Optimal. Leaf size=30 \[ \frac {2}{5} \left (x+\frac {1}{4} \left (-x+\frac {6+\log (x)}{1+\frac {e^5}{x}}\right )\right ) \]
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Rubi [B] time = 0.15, antiderivative size = 69, normalized size of antiderivative = 2.30, number of steps used = 12, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {27, 12, 6742, 43, 2314, 31} \begin {gather*} \frac {3 x}{10}-\frac {e^5 \left (7-6 e^5\right )}{10 \left (x+e^5\right )}-\frac {3 e^{10}}{5 \left (x+e^5\right )}+\frac {e^5}{10 \left (x+e^5\right )}+\frac {x \log (x)}{10 \left (x+e^5\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 31
Rule 43
Rule 2314
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{10}+x+3 x^2+e^5 (7+6 x)+e^5 \log (x)}{10 \left (e^5+x\right )^2} \, dx\\ &=\frac {1}{10} \int \frac {3 e^{10}+x+3 x^2+e^5 (7+6 x)+e^5 \log (x)}{\left (e^5+x\right )^2} \, dx\\ &=\frac {1}{10} \int \left (\frac {3 e^{10}}{\left (e^5+x\right )^2}+\frac {x}{\left (e^5+x\right )^2}+\frac {3 x^2}{\left (e^5+x\right )^2}+\frac {e^5 (7+6 x)}{\left (e^5+x\right )^2}+\frac {e^5 \log (x)}{\left (e^5+x\right )^2}\right ) \, dx\\ &=-\frac {3 e^{10}}{10 \left (e^5+x\right )}+\frac {1}{10} \int \frac {x}{\left (e^5+x\right )^2} \, dx+\frac {3}{10} \int \frac {x^2}{\left (e^5+x\right )^2} \, dx+\frac {1}{10} e^5 \int \frac {7+6 x}{\left (e^5+x\right )^2} \, dx+\frac {1}{10} e^5 \int \frac {\log (x)}{\left (e^5+x\right )^2} \, dx\\ &=-\frac {3 e^{10}}{10 \left (e^5+x\right )}+\frac {x \log (x)}{10 \left (e^5+x\right )}-\frac {1}{10} \int \frac {1}{e^5+x} \, dx+\frac {1}{10} \int \left (-\frac {e^5}{\left (e^5+x\right )^2}+\frac {1}{e^5+x}\right ) \, dx+\frac {3}{10} \int \left (1+\frac {e^{10}}{\left (e^5+x\right )^2}-\frac {2 e^5}{e^5+x}\right ) \, dx+\frac {1}{10} e^5 \int \left (\frac {7-6 e^5}{\left (e^5+x\right )^2}+\frac {6}{e^5+x}\right ) \, dx\\ &=\frac {3 x}{10}+\frac {e^5}{10 \left (e^5+x\right )}-\frac {3 e^{10}}{5 \left (e^5+x\right )}-\frac {e^5 \left (7-6 e^5\right )}{10 \left (e^5+x\right )}+\frac {x \log (x)}{10 \left (e^5+x\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 26, normalized size = 0.87 \begin {gather*} \frac {1}{10} \left (3 x+\log (x)-\frac {e^5 (6+\log (x))}{e^5+x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.07, size = 25, normalized size = 0.83 \begin {gather*} \frac {3 \, x^{2} + 3 \, {\left (x - 2\right )} e^{5} + x \log \relax (x)}{10 \, {\left (x + e^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 27, normalized size = 0.90 \begin {gather*} \frac {3 \, x^{2} + 3 \, x e^{5} + x \log \relax (x) - 6 \, e^{5}}{10 \, {\left (x + e^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.42, size = 29, normalized size = 0.97
method | result | size |
norman | \(\frac {\frac {3 x^{2}}{10}+\frac {x \ln \relax (x )}{10}-\frac {3 \,{\mathrm e}^{10}}{10}-\frac {3 \,{\mathrm e}^{5}}{5}}{{\mathrm e}^{5}+x}\) | \(29\) |
risch | \(-\frac {{\mathrm e}^{5} \ln \relax (x )}{10 \left ({\mathrm e}^{5}+x \right )}+\frac {{\mathrm e}^{5} \ln \relax (x )+3 x \,{\mathrm e}^{5}+x \ln \relax (x )+3 x^{2}-6 \,{\mathrm e}^{5}}{10 \,{\mathrm e}^{5}+10 x}\) | \(46\) |
default | \(\frac {3 x^{2}-3 \left ({\mathrm e}^{5}\right )^{2}-6 \,{\mathrm e}^{5}}{10 \,{\mathrm e}^{5}+10 x}+\frac {\ln \left ({\mathrm e}^{5}+x \right )}{10}+\frac {{\mathrm e}^{5} \ln \relax (x ) \ln \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}-\frac {{\mathrm e}^{5} \ln \relax (x ) \ln \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}+\frac {{\mathrm e}^{5} \dilog \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}-\frac {{\mathrm e}^{5} \dilog \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\) | \(237\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.37, size = 87, normalized size = 2.90 \begin {gather*} -\frac {1}{10} \, {\left (e^{\left (-5\right )} \log \left (x + e^{5}\right ) - e^{\left (-5\right )} \log \relax (x) + \frac {\log \relax (x)}{x + e^{5}}\right )} e^{5} + \frac {3}{5} \, {\left (\frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right )\right )} e^{5} - \frac {3}{5} \, e^{5} \log \left (x + e^{5}\right ) + \frac {3}{10} \, x - \frac {3 \, e^{10}}{5 \, {\left (x + e^{5}\right )}} - \frac {3 \, e^{5}}{5 \, {\left (x + e^{5}\right )}} + \frac {1}{10} \, \log \left (x + e^{5}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.72, size = 23, normalized size = 0.77 \begin {gather*} \frac {x\,\left (3\,x+3\,{\mathrm {e}}^5+\ln \relax (x)+6\right )}{10\,\left (x+{\mathrm {e}}^5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.27, size = 37, normalized size = 1.23 \begin {gather*} \frac {3 x}{10} + \frac {\log {\relax (x )}}{10} - \frac {e^{5} \log {\relax (x )}}{10 x + 10 e^{5}} - \frac {3 e^{5}}{5 x + 5 e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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