Optimal. Leaf size=23 \[ \frac {x \log (-16-x)}{\log ^2\left (\frac {1}{25} e^{\frac {8}{\log (3)}}\right )} \]
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Rubi [A] time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.96, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6742, 43, 2389, 2295} \begin {gather*} \frac {(x+16) \log ^2(3) \log (-x-16)}{(8-\log (3) \log (25))^2}-\frac {16 \log ^2(3) \log (x+16)}{(8-\log (3) \log (25))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 2295
Rule 2389
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\log ^2(3) \int \frac {x+(16+x) \log (-16-x)}{16+x} \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \left (\frac {x}{16+x}+\log (-16-x)\right ) \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \frac {x}{16+x} \, dx}{(8-\log (3) \log (25))^2}+\frac {\log ^2(3) \int \log (-16-x) \, dx}{(8-\log (3) \log (25))^2}\\ &=\frac {\log ^2(3) \int \left (1-\frac {16}{16+x}\right ) \, dx}{(8-\log (3) \log (25))^2}-\frac {\log ^2(3) \operatorname {Subst}(\int \log (x) \, dx,x,-16-x)}{(8-\log (3) \log (25))^2}\\ &=\frac {(16+x) \log ^2(3) \log (-16-x)}{(8-\log (3) \log (25))^2}-\frac {16 \log ^2(3) \log (16+x)}{(8-\log (3) \log (25))^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.02, size = 29, normalized size = 1.26 \begin {gather*} \frac {(16+x) \log (-16-x)-16 \log (16+x)}{\left (-\frac {8}{\log (3)}+\log (25)\right )^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.58, size = 32, normalized size = 1.39 \begin {gather*} \frac {x \log \relax (3)^{2} \log \left (-x - 16\right )}{4 \, {\left (\log \relax (5)^{2} \log \relax (3)^{2} - 8 \, \log \relax (5) \log \relax (3) + 16\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 20, normalized size = 0.87 \begin {gather*} \frac {x \log \left (-x - 16\right )}{\log \left (\frac {1}{25} \, e^{\frac {8}{\log \relax (3)}}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 22, normalized size = 0.96
method | result | size |
risch | \(\frac {\ln \left (-x -16\right ) x}{\left (-2 \ln \relax (5)+\frac {8}{\ln \relax (3)}\right )^{2}}\) | \(22\) |
norman | \(\frac {\ln \relax (3)^{2} x \ln \left (-x -16\right )}{4 \left (\ln \relax (3) \ln \relax (5)-4\right )^{2}}\) | \(23\) |
derivativedivides | \(\frac {-\left (-x -16\right ) \ln \left (-x -16\right )-16 \ln \left (-x -16\right )}{\ln \left (\frac {{\mathrm e}^{\frac {8}{\ln \relax (3)}}}{25}\right )^{2}}\) | \(38\) |
default | \(\frac {-\left (-x -16\right ) \ln \left (-x -16\right )-16 \ln \left (-x -16\right )}{\ln \left (\frac {{\mathrm e}^{\frac {8}{\ln \relax (3)}}}{25}\right )^{2}}\) | \(38\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.35, size = 47, normalized size = 2.04 \begin {gather*} \frac {8 \, \log \left (x + 16\right )^{2} + {\left (x - 16 \, \log \left (x + 16\right )\right )} \log \left (-x - 16\right ) + 8 \, \log \left (-x - 16\right )^{2}}{\log \left (\frac {1}{25} \, e^{\frac {8}{\log \relax (3)}}\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.21, size = 32, normalized size = 1.39 \begin {gather*} \frac {x\,{\ln \relax (3)}^2\,\ln \left (-x-16\right )}{4\,{\ln \relax (3)}^2\,{\ln \relax (5)}^2-32\,\ln \relax (3)\,\ln \relax (5)+64} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.20, size = 34, normalized size = 1.48 \begin {gather*} \frac {x \log {\relax (3 )}^{2} \log {\left (- x - 16 \right )}}{- 32 \log {\relax (3 )} \log {\relax (5 )} + 4 \log {\relax (3 )}^{2} \log {\relax (5 )}^{2} + 64} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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