Integrand size = 74, antiderivative size = 26 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=\log \left (\left (5+x \left (-x+\log \left (e^{-x} x\right )\right ) (x+2 \log (\log (3)))\right )^2\right ) \] Output:
ln((x*(ln(x/exp(x))-x)*(x+2*ln(ln(3)))+5)^2)
Leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(26)=52\).
Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.69 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \log \left (5-2 x^3+x^2 \log (x)+x^2 \left (x-\log (x)+\log \left (e^{-x} x\right )\right )-4 x^2 \log (\log (3))+2 x \log (x) \log (\log (3))+2 x \left (x-\log (x)+\log \left (e^{-x} x\right )\right ) \log (\log (3))\right ) \] Input:
Integrate[(2*x - 8*x^2 + 2*(2 - 6*x)*Log[Log[3]] + Log[x/E^x]*(4*x + 4*Log [Log[3]]))/(5 - x^3 - 2*x^2*Log[Log[3]] + Log[x/E^x]*(x^2 + 2*x*Log[Log[3] ])),x]
Output:
2*Log[5 - 2*x^3 + x^2*Log[x] + x^2*(x - Log[x] + Log[x/E^x]) - 4*x^2*Log[L og[3]] + 2*x*Log[x]*Log[Log[3]] + 2*x*(x - Log[x] + Log[x/E^x])*Log[Log[3] ]]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-8 x^2+2 x+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))+2 (2-6 x) \log (\log (3))}{-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )+5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (-2 x^4+x^3 (1-8 \log (\log (3)))+4 x^2 (1-2 \log (\log (3))) \log (\log (3))-2 x \left (5-2 \log ^2(\log (3))\right )-10 \log (\log (3))\right )}{x (x+2 \log (\log (3))) \left (-x^3+x^2 \log \left (e^{-x} x\right )-2 x^2 \log (\log (3))+2 x \log (\log (3)) \log \left (e^{-x} x\right )+5\right )}+\frac {4 (x+\log (\log (3)))}{x (x+2 \log (\log (3)))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \log (\log (3)) \int \frac {1}{x^3-\log \left (e^{-x} x\right ) x^2+2 \log (\log (3)) x^2-2 \log \left (e^{-x} x\right ) \log (\log (3)) x-5}dx+10 \int \frac {1}{x \left (x^3-\log \left (e^{-x} x\right ) x^2+2 \log (\log (3)) x^2-2 \log \left (e^{-x} x\right ) \log (\log (3)) x-5\right )}dx-2 (1-4 \log (\log (3))) \int \frac {x}{x^3-\log \left (e^{-x} x\right ) x^2+2 \log (\log (3)) x^2-2 \log \left (e^{-x} x\right ) \log (\log (3)) x-5}dx+4 \int \frac {x^2}{x^3-\log \left (e^{-x} x\right ) x^2+2 \log (\log (3)) x^2-2 \log \left (e^{-x} x\right ) \log (\log (3)) x-5}dx+10 \int \frac {1}{(x+2 \log (\log (3))) \left (x^3-\log \left (e^{-x} x\right ) x^2+2 \log (\log (3)) x^2-2 \log \left (e^{-x} x\right ) \log (\log (3)) x-5\right )}dx+2 \log (x)+2 \log (x+2 \log (\log (3)))\) |
Input:
Int[(2*x - 8*x^2 + 2*(2 - 6*x)*Log[Log[3]] + Log[x/E^x]*(4*x + 4*Log[Log[3 ]]))/(5 - x^3 - 2*x^2*Log[Log[3]] + Log[x/E^x]*(x^2 + 2*x*Log[Log[3]])),x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62
method | result | size |
norman | \(2 \ln \left (-2 \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\ln \left (3\right )\right ) x -\ln \left (x \,{\mathrm e}^{-x}\right ) x^{2}+2 x^{2} \ln \left (\ln \left (3\right )\right )+x^{3}-5\right )\) | \(42\) |
parallelrisch | \(2 \ln \left (-2 \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\ln \left (3\right )\right ) x -\ln \left (x \,{\mathrm e}^{-x}\right ) x^{2}+2 x^{2} \ln \left (\ln \left (3\right )\right )+x^{3}-5\right )\) | \(42\) |
risch | \(2 \ln \left (2 \ln \left (\ln \left (3\right )\right ) x +x^{2}\right )+2 \ln \left (\ln \left ({\mathrm e}^{x}\right )+\frac {-630+252 x^{2} \ln \left (\ln \left (3\right )\right )+126 i \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )-126 x^{2} \ln \left (x \right )+126 x^{3}+2 i x^{3}-252 \ln \left (\ln \left (3\right )\right ) x \ln \left (x \right )-\pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+2 \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+2 \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-63 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-63 i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+126 i \ln \left (\ln \left (3\right )\right ) \pi x \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+4 i x^{2} \ln \left (\ln \left (3\right )\right )-2 i x^{2} \ln \left (x \right )-\pi \,x^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+\pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}+\pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-2 \ln \left (\ln \left (3\right )\right ) \pi x \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}+63 i \pi \,x^{2} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}-4 i \ln \left (\ln \left (3\right )\right ) x \ln \left (x \right )-126 i \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-126 i \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-2 \ln \left (\ln \left (3\right )\right ) \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )+63 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )-10 i}{2 x \left (2 i \ln \left (\ln \left (3\right )\right )+i x +126 \ln \left (\ln \left (3\right )\right )+63 x \right )}\right )\) | \(494\) |
Input:
int(((4*ln(ln(3))+4*x)*ln(x/exp(x))+2*(-6*x+2)*ln(ln(3))-8*x^2+2*x)/((2*ln (ln(3))*x+x^2)*ln(x/exp(x))-2*x^2*ln(ln(3))-x^3+5),x,method=_RETURNVERBOSE )
Output:
2*ln(-2*ln(x/exp(x))*ln(ln(3))*x+2*x^2*ln(ln(3))-ln(x/exp(x))*x^2+x^3-5)
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (25) = 50\).
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \, \log \left (x^{2} + 2 \, x \log \left (\log \left (3\right )\right )\right ) + 2 \, \log \left (-\frac {x^{3} + 2 \, x^{2} \log \left (\log \left (3\right )\right ) - {\left (x^{2} + 2 \, x \log \left (\log \left (3\right )\right )\right )} \log \left (x e^{\left (-x\right )}\right ) - 5}{x^{2} + 2 \, x \log \left (\log \left (3\right )\right )}\right ) \] Input:
integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+ 2*x)/((2*log(log(3))*x+x^2)*log(x/exp(x))-2*x^2*log(log(3))-x^3+5),x, algo rithm="fricas")
Output:
2*log(x^2 + 2*x*log(log(3))) + 2*log(-(x^3 + 2*x^2*log(log(3)) - (x^2 + 2* x*log(log(3)))*log(x*e^(-x)) - 5)/(x^2 + 2*x*log(log(3))))
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \log {\left (x^{2} + 2 x \log {\left (\log {\left (3 \right )} \right )} \right )} + 2 \log {\left (\log {\left (x e^{- x} \right )} + \frac {- x^{3} - 2 x^{2} \log {\left (\log {\left (3 \right )} \right )} + 5}{x^{2} + 2 x \log {\left (\log {\left (3 \right )} \right )}} \right )} \] Input:
integrate(((4*ln(ln(3))+4*x)*ln(x/exp(x))+2*(-6*x+2)*ln(ln(3))-8*x**2+2*x) /((2*ln(ln(3))*x+x**2)*ln(x/exp(x))-2*x**2*ln(ln(3))-x**3+5),x)
Output:
2*log(x**2 + 2*x*log(log(3))) + 2*log(log(x*exp(-x)) + (-x**3 - 2*x**2*log (log(3)) + 5)/(x**2 + 2*x*log(log(3))))
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (25) = 50\).
Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \, \log \left (x + 2 \, \log \left (\log \left (3\right )\right )\right ) + 2 \, \log \left (x\right ) + 2 \, \log \left (-\frac {2 \, x^{3} + 4 \, x^{2} \log \left (\log \left (3\right )\right ) - {\left (x^{2} + 2 \, x \log \left (\log \left (3\right )\right )\right )} \log \left (x\right ) - 5}{x^{2} + 2 \, x \log \left (\log \left (3\right )\right )}\right ) \] Input:
integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+ 2*x)/((2*log(log(3))*x+x^2)*log(x/exp(x))-2*x^2*log(log(3))-x^3+5),x, algo rithm="maxima")
Output:
2*log(x + 2*log(log(3))) + 2*log(x) + 2*log(-(2*x^3 + 4*x^2*log(log(3)) - (x^2 + 2*x*log(log(3)))*log(x) - 5)/(x^2 + 2*x*log(log(3))))
Time = 0.13 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \, \log \left (2 \, x^{3} - x^{2} \log \left (x\right ) + 4 \, x^{2} \log \left (\log \left (3\right )\right ) - 2 \, x \log \left (x\right ) \log \left (\log \left (3\right )\right ) - 5\right ) \] Input:
integrate(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+ 2*x)/((2*log(log(3))*x+x^2)*log(x/exp(x))-2*x^2*log(log(3))-x^3+5),x, algo rithm="giac")
Output:
2*log(2*x^3 - x^2*log(x) + 4*x^2*log(log(3)) - 2*x*log(x)*log(log(3)) - 5)
Time = 4.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2\,\ln \left (\frac {x^2\,\left (4\,\ln \left (\ln \left (3\right )\right )-\ln \left (x\right )\right )+2\,x^3-2\,x\,\ln \left (\ln \left (3\right )\right )\,\ln \left (x\right )-5}{x^2+2\,\ln \left (\ln \left (3\right )\right )\,x}\right )+2\,\ln \left (x^2+2\,\ln \left (\ln \left (3\right )\right )\,x\right ) \] Input:
int(-(2*x + log(x*exp(-x))*(4*x + 4*log(log(3))) - 2*log(log(3))*(6*x - 2) - 8*x^2)/(2*x^2*log(log(3)) - log(x*exp(-x))*(2*x*log(log(3)) + x^2) + x^ 3 - 5),x)
Output:
2*log((x^2*(4*log(log(3)) - log(x)) + 2*x^3 - 2*x*log(log(3))*log(x) - 5)/ (2*x*log(log(3)) + x^2)) + 2*log(2*x*log(log(3)) + x^2)
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {2 x-8 x^2+2 (2-6 x) \log (\log (3))+\log \left (e^{-x} x\right ) (4 x+4 \log (\log (3)))}{5-x^3-2 x^2 \log (\log (3))+\log \left (e^{-x} x\right ) \left (x^2+2 x \log (\log (3))\right )} \, dx=2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (\mathrm {log}\left (3\right )\right ) \mathrm {log}\left (\frac {x}{e^{x}}\right ) x -2 \,\mathrm {log}\left (\mathrm {log}\left (3\right )\right ) x^{2}+\mathrm {log}\left (\frac {x}{e^{x}}\right ) x^{2}-x^{3}+5\right ) \] Input:
int(((4*log(log(3))+4*x)*log(x/exp(x))+2*(-6*x+2)*log(log(3))-8*x^2+2*x)/( (2*log(log(3))*x+x^2)*log(x/exp(x))-2*x^2*log(log(3))-x^3+5),x)
Output:
2*log(2*log(log(3))*log(x/e**x)*x - 2*log(log(3))*x**2 + log(x/e**x)*x**2 - x**3 + 5)