Optimal. Leaf size=31 \[ 2 x+\frac {6}{x+\frac {3 (5-x)}{\left (e^4-\log (2)\right ) \log (x)}} \]
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Rubi [F] time = 2.07, 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 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(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 {2 \left (9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)\right )}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx\\ &=2 \int \frac {9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx\\ &=2 \int \left (\frac {-3+x^2}{x^2}+\frac {9 (5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {9 (10-x)}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {-3+x^2}{x^2} \, dx+18 \int \frac {(5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+18 \int \frac {10-x}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx\\ &=2 \int \left (1-\frac {3}{x^2}\right ) \, dx+18 \int \left (-\frac {75}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {5 \left (3+e^4-\log (2)\right )}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}-\frac {e^4 \left (1-\frac {\log (2)}{e^4}\right )}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}\right ) \, dx+18 \int \left (\frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}+\frac {10}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx\\ &=\frac {6}{x}+2 x+18 \int \frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx+180 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx-1350 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx-\left (18 \left (e^4-\log (2)\right )\right ) \int \frac {1}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+\left (90 \left (3+e^4-\log (2)\right )\right ) \int \frac {1}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 38, normalized size = 1.23 \begin {gather*} 2 \left (\frac {3}{x}+x+\frac {9 (-5+x)}{x \left (15-3 x+e^4 x \log (x)-x \log (2) \log (x)\right )}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 53, normalized size = 1.71 \begin {gather*} -\frac {2 \, {\left (3 \, x^{2} - {\left ({\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} \log \relax (2)\right )} \log \relax (x) - 15 \, x\right )}}{{\left (x e^{4} - x \log \relax (2)\right )} \log \relax (x) - 3 \, x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.25, size = 60, normalized size = 1.94 \begin {gather*} \frac {2 \, {\left (x^{2} e^{4} \log \relax (x) - x^{2} \log \relax (2) \log \relax (x) - 3 \, x^{2} + 3 \, e^{4} \log \relax (x) - 3 \, \log \relax (2) \log \relax (x) + 15 \, x\right )}}{x e^{4} \log \relax (x) - x \log \relax (2) \log \relax (x) - 3 \, x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 40, normalized size = 1.29
method | result | size |
risch | \(\frac {2 x^{2}+6}{x}+\frac {18 x -90}{x \left (x \,{\mathrm e}^{4} \ln \relax (x )-x \ln \relax (2) \ln \relax (x )-3 x +15\right )}\) | \(40\) |
norman | \(\frac {\left (6 \,{\mathrm e}^{4}-6 \ln \relax (2)\right ) \ln \relax (x )+\left (2 \,{\mathrm e}^{4}-2 \ln \relax (2)\right ) x^{2} \ln \relax (x )+\left (10 \,{\mathrm e}^{4}-10 \ln \relax (2)\right ) x \ln \relax (x )-6 x^{2}+150}{x \,{\mathrm e}^{4} \ln \relax (x )-x \ln \relax (2) \ln \relax (x )-3 x +15}\) | \(69\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.15, size = 53, normalized size = 1.71 \begin {gather*} -\frac {2 \, {\left (3 \, x^{2} - {\left (x^{2} {\left (e^{4} - \log \relax (2)\right )} + 3 \, e^{4} - 3 \, \log \relax (2)\right )} \log \relax (x) - 15 \, x\right )}}{x {\left (e^{4} - \log \relax (2)\right )} \log \relax (x) - 3 \, x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {450\,x+\ln \relax (2)\,\left (18\,x-90\right )-{\ln \relax (x)}^2\,\left ({\mathrm {e}}^8\,\left (6\,x-2\,x^3\right )+{\ln \relax (2)}^2\,\left (6\,x-2\,x^3\right )-{\mathrm {e}}^4\,\ln \relax (2)\,\left (12\,x-4\,x^3\right )\right )+\ln \relax (x)\,\left ({\mathrm {e}}^4\,\left (-12\,x^3+60\,x^2+18\,x\right )-\ln \relax (2)\,\left (-12\,x^3+60\,x^2+18\,x\right )\right )-180\,x^2+18\,x^3-{\mathrm {e}}^4\,\left (18\,x-90\right )}{225\,x+{\ln \relax (x)}^2\,\left (x^3\,{\ln \relax (2)}^2+x^3\,{\mathrm {e}}^8-2\,x^3\,{\mathrm {e}}^4\,\ln \relax (2)\right )+\ln \relax (x)\,\left ({\mathrm {e}}^4\,\left (30\,x^2-6\,x^3\right )-\ln \relax (2)\,\left (30\,x^2-6\,x^3\right )\right )-90\,x^2+9\,x^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.36, size = 36, normalized size = 1.16 \begin {gather*} 2 x + \frac {18 x - 90}{- 3 x^{2} + 15 x + \left (- x^{2} \log {\relax (2 )} + x^{2} e^{4}\right ) \log {\relax (x )}} + \frac {6}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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