Optimal. Leaf size=33 \[ -e^x+x-4 x^2+\frac {1}{3} \left (1+\frac {3}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )}\right ) \]
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Rubi [A] time = 0.78, antiderivative size = 25, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 6, integrand size = 83, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {2561, 6688, 2194, 2390, 2302, 30} \begin {gather*} -4 x^2+\frac {1}{\log \left (\log \left (\frac {x^2}{16}\right )+5\right )}+x-e^x \end {gather*}
Antiderivative was successfully verified.
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Rule 30
Rule 2194
Rule 2302
Rule 2390
Rule 2561
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+\left (5 x-5 e^x x-40 x^2+\left (x-e^x x-8 x^2\right ) \log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx\\ &=\int \left (1-e^x-8 x-\frac {2}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )}\right ) \, dx\\ &=x-4 x^2-2 \int \frac {1}{x \left (5+\log \left (\frac {x^2}{16}\right )\right ) \log ^2\left (5+\log \left (\frac {x^2}{16}\right )\right )} \, dx-\int e^x \, dx\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{(5+x) \log ^2(5+x)} \, dx,x,\log \left (\frac {x^2}{16}\right )\right )\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5+\log \left (\frac {x^2}{16}\right )\right )\\ &=-e^x+x-4 x^2-\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (5+\log \left (\frac {x^2}{16}\right )\right )\right )\\ &=-e^x+x-4 x^2+\frac {1}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.03, size = 25, normalized size = 0.76 \begin {gather*} -e^x+x-4 x^2+\frac {1}{\log \left (5+\log \left (\frac {x^2}{16}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.78, size = 36, normalized size = 1.09 \begin {gather*} -\frac {{\left (4 \, x^{2} - x + e^{x}\right )} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 1}{\log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.54, size = 186, normalized size = 5.64 \begin {gather*} -\frac {16 \, x^{2} \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x^{2} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 20 \, x^{2} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 4 \, x \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 4 \, e^{x} \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + x \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - e^{x} \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + 5 \, x \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, e^{x} \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) + \log \left (\frac {1}{16} \, x^{2}\right ) + 5}{4 \, \log \relax (2) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - \log \left (x^{2}\right ) \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right ) - 5 \, \log \left (\log \left (\frac {1}{16} \, x^{2}\right ) + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 23, normalized size = 0.70
method | result | size |
default | \(-{\mathrm e}^{x}-4 x^{2}+x +\frac {1}{\ln \left (\ln \left (\frac {x^{2}}{16}\right )+5\right )}\) | \(23\) |
risch | \(-4 x^{2}+x -{\mathrm e}^{x}+\frac {1}{\ln \left (-4 \ln \relax (2)+2 \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}+5\right )}\) | \(53\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 24, normalized size = 0.73 \begin {gather*} -4 \, x^{2} + x + \frac {1}{\log \left (-4 \, \log \relax (2) + 2 \, \log \relax (x) + 5\right )} - e^{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.20, size = 22, normalized size = 0.67 \begin {gather*} x-{\mathrm {e}}^x+\frac {1}{\ln \left (\ln \left (\frac {x^2}{16}\right )+5\right )}-4\,x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.47, size = 20, normalized size = 0.61 \begin {gather*} - 4 x^{2} + x - e^{x} + \frac {1}{\log {\left (\log {\left (\frac {x^{2}}{16} \right )} + 5 \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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