Optimal. Leaf size=30 \[ 5+\frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \]
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Rubi [A] time = 0.02, antiderivative size = 29, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6, 12, 14} \begin {gather*} \frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 14
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 e^{2 e^2}-5 e^{10} x^4+\left (-80+40 e^5\right ) x^4+5 x^5}{5 x^5} \, dx\\ &=\int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{5 x^5} \, dx\\ &=\frac {1}{5} \int \frac {-16 e^{2 e^2}+\left (-80+40 e^5-5 e^{10}\right ) x^4+5 x^5}{x^5} \, dx\\ &=\frac {1}{5} \int \left (5-\frac {16 e^{2 e^2}}{x^5}-\frac {5 \left (-4+e^5\right )^2}{x}\right ) \, dx\\ &=\frac {4 e^{2 e^2}}{5 x^4}+x-\left (4-e^5\right )^2 \log (x)\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 34, normalized size = 1.13 \begin {gather*} \frac {4 e^{2 e^2}}{5 x^4}+x-16 \log (x)+8 e^5 \log (x)-e^{10} \log (x) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 41, normalized size = 1.37 \begin {gather*} \frac {5 \, x^{5} - 5 \, {\left (x^{4} e^{10} - 8 \, x^{4} e^{5} + 16 \, x^{4}\right )} \log \relax (x) + 4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 25, normalized size = 0.83 \begin {gather*} -{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \left ({\left | x \right |}\right ) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 27, normalized size = 0.90
method | result | size |
default | \(x +\frac {\left (40 \,{\mathrm e}^{5}-5 \,{\mathrm e}^{10}-80\right ) \ln \relax (x )}{5}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}\) | \(27\) |
risch | \(x +\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5 x^{4}}+8 \,{\mathrm e}^{5} \ln \relax (x )-\ln \relax (x ) {\mathrm e}^{10}-16 \ln \relax (x )\) | \(29\) |
norman | \(\frac {x^{5}+\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{2}}}{5}}{x^{4}}+\left (8 \,{\mathrm e}^{5}-{\mathrm e}^{10}-16\right ) \ln \relax (x )\) | \(32\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.36, size = 24, normalized size = 0.80 \begin {gather*} -{\left (e^{10} - 8 \, e^{5} + 16\right )} \log \relax (x) + x + \frac {4 \, e^{\left (2 \, e^{2}\right )}}{5 \, x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.89, size = 24, normalized size = 0.80 \begin {gather*} x+\frac {4\,{\mathrm {e}}^{2\,{\mathrm {e}}^2}}{5\,x^4}-\ln \relax (x)\,\left ({\mathrm {e}}^{10}-8\,{\mathrm {e}}^5+16\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.32, size = 24, normalized size = 0.80 \begin {gather*} x - \left (-4 + e^{5}\right )^{2} \log {\relax (x )} + \frac {4 e^{2 e^{2}}}{5 x^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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