Optimal. Leaf size=23 \[ \log (4+x) \log \left (x \left (2-x-\frac {1}{5} \log \left (\frac {5}{2}\right )\right )^2\right ) \]
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Rubi [A] time = 0.51, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 7, integrand size = 110, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6688, 2418, 2392, 2391, 2394, 2393, 2494} \begin {gather*} \log (x+4) \log \left (\frac {1}{25} x \left (-5 x+10-\log \left (\frac {5}{2}\right )\right )^2\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2391
Rule 2392
Rule 2393
Rule 2394
Rule 2418
Rule 2494
Rule 6688
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {\left (-10+15 x+\log \left (\frac {5}{2}\right )\right ) \log (4+x)}{x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )}+\frac {\log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right )}{4+x}\right ) \, dx\\ &=\int \frac {\left (-10+15 x+\log \left (\frac {5}{2}\right )\right ) \log (4+x)}{x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )} \, dx+\int \frac {\log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right )}{4+x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )-10 \int \frac {\log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )} \, dx-\int \frac {\log (4+x)}{x} \, dx+\int \left (\frac {\log (4+x)}{x}+\frac {10 \log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )}\right ) \, dx\\ &=-\log (4) \log (x)-2 \log (4+x) \log \left (\frac {10-5 x-\log \left (\frac {5}{2}\right )}{30-\log \left (\frac {5}{2}\right )}\right )+\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )+2 \int \frac {\log \left (\frac {-10+5 x+\log \left (\frac {5}{2}\right )}{-30+\log \left (\frac {5}{2}\right )}\right )}{4+x} \, dx+10 \int \frac {\log (4+x)}{-10+5 x+\log \left (\frac {5}{2}\right )} \, dx-\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx+\int \frac {\log (4+x)}{x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )+\text {Li}_2\left (-\frac {x}{4}\right )-2 \int \frac {\log \left (\frac {-10+5 x+\log \left (\frac {5}{2}\right )}{-30+\log \left (\frac {5}{2}\right )}\right )}{4+x} \, dx+2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {5 x}{-30+\log \left (\frac {5}{2}\right )}\right )}{x} \, dx,x,4+x\right )+\int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )-2 \text {Li}_2\left (\frac {5 (4+x)}{30-\log \left (\frac {5}{2}\right )}\right )-2 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {5 x}{-30+\log \left (\frac {5}{2}\right )}\right )}{x} \, dx,x,4+x\right )\\ &=\log (4+x) \log \left (\frac {1}{25} x \left (10-5 x-\log \left (\frac {5}{2}\right )\right )^2\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.05, size = 22, normalized size = 0.96 \begin {gather*} \log (4+x) \log \left (\frac {1}{25} x \left (-10+5 x+\log \left (\frac {5}{2}\right )\right )^2\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.79, size = 36, normalized size = 1.57 \begin {gather*} \log \left (x^{3} + \frac {1}{25} \, x \log \left (\frac {5}{2}\right )^{2} - 4 \, x^{2} + \frac {2}{5} \, {\left (x^{2} - 2 \, x\right )} \log \left (\frac {5}{2}\right ) + 4 \, x\right ) \log \left (x + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.74, size = 47, normalized size = 2.04 \begin {gather*} -2 \, \log \relax (5) \log \left (x + 4\right ) + \log \left (25 \, x^{3} + 10 \, x^{2} \log \left (\frac {5}{2}\right ) + x \log \left (\frac {5}{2}\right )^{2} - 100 \, x^{2} - 20 \, x \log \left (\frac {5}{2}\right ) + 100 \, x\right ) \log \left (x + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.18, size = 48, normalized size = 2.09
method | result | size |
default | \(\ln \left (4+x \right ) \ln \left (x \ln \left (\frac {5}{2}\right )^{2}+10 x^{2} \ln \left (\frac {5}{2}\right )-20 x \ln \left (\frac {5}{2}\right )+25 x^{3}-100 x^{2}+100 x \right )-2 \ln \left (4+x \right ) \ln \relax (5)\) | \(48\) |
risch | \(2 \ln \left (4+x \right ) \ln \left (\ln \relax (5)-\ln \relax (2)+5 x -10\right )+\ln \relax (x ) \ln \left (4+x \right )-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )}{2}+\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}}{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )\right )^{2} \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )}{2}+i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )\right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{3}}{2}+\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{2}}{2}-\frac {i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i x \left (\ln \left (\frac {5}{2}\right )+5 x -10\right )^{2}\right )^{3}}{2}-2 \ln \left (4+x \right ) \ln \relax (5)\) | \(256\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.63, size = 34, normalized size = 1.48 \begin {gather*} -{\left (2 \, \log \relax (5) - \log \relax (x)\right )} \log \left (x + 4\right ) + 2 \, \log \left (5 \, x + \log \relax (5) - \log \relax (2) - 10\right ) \log \left (x + 4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\ln \left (4\,x-\frac {\ln \left (\frac {5}{2}\right )\,\left (20\,x-10\,x^2\right )}{25}+\frac {x\,{\ln \left (\frac {5}{2}\right )}^2}{25}-4\,x^2+x^3\right )\,\left (x\,\ln \left (\frac {5}{2}\right )-10\,x+5\,x^2\right )+\ln \left (x+4\right )\,\left (50\,x+\ln \left (\frac {5}{2}\right )\,\left (x+4\right )+15\,x^2-40\right )}{10\,x^2-40\,x+5\,x^3+\ln \left (\frac {5}{2}\right )\,\left (x^2+4\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.64, size = 44, normalized size = 1.91 \begin {gather*} \log {\left (x + 4 \right )} \log {\left (x^{3} - 4 x^{2} + \frac {x \log {\left (\frac {5}{2} \right )}^{2}}{25} + 4 x + \left (\frac {2 x^{2}}{5} - \frac {4 x}{5}\right ) \log {\left (\frac {5}{2} \right )} \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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