Optimal. Leaf size=26 \[ e-\frac {9 \left (e^{10} \log (4)-\frac {\log (5)}{x+\log (x)}\right )}{3+x} \]
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Rubi [F] time = 0.92, 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 {9 e^{10} x^3 \log (4)+\left (-27-36 x-18 x^2\right ) \log (5)+\left (18 e^{10} x^2 \log (4)-9 x \log (5)\right ) \log (x)+9 e^{10} x \log (4) \log ^2(x)}{9 x^3+6 x^4+x^5+\left (18 x^2+12 x^3+2 x^4\right ) \log (x)+\left (9 x+6 x^2+x^3\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 {9 \left (e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)\right )}{x (3+x)^2 (x+\log (x))^2} \, dx\\ &=9 \int \frac {e^{10} x^3 \log (4)-\left (3+4 x+2 x^2\right ) \log (5)+x \left (-\log (5)+e^{10} x \log (16)\right ) \log (x)+e^{10} x \log (4) \log ^2(x)}{x (3+x)^2 (x+\log (x))^2} \, dx\\ &=9 \int \left (\frac {e^{10} \log (4)}{(3+x)^2}-\frac {(1+x) \log (5)}{x (3+x) (x+\log (x))^2}-\frac {\log (5)}{(3+x)^2 (x+\log (x))}\right ) \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1+x}{x (3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx-(9 \log (5)) \int \left (\frac {1}{3 x (x+\log (x))^2}+\frac {2}{3 (3+x) (x+\log (x))^2}\right ) \, dx\\ &=-\frac {9 e^{10} \log (4)}{3+x}-(3 \log (5)) \int \frac {1}{x (x+\log (x))^2} \, dx-(6 \log (5)) \int \frac {1}{(3+x) (x+\log (x))^2} \, dx-(9 \log (5)) \int \frac {1}{(3+x)^2 (x+\log (x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.60, size = 33, normalized size = 1.27 \begin {gather*} \frac {9 \left (-e^{10} x \log (4)+\log (5)-e^{10} \log (4) \log (x)\right )}{(3+x) (x+\log (x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.18, size = 37, normalized size = 1.42 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + {\left (x + 3\right )} \log \relax (x) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 39, normalized size = 1.50 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + x \log \relax (x) + 3 \, x + 3 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.31, size = 28, normalized size = 1.08
method | result | size |
risch | \(-\frac {18 \,{\mathrm e}^{10} \ln \relax (2)}{3+x}+\frac {9 \ln \relax (5)}{\left (3+x \right ) \left (x +\ln \relax (x )\right )}\) | \(28\) |
default | \(-\frac {18 \,{\mathrm e}^{10} \ln \relax (2)}{3+x}+\frac {9 \ln \relax (5)}{x \ln \relax (x )+x^{2}+3 \ln \relax (x )+3 x}\) | \(36\) |
norman | \(\frac {-18 x \,{\mathrm e}^{10} \ln \relax (2)-18 \,{\mathrm e}^{10} \ln \relax (2) \ln \relax (x )+9 \ln \relax (5)}{x \ln \relax (x )+x^{2}+3 \ln \relax (x )+3 x}\) | \(43\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 37, normalized size = 1.42 \begin {gather*} -\frac {9 \, {\left (2 \, x e^{10} \log \relax (2) + 2 \, e^{10} \log \relax (2) \log \relax (x) - \log \relax (5)\right )}}{x^{2} + {\left (x + 3\right )} \log \relax (x) + 3 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.13, size = 27, normalized size = 1.04 \begin {gather*} \frac {9\,\ln \relax (5)}{\left (x+\ln \relax (x)\right )\,\left (x+3\right )}-\frac {18\,{\mathrm {e}}^{10}\,\ln \relax (2)}{x+3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.17, size = 29, normalized size = 1.12 \begin {gather*} \frac {9 \log {\relax (5 )}}{x^{2} + 3 x + \left (x + 3\right ) \log {\relax (x )}} - \frac {18 e^{10} \log {\relax (2 )}}{x + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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