Optimal. Leaf size=27 \[ \frac {2 x \log \left (9 \left (-x+\frac {1}{15} \left (-5+e^5-x^3\right )\right )\right )}{e^5} \]
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Rubi [A] time = 8.23, antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 23, number of rules used = 10, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6688, 12, 6742, 2079, 800, 634, 618, 204, 628, 2523} \begin {gather*} \frac {2 x \log \left (-\frac {3}{5} \left (x^3+15 x-e^5+5\right )\right )}{e^5} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 628
Rule 634
Rule 800
Rule 2079
Rule 2523
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (\frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3}+\log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )\right )}{e^5} \, dx\\ &=\frac {2 \int \left (\frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3}+\log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )\right ) \, dx}{e^5}\\ &=\frac {2 \int \log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right ) \, dx}{e^5}+\frac {6 \int \frac {x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}\\ &=\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}-\frac {2 \int \frac {3 x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}+\frac {6 \int \left (1+\frac {-5+e^5-10 x}{5-e^5+15 x+x^3}\right ) \, dx}{e^5}\\ &=\frac {6 x}{e^5}+\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}+\frac {6 \int \frac {-5+e^5-10 x}{5-e^5+15 x+x^3} \, dx}{e^5}-\frac {6 \int \frac {x \left (5+x^2\right )}{5-e^5+15 x+x^3} \, dx}{e^5}\\ &=\frac {6 x}{e^5}+\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}+\frac {6 \int \frac {-5+e^5-10 x}{\left (5 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-\sqrt [3]{\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )}+x\right ) \left (5+25 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}+\left (\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}-\left (5 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-\sqrt [3]{\frac {1}{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )}\right ) x+x^2\right )} \, dx}{e^5}-\frac {6 \int \left (1+\frac {-5+e^5-10 x}{5-e^5+15 x+x^3}\right ) \, dx}{e^5}\\ &=\frac {2 x \log \left (-\frac {3}{5} \left (5-e^5+15 x+x^3\right )\right )}{e^5}-\frac {6 \int \frac {-5+e^5-10 x}{5-e^5+15 x+x^3} \, dx}{e^5}+\frac {6 \int \left (\frac {4 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} \left (5-e^5-50 \sqrt [3]{\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}+5\ 2^{2/3} \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )}{3 \left (-10+50 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}+\sqrt [3]{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )^{2/3}\right ) \left (-10 \sqrt [3]{2}+\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}-2 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} x\right )}+\frac {4 \left (e^5 \left (5 \sqrt [3]{2}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )-5 \left (5 \sqrt [3]{2}+\sqrt [3]{2} \sqrt {525-10 e^5+e^{10}}+\frac {50\ 2^{2/3}}{\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}+10 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )-\left (e^5 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}+5 \left (10 \sqrt [3]{2}-\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right )\right ) x\right )}{3 \left (10-50 \left (\frac {2}{-5+e^5+\sqrt {525-10 e^5+e^{10}}}\right )^{2/3}-\sqrt [3]{2} \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )^{2/3}\right ) \left (5 \sqrt [3]{2}-\sqrt [3]{2} e^5-\sqrt [3]{2} \sqrt {525-10 e^5+e^{10}}-\frac {50\ 2^{2/3}}{\sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}}-10 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}}+\left (10 \sqrt [3]{2}-\left (2 \left (-5+e^5+\sqrt {525-10 e^5+e^{10}}\right )\right )^{2/3}\right ) x-2 \sqrt [3]{-5+e^5+\sqrt {525-10 e^5+e^{10}}} x^2\right )}\right ) \, dx}{e^5}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 x \log \left (\frac {3}{5} \left (-5+e^5-15 x-x^3\right )\right )}{e^5} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 20, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (-5\right )} \log \left (-\frac {3}{5} \, x^{3} - 9 \, x + \frac {3}{5} \, e^{5} - 3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 20, normalized size = 0.74 \begin {gather*} 2 \, x e^{\left (-5\right )} \log \left (-\frac {3}{5} \, x^{3} - 9 \, x + \frac {3}{5} \, e^{5} - 3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 21, normalized size = 0.78
method | result | size |
risch | \(2 x \ln \left (\frac {3 \,{\mathrm e}^{5}}{5}-\frac {3 x^{3}}{5}-9 x -3\right ) {\mathrm e}^{-5}\) | \(21\) |
norman | \(2 x \ln \left (\frac {3 \,{\mathrm e}^{5}}{5}-\frac {3 x^{3}}{5}-9 x -3\right ) {\mathrm e}^{-5}\) | \(23\) |
default | \(2 \,{\mathrm e}^{-5} \ln \relax (3) x +2 \,{\mathrm e}^{-5} x \ln \left (-x^{3}+{\mathrm e}^{5}-15 x -5\right )+2 \,{\mathrm e}^{-5} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-{\mathrm e}^{5}+15 \textit {\_Z} +5\right )}{\sum }\frac {\left (10 \textit {\_R} +5-{\mathrm e}^{5}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}+5}\right )-2 x \ln \relax (5) {\mathrm e}^{-5}+2 \,{\mathrm e}^{-5} \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3}-{\mathrm e}^{5}+15 \textit {\_Z} +5\right )}{\sum }\frac {\left (-10 \textit {\_R} -5+{\mathrm e}^{5}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}+5}\right )\) | \(128\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.48, size = 32, normalized size = 1.19 \begin {gather*} 2 \, {\left ({\left (i \, \pi - \log \relax (5) + \log \relax (3)\right )} x + x \log \left (x^{3} + 15 \, x - e^{5} + 5\right )\right )} e^{\left (-5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.55, size = 21, normalized size = 0.78 \begin {gather*} 2\,x\,{\mathrm {e}}^{-5}\,\left (\ln \left (-x^3-15\,x+{\mathrm {e}}^5-5\right )+\ln \left (\frac {3}{5}\right )\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.20, size = 26, normalized size = 0.96 \begin {gather*} \frac {2 x \log {\left (- \frac {3 x^{3}}{5} - 9 x - 3 + \frac {3 e^{5}}{5} \right )}}{e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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