Optimal. Leaf size=30 \[ \frac {x}{x (2+x)-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \]
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Rubi [F] time = 3.59, 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 {4 e^5-5 x+\left (-e^5 x^2+x^3\right ) \log \left (9 e^5 x^4-9 x^5\right )+\left (-e^5+x\right ) \log \left (9 e^5 x^4-9 x^5\right ) \log \left (\frac {1}{5} \log \left (9 e^5 x^4-9 x^5\right )\right )}{\left (-4 x^3-4 x^4-x^5+e^5 \left (4 x^2+4 x^3+x^4\right )\right ) \log \left (9 e^5 x^4-9 x^5\right )+\left (4 x^2+2 x^3+e^5 \left (-4 x-2 x^2\right )\right ) \log \left (9 e^5 x^4-9 x^5\right ) \log \left (\frac {1}{5} \log \left (9 e^5 x^4-9 x^5\right )\right )+\left (e^5-x\right ) \log \left (9 e^5 x^4-9 x^5\right ) \log ^2\left (\frac {1}{5} \log \left (9 e^5 x^4-9 x^5\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^5-5 x-\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (x^2+\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (x (2+x)-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx\\ &=\int \left (\frac {4 e^5-5 x-2 e^5 x \log \left (9 \left (e^5-x\right ) x^4\right )+2 \left (1-e^5\right ) x^2 \log \left (9 \left (e^5-x\right ) x^4\right )+2 x^3 \log \left (9 \left (e^5-x\right ) x^4\right )}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )}\right ) \, dx\\ &=\int \frac {4 e^5-5 x-2 e^5 x \log \left (9 \left (e^5-x\right ) x^4\right )+2 \left (1-e^5\right ) x^2 \log \left (9 \left (e^5-x\right ) x^4\right )+2 x^3 \log \left (9 \left (e^5-x\right ) x^4\right )}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx+\int \frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \, dx\\ &=\int \frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \, dx+\int \frac {4 e^5-5 x-2 \left (e^5-x\right ) x (1+x) \log \left (9 \left (e^5-x\right ) x^4\right )}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (x (2+x)-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx\\ &=\int \left (-\frac {2 e^5 x}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {2 \left (1-e^5\right ) x^2}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {2 x^3}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {4 e^5}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}-\frac {5 x}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}\right ) \, dx+\int \frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \, dx\\ &=2 \int \frac {x^3}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx-5 \int \frac {x}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx-\left (2 e^5\right ) \int \frac {x}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx+\left (4 e^5\right ) \int \frac {1}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx+\left (2 \left (1-e^5\right )\right ) \int \frac {x^2}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx+\int \frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \, dx\\ &=2 \int \left (-\frac {e^{10}}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {e^{15}}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}-\frac {e^5 x}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}-\frac {x^2}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}\right ) \, dx-5 \int \left (-\frac {1}{\log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {e^5}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}\right ) \, dx-\left (2 e^5\right ) \int \left (-\frac {1}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {e^5}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}\right ) \, dx+\left (4 e^5\right ) \int \frac {1}{\left (e^5-x\right ) \log \left (9 \left (e^5-x\right ) x^4\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2} \, dx+\left (2 \left (1-e^5\right )\right ) \int \left (-\frac {e^5}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}+\frac {e^{10}}{\left (e^5-x\right ) \left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}-\frac {x}{\left (2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )\right )^2}\right ) \, dx+\int \frac {1}{2 x+x^2-\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 32, normalized size = 1.07 \begin {gather*} -\frac {x}{-2 x-x^2+\log \left (\frac {1}{5} \log \left (9 \left (e^5-x\right ) x^4\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 30, normalized size = 1.00 \begin {gather*} \frac {x}{x^{2} + 2 \, x - \log \left (\frac {1}{5} \, \log \left (-9 \, x^{5} + 9 \, x^{4} e^{5}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 2.25, size = 30, normalized size = 1.00 \begin {gather*} \frac {x}{x^{2} + 2 \, x + \log \relax (5) - \log \left (\log \left (-9 \, x^{5} + 9 \, x^{4} e^{5}\right )\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (-{\mathrm e}^{5}+x \right ) \ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right ) \ln \left (\frac {\ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right )}{5}\right )+\left (-x^{2} {\mathrm e}^{5}+x^{3}\right ) \ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right )+4 \,{\mathrm e}^{5}-5 x}{\left ({\mathrm e}^{5}-x \right ) \ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right ) \ln \left (\frac {\ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right )}{5}\right )^{2}+\left (\left (-2 x^{2}-4 x \right ) {\mathrm e}^{5}+2 x^{3}+4 x^{2}\right ) \ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right ) \ln \left (\frac {\ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right )}{5}\right )+\left (\left (x^{4}+4 x^{3}+4 x^{2}\right ) {\mathrm e}^{5}-x^{5}-4 x^{4}-4 x^{3}\right ) \ln \left (9 x^{4} {\mathrm e}^{5}-9 x^{5}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.57, size = 35, normalized size = 1.17 \begin {gather*} \frac {x}{x^{2} + 2 \, x + \log \relax (5) - \log \left (i \, \pi + 2 \, \log \relax (3) + \log \left (x - e^{5}\right ) + 4 \, \log \relax (x)\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.46, size = 27, normalized size = 0.90 \begin {gather*} - \frac {x}{- x^{2} - 2 x + \log {\left (\frac {\log {\left (- 9 x^{5} + 9 x^{4} e^{5} \right )}}{5} \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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