Optimal. Leaf size=24 \[ -5-\frac {5}{(4-x)^2}+25 \left (e^{e^3}+\log (\log (x))\right )^2 \]
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Rubi [A] time = 0.34, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 12, 2301} \begin {gather*} 25 \left (\log (\log (x))+e^{e^3}\right )^2-\frac {5}{(4-x)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2301
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int 10 \left (\frac {1}{(-4+x)^3}+\frac {5 \left (e^{e^3}+\log (\log (x))\right )}{x \log (x)}\right ) \, dx\\ &=10 \int \left (\frac {1}{(-4+x)^3}+\frac {5 \left (e^{e^3}+\log (\log (x))\right )}{x \log (x)}\right ) \, dx\\ &=-\frac {5}{(4-x)^2}+50 \int \frac {e^{e^3}+\log (\log (x))}{x \log (x)} \, dx\\ &=-\frac {5}{(4-x)^2}+50 \operatorname {Subst}\left (\int \frac {e^{e^3}+\log (x)}{x} \, dx,x,\log (x)\right )\\ &=-\frac {5}{(4-x)^2}+25 \left (e^{e^3}+\log (\log (x))\right )^2\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 27, normalized size = 1.12 \begin {gather*} 10 \left (-\frac {1}{2 (-4+x)^2}+\frac {5}{2} \left (e^{e^3}+\log (\log (x))\right )^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.82, size = 45, normalized size = 1.88 \begin {gather*} \frac {5 \, {\left (10 \, {\left (x^{2} - 8 \, x + 16\right )} e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 5 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (\log \relax (x)\right )^{2} - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.13, size = 67, normalized size = 2.79 \begin {gather*} \frac {5 \, {\left (10 \, x^{2} e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 5 \, x^{2} \log \left (\log \relax (x)\right )^{2} - 80 \, x e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) - 40 \, x \log \left (\log \relax (x)\right )^{2} + 160 \, e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 80 \, \log \left (\log \relax (x)\right )^{2} - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 24, normalized size = 1.00
method | result | size |
default | \(25 \ln \left (\ln \relax (x )\right )^{2}+50 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \left (\ln \relax (x )\right )-\frac {5}{\left (x -4\right )^{2}}\) | \(24\) |
risch | \(25 \ln \left (\ln \relax (x )\right )^{2}+\frac {50 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{3}} x^{2}-400 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{3}} x +800 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \left (\ln \relax (x )\right )-5}{x^{2}-8 x +16}\) | \(51\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.40, size = 53, normalized size = 2.21 \begin {gather*} \frac {5 \, {\left (5 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (\log \relax (x)\right )^{2} + 10 \, {\left (x^{2} e^{\left (e^{3}\right )} - 8 \, x e^{\left (e^{3}\right )} + 16 \, e^{\left (e^{3}\right )}\right )} \log \left (\log \relax (x)\right ) - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.20, size = 28, normalized size = 1.17 \begin {gather*} 50\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{{\mathrm {e}}^3}-\frac {5}{x^2-8\,x+16}+25\,{\ln \left (\ln \relax (x)\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 31, normalized size = 1.29 \begin {gather*} 25 \log {\left (\log {\relax (x )} \right )}^{2} + 50 e^{e^{3}} \log {\left (\log {\relax (x )} \right )} - \frac {10}{2 x^{2} - 16 x + 32} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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