log (x)-5 \log ^2(x)\right )}{x \log (x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx\\ &=2 \int \frac {-\left (\left (-12+e^4\right ) (4+x)\right )-4 \left (-12+e^4\right ) \log (x)-5 \log ^2(x)}{x \log (x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx\\ &=2 \int \left (-\frac {5}{x (10+3 x)}-\frac {1}{2 x \log (x)}+\frac {-10-3 x}{2 x \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )}+\frac {2 \left (-12+e^4\right )}{(10+3 x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )}\right ) \, dx\\ &=-\left (10 \int \frac {1}{x (10+3 x)} \, dx\right )-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(10+3 x) \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx-\int \frac {1}{x \log (x)} \, dx+\int \frac {-10-3 x}{x \left (96 \left (1-\frac {e^4}{12}\right )+24 \left (1-\frac {e^4}{12}\right ) x-10 \log (x)-3 x \log (x)\right )} \, dx\\ &=3 \int \frac {1}{10+3 x} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\int \frac {1}{x} \, dx+\int \frac {10+3 x}{x \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\log (x)+\log (10+3 x)-\log (\log (x))-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx+\int \left (\frac {3}{-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)}+\frac {10}{x \left (-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)\right )}\right ) \, dx\\ &=-\log (x)+\log (10+3 x)-\log (\log (x))+3 \int \frac {1}{-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)} \, dx+10 \int \frac {1}{x \left (-96 \left (1-\frac {e^4}{12}\right )-24 \left (1-\frac {e^4}{12}\right ) x+10 \log (x)+3 x \log (x)\right )} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx\\ &=-\log (x)+\log (10+3 x)-\log (\log (x))+3 \int \frac {1}{2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)} \, dx+10 \int \frac {1}{x \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx-\left (4 \left (12-e^4\right )\right ) \int \frac {1}{(-10-3 x) \left (2 \left (-12+e^4\right ) (4+x)+(10+3 x) \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 3.15, size = 70, normalized size = 2.41 \begin {gather*} -2 \left (\frac {\log (x)}{2}-\frac {1}{2} \log (10+3 x)+\frac {1}{2} \log (\log (x))\right )-2 \left (\frac {1}{2} \log (10+3 x)-\frac {1}{2} \log \left (96-8 e^4+24 x-2 e^4 x-10 \log (x)-3 x \log (x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 45, normalized size = 1.55 \begin {gather*} \log \left (3 \, x + 10\right ) - \log \relax (x) + \log \left (\frac {2 \, {\left (x + 4\right )} e^{4} + {\left (3 \, x + 10\right )} \log \relax (x) - 24 \, x - 96}{3 \, x + 10}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 39, normalized size = 1.34
method | result | size |
norman | \(-\ln \relax (x )-\ln \left (\ln \relax (x )\right )+\ln \left (2 x \,{\mathrm e}^{4}+8 \,{\mathrm e}^{4}+3 x \ln \relax (x )+10 \ln \relax (x )-24 x -96\right )\) | \(39\) |
risch | \(-\ln \relax (x )+\ln \left (3 x +10\right )+\ln \left (\ln \relax (x )+\frac {2 x \,{\mathrm e}^{4}+8 \,{\mathrm e}^{4}-24 x -96}{3 x +10}\right )-\ln \left (\ln \relax (x )\right )\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.55, size = 46, normalized size = 1.59 \begin {gather*} \log \left (3 \, x + 10\right ) - \log \relax (x) + \log \left (\frac {2 \, x {\left (e^{4} - 12\right )} + {\left (3 \, x + 10\right )} \log \relax (x) + 8 \, e^{4} - 96}{3 \, x + 10}\right ) - \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.36, size = 34, normalized size = 1.17 \begin {gather*} \ln \left (8\,{\mathrm {e}}^4-24\,x+10\,\ln \relax (x)+2\,x\,{\mathrm {e}}^4+3\,x\,\ln \relax (x)-96\right )-\ln \left (\ln \relax (x)\right )-\ln \relax (x) \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: PolynomialError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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