Optimal. Leaf size=33 \[ \frac {2}{e^4 \left (\frac {e^{10}}{x^4}+2 x\right ) \log \left (x \left (3-x-x^2\right )\right )} \]
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Rubi [F] time = 8.82, 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 {12 x^8-8 x^9-12 x^{10}+e^{10} \left (6 x^3-4 x^4-6 x^5\right )+\left (12 x^8-4 x^9-4 x^{10}+e^{10} \left (-24 x^3+8 x^4+8 x^5\right )\right ) \log \left (3 x-x^2-x^3\right )}{\left (e^{24} \left (-3+x+x^2\right )+e^{14} \left (-12 x^5+4 x^6+4 x^7\right )+e^4 \left (-12 x^{10}+4 x^{11}+4 x^{12}\right )\right ) \log ^2\left (3 x-x^2-x^3\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 {2 x^3 \left (\left (-3+2 x+3 x^2\right ) \left (e^{10}+2 x^5\right )+2 \left (-3+x+x^2\right ) \left (-2 e^{10}+x^5\right ) \log \left (-x \left (-3+x+x^2\right )\right )\right )}{e^4 \left (3-x-x^2\right ) \left (e^{10}+2 x^5\right )^2 \log ^2\left (x \left (3-x-x^2\right )\right )} \, dx\\ &=\frac {2 \int \frac {x^3 \left (\left (-3+2 x+3 x^2\right ) \left (e^{10}+2 x^5\right )+2 \left (-3+x+x^2\right ) \left (-2 e^{10}+x^5\right ) \log \left (-x \left (-3+x+x^2\right )\right )\right )}{\left (3-x-x^2\right ) \left (e^{10}+2 x^5\right )^2 \log ^2\left (x \left (3-x-x^2\right )\right )} \, dx}{e^4}\\ &=\frac {2 \int \left (\frac {x^3 \left (-3+2 x+3 x^2\right )}{\left (3-x-x^2\right ) \left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {2 x^3 \left (2 e^{10}-x^5\right )}{\left (e^{10}+2 x^5\right )^2 \log \left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4}\\ &=\frac {2 \int \frac {x^3 \left (-3+2 x+3 x^2\right )}{\left (3-x-x^2\right ) \left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )} \, dx}{e^4}+\frac {4 \int \frac {x^3 \left (2 e^{10}-x^5\right )}{\left (e^{10}+2 x^5\right )^2 \log \left (x \left (3-x-x^2\right )\right )} \, dx}{e^4}\\ &=\frac {2 \int \left (\frac {6 \left (189+5 e^{10}\right )+\left (162-31 e^{10}\right ) x}{\left (972+122 e^{10}-e^{20}\right ) \left (3-x-x^2\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {-2 e^{10} \left (189+5 e^{10}\right )-e^{10} \left (180-7 e^{10}\right ) x-e^{10} \left (186+e^{10}\right ) x^2-\left (972+244 e^{10}-3 e^{20}\right ) x^3+2 \left (162-31 e^{10}\right ) x^4}{\left (972+122 e^{10}-e^{20}\right ) \left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4}+\frac {4 \int \left (\frac {x^3}{2 \left (-e^{10}-2 x^5\right ) \log \left (x \left (3-x-x^2\right )\right )}+\frac {5 e^{10} x^3}{2 \left (e^{10}+2 x^5\right )^2 \log \left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4}\\ &=\frac {2 \int \frac {x^3}{\left (-e^{10}-2 x^5\right ) \log \left (x \left (3-x-x^2\right )\right )} \, dx}{e^4}+\left (10 e^6\right ) \int \frac {x^3}{\left (e^{10}+2 x^5\right )^2 \log \left (x \left (3-x-x^2\right )\right )} \, dx+\frac {2 \int \frac {6 \left (189+5 e^{10}\right )+\left (162-31 e^{10}\right ) x}{\left (3-x-x^2\right ) \log ^2\left (x \left (3-x-x^2\right )\right )} \, dx}{e^4 \left (972+122 e^{10}-e^{20}\right )}+\frac {2 \int \frac {-2 e^{10} \left (189+5 e^{10}\right )-e^{10} \left (180-7 e^{10}\right ) x-e^{10} \left (186+e^{10}\right ) x^2-\left (972+244 e^{10}-3 e^{20}\right ) x^3+2 \left (162-31 e^{10}\right ) x^4}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )} \, dx}{e^4 \left (972+122 e^{10}-e^{20}\right )}\\ &=\frac {2 \int \left (\frac {(-1)^{2/5}}{5\ 2^{3/5} e^2 \left (e^2-\sqrt [5]{-2} x\right ) \log \left (x \left (3-x-x^2\right )\right )}+\frac {1}{5\ 2^{3/5} e^2 \left (e^2+\sqrt [5]{2} x\right ) \log \left (x \left (3-x-x^2\right )\right )}+\frac {(-1)^{4/5}}{5\ 2^{3/5} e^2 \left (e^2+(-1)^{2/5} \sqrt [5]{2} x\right ) \log \left (x \left (3-x-x^2\right )\right )}-\frac {\sqrt [5]{-1}}{5\ 2^{3/5} e^2 \left (e^2-(-1)^{3/5} \sqrt [5]{2} x\right ) \log \left (x \left (3-x-x^2\right )\right )}-\frac {\left (-\frac {1}{2}\right )^{3/5}}{5 e^2 \left (e^2+(-1)^{4/5} \sqrt [5]{2} x\right ) \log \left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4}+\left (10 e^6\right ) \int \frac {x^3}{\left (e^{10}+2 x^5\right )^2 \log \left (x \left (3-x-x^2\right )\right )} \, dx+\frac {2 \int \left (\frac {6 \left (189+5 e^{10}\right )}{\left (3-x-x^2\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {\left (162-31 e^{10}\right ) x}{\left (3-x-x^2\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4 \left (972+122 e^{10}-e^{20}\right )}+\frac {2 \int \left (\frac {2 e^{10} \left (-189-5 e^{10}\right )}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {e^{10} \left (-180+7 e^{10}\right ) x}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {e^{10} \left (-186-e^{10}\right ) x^2}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {\left (-972-244 e^{10}+3 e^{20}\right ) x^3}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}+\frac {2 \left (162-31 e^{10}\right ) x^4}{\left (e^{10}+2 x^5\right ) \log ^2\left (x \left (3-x-x^2\right )\right )}\right ) \, dx}{e^4 \left (972+122 e^{10}-e^{20}\right )}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 31, normalized size = 0.94 \begin {gather*} \frac {2 x^4}{e^4 \left (e^{10}+2 x^5\right ) \log \left (-x \left (-3+x+x^2\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.97, size = 34, normalized size = 1.03 \begin {gather*} \frac {2 \, x^{4}}{{\left (2 \, x^{5} e^{4} + e^{14}\right )} \log \left (-x^{3} - x^{2} + 3 \, 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 = 0.22, size = 35, normalized size = 1.06
method | result | size |
risch | \(\frac {2 x^{4} {\mathrm e}^{-4}}{\left ({\mathrm e}^{10}+2 x^{5}\right ) \ln \left (-x^{3}-x^{2}+3 x \right )}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 43, normalized size = 1.30 \begin {gather*} \frac {2 \, x^{4}}{{\left (2 \, x^{5} e^{4} + e^{14}\right )} \log \left (-x^{2} - x + 3\right ) + {\left (2 \, x^{5} e^{4} + e^{14}\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.52, size = 34, normalized size = 1.03 \begin {gather*} \frac {2\,x^4\,{\mathrm {e}}^{-4}}{\ln \left (-x^3-x^2+3\,x\right )\,\left (2\,x^5+{\mathrm {e}}^{10}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.24, size = 27, normalized size = 0.82 \begin {gather*} \frac {2 x^{4}}{\left (2 x^{5} e^{4} + e^{14}\right ) \log {\left (- x^{3} - x^{2} + 3 x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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