Integrand size = 182, antiderivative size = 28 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10}}{x \left (4-\frac {3}{1-\log \left (3+\frac {1}{x}+\log (\log (3))\right )}\right )} \] Output:
exp(5)^2/x/(4-3/(1-ln(3+1/x+ln(ln(3)))))
Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=-\frac {e^{10} \left (1-\log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}{x \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )} \] Input:
Integrate[(E^10*(-4 - 3*x) - E^10*x*Log[Log[3]] + (E^10*(5 + 15*x) + 5*E^1 0*x*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x] + (E^10*(-4 - 12*x) - 4* E^10*x*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2)/(x^2 + 3*x^3 + x^3 *Log[Log[3]] + (-8*x^2 - 24*x^3 - 8*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[ Log[3]])/x] + (16*x^2 + 48*x^3 + 16*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[ Log[3]])/x]^2),x]
Output:
-((E^10*(1 - Log[3 + x^(-1) + Log[Log[3]]]))/(x*(-1 + 4*Log[3 + x^(-1) + L og[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.
| \(\displaystyle \int \frac {e^{10} (-3 x-4)+\left (e^{10} (-12 x-4)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {3 x+x \log (\log (3))+1}{x}\right )+\left (e^{10} (15 x+5)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {3 x+x \log (\log (3))+1}{x}\right )-e^{10} x \log (\log (3))}{3 x^3+x^3 \log (\log (3))+x^2+\left (48 x^3+16 x^3 \log (\log (3))+16 x^2\right ) \log ^2\left (\frac {3 x+x \log (\log (3))+1}{x}\right )+\left (-24 x^3-8 x^3 \log (\log (3))-8 x^2\right ) \log \left (\frac {3 x+x \log (\log (3))+1}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {e^{10} (-3 x-4)+\left (e^{10} (-12 x-4)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {3 x+x \log (\log (3))+1}{x}\right )+\left (e^{10} (15 x+5)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {3 x+x \log (\log (3))+1}{x}\right )-e^{10} x \log (\log (3))}{x^3 (3+\log (\log (3)))+x^2+\left (48 x^3+16 x^3 \log (\log (3))+16 x^2\right ) \log ^2\left (\frac {3 x+x \log (\log (3))+1}{x}\right )+\left (-24 x^3-8 x^3 \log (\log (3))-8 x^2\right ) \log \left (\frac {3 x+x \log (\log (3))+1}{x}\right )}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{10} \left (-4 (x (3+\log (\log (3)))+1) \log ^2\left (\frac {1}{x}+3+\log (\log (3))\right )+5 (x (3+\log (\log (3)))+1) \log \left (\frac {1}{x}+3+\log (\log (3))\right )-x (3+\log (\log (3)))-4\right )}{(x (3+\log (\log (3)))+1) \left (x-4 x \log \left (\frac {1}{x}+3+\log (\log (3))\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^{10} \int -\frac {4 ((3+\log (\log (3))) x+1) \log ^2\left (\log (\log (3))+3+\frac {1}{x}\right )-5 ((3+\log (\log (3))) x+1) \log \left (\log (\log (3))+3+\frac {1}{x}\right )+x (3+\log (\log (3)))+4}{((3+\log (\log (3))) x+1) \left (x-4 x \log \left (\log (\log (3))+3+\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -e^{10} \int \frac {4 ((3+\log (\log (3))) x+1) \log ^2\left (\log (\log (3))+3+\frac {1}{x}\right )-5 ((3+\log (\log (3))) x+1) \log \left (\log (\log (3))+3+\frac {1}{x}\right )+x (3+\log (\log (3)))+4}{((3+\log (\log (3))) x+1) \left (x-4 x \log \left (\log (\log (3))+3+\frac {1}{x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -e^{10} \int \left (-\frac {3}{4 x^2 \left (4 \log \left (\log (\log (3))+3+\frac {1}{x}\right )-1\right )}+\frac {1}{4 x^2}+\frac {3}{x^2 \left (1-4 \log \left (\log (\log (3))+3+\frac {1}{x}\right )\right )^2 ((3+\log (\log (3))) x+1)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -e^{10} \left (-3 (3+\log (\log (3))) \int \frac {1}{x \left (1-4 \log \left (\log (\log (3))+3+\frac {1}{x}\right )\right )^2}dx+3 (3+\log (\log (3)))^2 \int \frac {1}{((3+\log (\log (3))) x+1) \left (1-4 \log \left (\log (\log (3))+3+\frac {1}{x}\right )\right )^2}dx-\frac {1}{4 x}-\frac {3 \left (\frac {1}{x}+3+\log (\log (3))\right )}{4 \left (1-4 \log \left (\frac {1}{x}+3+\log (\log (3))\right )\right )}\right )\) |
Input:
Int[(E^10*(-4 - 3*x) - E^10*x*Log[Log[3]] + (E^10*(5 + 15*x) + 5*E^10*x*Lo g[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x] + (E^10*(-4 - 12*x) - 4*E^10*x *Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2)/(x^2 + 3*x^3 + x^3*Log[L og[3]] + (-8*x^2 - 24*x^3 - 8*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3] ])/x] + (16*x^2 + 48*x^3 + 16*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3] ])/x]^2),x]
Output:
$Aborted
Time = 1.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
| method | result | size |
| risch | \(\frac {{\mathrm e}^{10}}{4 x}-\frac {3 \,{\mathrm e}^{10}}{4 x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\) | \(37\) |
| norman | \(\frac {\ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right ) {\mathrm e}^{10}-{\mathrm e}^{10}}{x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\) | \(53\) |
| derivativedivides | \(\frac {3 \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}+{\mathrm e}^{10} \left (\frac {9}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}-\frac {3 \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )}{16 \left (\ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-\frac {1}{4}\right )}+\frac {\ln \left (\ln \left (3\right )\right )}{4}+\frac {3}{4}+\frac {1}{4 x}\right )\) | \(83\) |
| default | \(\frac {3 \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}+{\mathrm e}^{10} \left (\frac {9}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}-\frac {3 \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )}{16 \left (\ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-\frac {1}{4}\right )}+\frac {\ln \left (\ln \left (3\right )\right )}{4}+\frac {3}{4}+\frac {1}{4 x}\right )\) | \(83\) |
| parallelrisch | \(-\frac {8 \ln \left (\ln \left (3\right )\right ) \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right ) {\mathrm e}^{10} x -2 x \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )+24 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right ) {\mathrm e}^{10} x -6 \,{\mathrm e}^{10} x -4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right ) {\mathrm e}^{10}+4 \,{\mathrm e}^{10}}{4 x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\) | \(119\) |
Input:
int(((-4*x*exp(5)^2*ln(ln(3))+(-12*x-4)*exp(5)^2)*ln((ln(ln(3))*x+3*x+1)/x )^2+(5*x*exp(5)^2*ln(ln(3))+(15*x+5)*exp(5)^2)*ln((ln(ln(3))*x+3*x+1)/x)-x *exp(5)^2*ln(ln(3))+(-3*x-4)*exp(5)^2)/((16*x^3*ln(ln(3))+48*x^3+16*x^2)*l n((ln(ln(3))*x+3*x+1)/x)^2+(-8*x^3*ln(ln(3))-24*x^3-8*x^2)*ln((ln(ln(3))*x +3*x+1)/x)+x^3*ln(ln(3))+3*x^3+x^2),x,method=_RETURNVERBOSE)
Output:
1/4*exp(10)/x-3/4/x*exp(10)/(-1+4*ln((ln(ln(3))*x+3*x+1)/x))
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (\frac {x \log \left (\log \left (3\right )\right ) + 3 \, x + 1}{x}\right ) - e^{10}}{4 \, x \log \left (\frac {x \log \left (\log \left (3\right )\right ) + 3 \, x + 1}{x}\right ) - x} \] Input:
integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3)) *x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log(3))+(15*x+5)*exp(5)^2)*log((log(log(3 ))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(log( 3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x ^3-8*x^2)*log((log(log(3))*x+3*x+1)/x)+x^3*log(log(3))+3*x^3+x^2),x, algor ithm="fricas")
Output:
(e^10*log((x*log(log(3)) + 3*x + 1)/x) - e^10)/(4*x*log((x*log(log(3)) + 3 *x + 1)/x) - x)
Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=- \frac {3 e^{10}}{16 x \log {\left (\frac {x \log {\left (\log {\left (3 \right )} \right )} + 3 x + 1}{x} \right )} - 4 x} + \frac {e^{10}}{4 x} \] Input:
integrate(((-4*x*exp(5)**2*ln(ln(3))+(-12*x-4)*exp(5)**2)*ln((ln(ln(3))*x+ 3*x+1)/x)**2+(5*x*exp(5)**2*ln(ln(3))+(15*x+5)*exp(5)**2)*ln((ln(ln(3))*x+ 3*x+1)/x)-x*exp(5)**2*ln(ln(3))+(-3*x-4)*exp(5)**2)/((16*x**3*ln(ln(3))+48 *x**3+16*x**2)*ln((ln(ln(3))*x+3*x+1)/x)**2+(-8*x**3*ln(ln(3))-24*x**3-8*x **2)*ln((ln(ln(3))*x+3*x+1)/x)+x**3*ln(ln(3))+3*x**3+x**2),x)
Output:
-3*exp(10)/(16*x*log((x*log(log(3)) + 3*x + 1)/x) - 4*x) + exp(10)/(4*x)
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (x {\left (\log \left (\log \left (3\right )\right ) + 3\right )} + 1\right ) - e^{10} \log \left (x\right ) - e^{10}}{4 \, x \log \left (x {\left (\log \left (\log \left (3\right )\right ) + 3\right )} + 1\right ) - 4 \, x \log \left (x\right ) - x} \] Input:
integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3)) *x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log(3))+(15*x+5)*exp(5)^2)*log((log(log(3 ))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(log( 3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x ^3-8*x^2)*log((log(log(3))*x+3*x+1)/x)+x^3*log(log(3))+3*x^3+x^2),x, algor ithm="maxima")
Output:
(e^10*log(x*(log(log(3)) + 3) + 1) - e^10*log(x) - e^10)/(4*x*log(x*(log(l og(3)) + 3) + 1) - 4*x*log(x) - x)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (x \log \left (\log \left (3\right )\right ) + 3 \, x + 1\right ) - e^{10} \log \left (x\right ) - e^{10}}{4 \, x \log \left (x \log \left (\log \left (3\right )\right ) + 3 \, x + 1\right ) - 4 \, x \log \left (x\right ) - x} \] Input:
integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3)) *x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log(3))+(15*x+5)*exp(5)^2)*log((log(log(3 ))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(log( 3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x ^3-8*x^2)*log((log(log(3))*x+3*x+1)/x)+x^3*log(log(3))+3*x^3+x^2),x, algor ithm="giac")
Output:
(e^10*log(x*log(log(3)) + 3*x + 1) - e^10*log(x) - e^10)/(4*x*log(x*log(lo g(3)) + 3*x + 1) - 4*x*log(x) - x)
Time = 3.65 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.71 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {{\mathrm {e}}^{10}\,\left (36\,x+12\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{64\,x}-\frac {\frac {{\mathrm {e}}^{10}\,\left (9\,x+3\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{16}-\frac {{\mathrm {e}}^{10}\,\left (36\,x+12\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{64}}{x\,\left (4\,\ln \left (\frac {3\,x+x\,\ln \left (\ln \left (3\right )\right )+1}{x}\right )-1\right )} \] Input:
int(-(log((3*x + x*log(log(3)) + 1)/x)^2*(exp(10)*(12*x + 4) + 4*x*exp(10) *log(log(3))) - log((3*x + x*log(log(3)) + 1)/x)*(exp(10)*(15*x + 5) + 5*x *exp(10)*log(log(3))) + exp(10)*(3*x + 4) + x*exp(10)*log(log(3)))/(x^3*lo g(log(3)) + log((3*x + x*log(log(3)) + 1)/x)^2*(16*x^3*log(log(3)) + 16*x^ 2 + 48*x^3) - log((3*x + x*log(log(3)) + 1)/x)*(8*x^3*log(log(3)) + 8*x^2 + 24*x^3) + x^2 + 3*x^3),x)
Output:
(exp(10)*(36*x + 12*x*log(log(3)) + 16))/(64*x) - ((exp(10)*(9*x + 3*x*log (log(3)) + 16))/16 - (exp(10)*(36*x + 12*x*log(log(3)) + 16))/64)/(x*(4*lo g((3*x + x*log(log(3)) + 1)/x) - 1))
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \left (\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (3\right )\right ) x +3 x +1}{x}\right )-1\right )}{x \left (4 \,\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (3\right )\right ) x +3 x +1}{x}\right )-1\right )} \] Input:
int(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3))*x+3*x +1)/x)^2+(5*x*exp(5)^2*log(log(3))+(15*x+5)*exp(5)^2)*log((log(log(3))*x+3 *x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(log(3))+48 *x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x^3-8*x ^2)*log((log(log(3))*x+3*x+1)/x)+x^3*log(log(3))+3*x^3+x^2),x)
Output:
(e**10*(log((log(log(3))*x + 3*x + 1)/x) - 1))/(x*(4*log((log(log(3))*x + 3*x + 1)/x) - 1))