Optimal. Leaf size=23 \[ 7+\frac {x^2}{6 (5-x)^2}+\log ((4+x) \log (x)) \]
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Rubi [A] time = 0.20, antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 4, integrand size = 73, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6688, 1620, 2302, 29} \begin {gather*} -\frac {5}{3 (5-x)}+\frac {25}{6 (5-x)^2}+\log (x+4)+\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 1620
Rule 2302
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-375+205 x-50 x^2+3 x^3}{3 (-5+x)^3 (4+x)}+\frac {1}{x \log (x)}\right ) \, dx\\ &=\frac {1}{3} \int \frac {-375+205 x-50 x^2+3 x^3}{(-5+x)^3 (4+x)} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\frac {1}{3} \int \left (-\frac {25}{(-5+x)^3}-\frac {5}{(-5+x)^2}+\frac {3}{4+x}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\frac {25}{6 (5-x)^2}-\frac {5}{3 (5-x)}+\log (4+x)+\log (\log (x))\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.04, size = 26, normalized size = 1.13 \begin {gather*} \frac {25}{6 (-5+x)^2}+\frac {5}{3 (-5+x)}+\log (4+x)+\log (\log (x)) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.92, size = 44, normalized size = 1.91 \begin {gather*} \frac {6 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (x + 4\right ) + 6 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (\log \relax (x)\right ) + 10 \, x - 25}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 25, normalized size = 1.09 \begin {gather*} \frac {5 \, {\left (2 \, x - 5\right )}}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} + \log \left (x + 4\right ) + \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 19, normalized size = 0.83
method | result | size |
norman | \(\frac {x^{2}}{6 \left (x -5\right )^{2}}+\ln \left (\ln \relax (x )\right )+\ln \left (4+x \right )\) | \(19\) |
default | \(\ln \left (4+x \right )+\frac {25}{6 \left (x -5\right )^{2}}+\frac {5}{3 \left (x -5\right )}+\ln \left (\ln \relax (x )\right )\) | \(23\) |
risch | \(\frac {6 x^{2} \ln \left (4+x \right )-60 x \ln \left (4+x \right )+150 \ln \left (4+x \right )+10 x -25}{6 x^{2}-60 x +150}+\ln \left (\ln \relax (x )\right )\) | \(44\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 25, normalized size = 1.09 \begin {gather*} \frac {5 \, {\left (2 \, x - 5\right )}}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} + \log \left (x + 4\right ) + \log \left (\log \relax (x)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 2.32, size = 19, normalized size = 0.83 \begin {gather*} \ln \left (x+4\right )+\ln \left (\ln \relax (x)\right )+\frac {\frac {5\,x}{3}-\frac {25}{6}}{{\left (x-5\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.19, size = 24, normalized size = 1.04 \begin {gather*} \frac {10 x - 25}{6 x^{2} - 60 x + 150} + \log {\left (x + 4 \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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