Optimal. Leaf size=25 \[ \frac {2 \left (8 x-\log ^2(5)\right )}{e^5-x+3 \log (x)} \]
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Rubi [F] time = 0.71, 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 {-48 x+16 e^5 x+(6-2 x) \log ^2(5)+48 x \log (x)}{e^{10} x-2 e^5 x^2+x^3+\left (6 e^5 x-6 x^2\right ) \log (x)+9 x \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 {\left (-48+16 e^5\right ) x+(6-2 x) \log ^2(5)+48 x \log (x)}{e^{10} x-2 e^5 x^2+x^3+\left (6 e^5 x-6 x^2\right ) \log (x)+9 x \log ^2(x)} \, dx\\ &=\int \frac {2 \left (3 \log ^2(5)-24 x \left (1+\frac {1}{24} \left (-8 e^5+\log ^2(5)\right )\right )+24 x \log (x)\right )}{x \left (e^5-x+3 \log (x)\right )^2} \, dx\\ &=2 \int \frac {3 \log ^2(5)-24 x \left (1+\frac {1}{24} \left (-8 e^5+\log ^2(5)\right )\right )+24 x \log (x)}{x \left (e^5-x+3 \log (x)\right )^2} \, dx\\ &=2 \int \left (\frac {(-3+x) \left (8 x-\log ^2(5)\right )}{x \left (-e^5+x-3 \log (x)\right )^2}+\frac {8}{e^5-x+3 \log (x)}\right ) \, dx\\ &=2 \int \frac {(-3+x) \left (8 x-\log ^2(5)\right )}{x \left (-e^5+x-3 \log (x)\right )^2} \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx\\ &=2 \int \left (\frac {3 \log ^2(5)}{x \left (-e^5+x-3 \log (x)\right )^2}+\frac {8 x}{\left (e^5-x+3 \log (x)\right )^2}-\frac {24 \left (1+\frac {\log ^2(5)}{24}\right )}{\left (e^5-x+3 \log (x)\right )^2}\right ) \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx\\ &=16 \int \frac {x}{\left (e^5-x+3 \log (x)\right )^2} \, dx+16 \int \frac {1}{e^5-x+3 \log (x)} \, dx+\left (6 \log ^2(5)\right ) \int \frac {1}{x \left (-e^5+x-3 \log (x)\right )^2} \, dx-\left (2 \left (24+\log ^2(5)\right )\right ) \int \frac {1}{\left (e^5-x+3 \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.33, size = 25, normalized size = 1.00 \begin {gather*} \frac {2 \left (8 x-\log ^2(5)\right )}{e^5-x+3 \log (x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.35, size = 23, normalized size = 0.92
method | result | size |
risch | \(-\frac {2 \left (\ln \relax (5)^{2}-8 x \right )}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\) | \(23\) |
norman | \(\frac {48 \ln \relax (x )-2 \ln \relax (5)^{2}+16 \,{\mathrm e}^{5}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\) | \(29\) |
default | \(\frac {48 \ln \relax (x )+16 \,{\mathrm e}^{5}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}-\frac {2 \ln \relax (5)^{2}}{3 \ln \relax (x )+{\mathrm e}^{5}-x}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 22, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (\log \relax (5)^{2} - 8 \, x\right )}}{x - e^{5} - 3 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.50, size = 47, normalized size = 1.88 \begin {gather*} \frac {2\,{\mathrm {e}}^{-5}\,\left (8\,x\,{\mathrm {e}}^5-x\,{\ln \relax (5)}^2\right )+6\,{\mathrm {e}}^{-5}\,{\ln \relax (5)}^2\,\left (\frac {x}{3}-\frac {{\mathrm {e}}^5}{3}\right )}{{\mathrm {e}}^5-x+3\,\ln \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.15, size = 19, normalized size = 0.76 \begin {gather*} \frac {16 x - 2 \log {\relax (5 )}^{2}}{- x + 3 \log {\relax (x )} + e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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