Optimal. Leaf size=34 \[ \frac {\log ^2\left (\frac {3}{-5+5 \left (\frac {3}{4}-x\right )}\right )}{-e^3-\frac {x}{12}+\log (4)} \]
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Rubi [C] time = 1.25, antiderivative size = 232, normalized size of antiderivative = 6.82, number of steps used = 15, number of rules used = 16, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6, 6688, 12, 6742, 36, 31, 2395, 2411, 2343, 2333, 2316, 2315, 2397, 2394, 2393, 2391} \begin {gather*} -\frac {96 \text {Li}_2\left (\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{4 x+1}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {4 x+1}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-\frac {12 (4 x+1) \log ^2\left (-\frac {12}{5 (4 x+1)}\right )}{\left (1-48 e^3+48 \log (4)\right ) \left (x+12 \left (e^3-\log (4)\right )\right )}-\frac {96 \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right ) \log \left (-\frac {12}{5 (4 x+1)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right ) \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{4 x+1}\right )}{1-48 e^3+48 \log (4)} \end {gather*}
Antiderivative was successfully verified.
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Rule 6
Rule 12
Rule 31
Rule 36
Rule 2315
Rule 2316
Rule 2333
Rule 2343
Rule 2391
Rule 2393
Rule 2394
Rule 2395
Rule 2397
Rule 2411
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (1152 e^3+96 x-1152 \log (4)\right ) \log \left (-\frac {12}{5+20 x}\right )+(12+48 x) \log ^2\left (-\frac {12}{5+20 x}\right )}{x^2+4 x^3+e^3 \left (24 x+96 x^2\right )+\left (e^3 (-288-1152 x)-24 x-96 x^2\right ) \log (4)+(144+576 x) \left (e^6+\log ^2(4)\right )} \, dx\\ &=\int \frac {12 \log \left (-\frac {12}{5+20 x}\right ) \left (8 \left (12 e^3+x-12 \log (4)\right )+(1+4 x) \log \left (-\frac {12}{5+20 x}\right )\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )^2} \, dx\\ &=12 \int \frac {\log \left (-\frac {12}{5+20 x}\right ) \left (8 \left (12 e^3+x-12 \log (4)\right )+(1+4 x) \log \left (-\frac {12}{5+20 x}\right )\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )^2} \, dx\\ &=12 \int \left (\frac {8 \log \left (-\frac {12}{5+20 x}\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )}+\frac {\log ^2\left (-\frac {12}{5+20 x}\right )}{\left (12 e^3+x-12 \log (4)\right )^2}\right ) \, dx\\ &=12 \int \frac {\log ^2\left (-\frac {12}{5+20 x}\right )}{\left (12 e^3+x-12 \log (4)\right )^2} \, dx+96 \int \frac {\log \left (-\frac {12}{5+20 x}\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )} \, dx\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {24}{5} \operatorname {Subst}\left (\int \frac {5 \log \left (-\frac {12}{x}\right )}{x \left (\frac {x}{20}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,5+20 x\right )-\frac {96 \int \frac {\log \left (-\frac {12}{5+20 x}\right )}{12 e^3+x-12 \log (4)} \, dx}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+24 \operatorname {Subst}\left (\int \frac {\log \left (-\frac {12}{x}\right )}{x \left (\frac {x}{20}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,5+20 x\right )-\frac {1920 \int \frac {\log \left (\frac {20 \left (12 e^3+x-12 \log (4)\right )}{-5+20 \left (12 e^3-12 \log (4)\right )}\right )}{5+20 x} \, dx}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log (-12 x)}{x \left (\frac {1}{20 x}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )-\frac {96 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-5+20 \left (12 e^3-12 \log (4)\right )}\right )}{x} \, dx,x,5+20 x\right )}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log (-12 x)}{\frac {1}{20}+\frac {1}{20} x \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {96 \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{1+4 x}\right ) \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log \left (x \left (5-20 \left (12 e^3-12 \log (4)\right )\right )\right )}{\frac {1}{20}+\frac {1}{20} x \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {96 \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{1+4 x}\right ) \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \text {Li}_2\left (1-\frac {1-48 e^3+48 \log (4)}{1+4 x}\right )}{1-48 e^3+48 \log (4)}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.12, size = 27, normalized size = 0.79 \begin {gather*} -\frac {12 \log ^2\left (-\frac {12}{5+20 x}\right )}{12 e^3+x-12 \log (4)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 26, normalized size = 0.76 \begin {gather*} -\frac {12 \, \log \left (-\frac {12}{5 \, {\left (4 \, x + 1\right )}}\right )^{2}}{x + 12 \, e^{3} - 24 \, \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.59, size = 56, normalized size = 1.65 \begin {gather*} -\frac {48 \, \log \left (-\frac {12}{5 \, {\left (4 \, x + 1\right )}}\right )^{2}}{{\left (4 \, x + 1\right )} {\left (\frac {48 \, e^{3}}{4 \, x + 1} - \frac {96 \, \log \relax (2)}{4 \, x + 1} - \frac {1}{4 \, x + 1} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 8.10, size = 27, normalized size = 0.79
method | result | size |
norman | \(-\frac {12 \ln \left (-\frac {12}{20 x +5}\right )^{2}}{x +12 \,{\mathrm e}^{3}-24 \ln \relax (2)}\) | \(27\) |
risch | \(-\frac {12 \ln \left (-\frac {12}{20 x +5}\right )^{2}}{x +12 \,{\mathrm e}^{3}-24 \ln \relax (2)}\) | \(27\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.65, size = 73, normalized size = 2.15 \begin {gather*} -\frac {12 \, {\left (\log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (3) + \log \relax (3)^{2} - 4 \, {\left (\log \relax (5) - \log \relax (3)\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} + 2 \, {\left (\log \relax (5) - \log \relax (3) - 2 \, \log \relax (2)\right )} \log \left (-4 \, x - 1\right ) + \log \left (-4 \, x - 1\right )^{2}\right )}}{x + 12 \, e^{3} - 24 \, \log \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.66, size = 26, normalized size = 0.76 \begin {gather*} -\frac {12\,{\ln \left (-\frac {12}{20\,x+5}\right )}^2}{x+12\,{\mathrm {e}}^3-24\,\ln \relax (2)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.25, size = 26, normalized size = 0.76 \begin {gather*} - \frac {12 \log {\left (- \frac {12}{20 x + 5} \right )}^{2}}{x - 24 \log {\relax (2 )} + 12 e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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