Integrand size = 339, antiderivative size = 23 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=4 \left (x+x^2+\frac {1}{x+\log \left (-2+\frac {1}{-x+\log (\log (5))}\right )}\right ) \]
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=4 \left (x+x^2+\frac {1}{x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}\right ) \]
Integrate[(4 - 4*x - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + (4 + 16*x - 4*x^2 - 24*x^3 - 32*x^4)*Log[Log[5]] + (-8 + 8*x^2 + 16*x^3)*Log[Log[5]]^2 + (8*x ^2 + 32*x^3 + 32*x^4 + (-8*x - 48*x^2 - 64*x^3)*Log[Log[5]] + (16*x + 32*x ^2)*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (4* x + 16*x^2 + 16*x^3 + (-4 - 24*x - 32*x^2)*Log[Log[5]] + (8 + 16*x)*Log[Lo g[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2)/(x^3 + 2*x^4 + (-x^2 - 4*x^3)*Log[Log[5]] + 2*x^2*Log[Log[5]]^2 + (2*x^2 + 4*x^3 + (-2 *x - 8*x^2)*Log[Log[5]] + 4*x*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]]) /(-x + Log[Log[5]])] + (x + 2*x^2 + (-1 - 4*x)*Log[Log[5]] + 2*Log[Log[5]] ^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2),x]
Time = 1.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {7292, 7279, 2009}
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.
| \(\displaystyle \int \frac {16 x^5+16 x^4+4 x^3-8 x^2+\left (16 x^3+16 x^2+\left (-32 x^2-24 x-4\right ) \log (\log (5))+4 x+(16 x+8) \log ^2(\log (5))\right ) \log ^2\left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (16 x^3+8 x^2-8\right ) \log ^2(\log (5))+\left (32 x^4+32 x^3+8 x^2+\left (32 x^2+16 x\right ) \log ^2(\log (5))+\left (-64 x^3-48 x^2-8 x\right ) \log (\log (5))\right ) \log \left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (-32 x^4-24 x^3-4 x^2+16 x+4\right ) \log (\log (5))-4 x+4}{2 x^4+x^3+2 x^2 \log ^2(\log (5))+\left (2 x^2+x+(-4 x-1) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (4 x^3+2 x^2+\left (-8 x^2-2 x\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (-4 x^3-x^2\right ) \log (\log (5))} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {16 x^5+16 x^4+4 x^3-8 x^2+\left (16 x^3+16 x^2+\left (-32 x^2-24 x-4\right ) \log (\log (5))+4 x+(16 x+8) \log ^2(\log (5))\right ) \log ^2\left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (16 x^3+8 x^2-8\right ) \log ^2(\log (5))+\left (32 x^4+32 x^3+8 x^2+\left (32 x^2+16 x\right ) \log ^2(\log (5))+\left (-64 x^3-48 x^2-8 x\right ) \log (\log (5))\right ) \log \left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )+\left (-32 x^4-24 x^3-4 x^2+16 x+4\right ) \log (\log (5))-4 x+4}{\left (2 x^2+x (1-4 \log (\log (5)))-(1-2 \log (\log (5))) \log (\log (5))\right ) \left (x+\log \left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {4 \left (-2 x^2-x (1-4 \log (\log (5)))+(1-\log (\log (5))) (1+2 \log (\log (5)))\right )}{(2 x+1-2 \log (\log (5))) (x-\log (\log (5))) \left (x+\log \left (\frac {2 x+1-2 \log (\log (5))}{\log (\log (5))-x}\right )\right )^2}+4 (2 x+1)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle (2 x+1)^2+\frac {4}{x+\log \left (-\frac {2 x+1-2 \log (\log (5))}{x-\log (\log (5))}\right )}\) |
Int[(4 - 4*x - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + (4 + 16*x - 4*x^2 - 24*x^ 3 - 32*x^4)*Log[Log[5]] + (-8 + 8*x^2 + 16*x^3)*Log[Log[5]]^2 + (8*x^2 + 3 2*x^3 + 32*x^4 + (-8*x - 48*x^2 - 64*x^3)*Log[Log[5]] + (16*x + 32*x^2)*Lo g[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (4*x + 16 *x^2 + 16*x^3 + (-4 - 24*x - 32*x^2)*Log[Log[5]] + (8 + 16*x)*Log[Log[5]]^ 2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2)/(x^3 + 2*x^4 + (-x ^2 - 4*x^3)*Log[Log[5]] + 2*x^2*Log[Log[5]]^2 + (2*x^2 + 4*x^3 + (-2*x - 8 *x^2)*Log[Log[5]] + 4*x*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (x + 2*x^2 + (-1 - 4*x)*Log[Log[5]] + 2*Log[Log[5]]^2)*Lo g[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2),x]
3.14.48.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.94 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61
| method | result | size |
| risch | \(4 x^{2}+4 x +\frac {4}{x +\ln \left (\frac {-2 \ln \left (\ln \left (5\right )\right )+2 x +1}{\ln \left (\ln \left (5\right )\right )-x}\right )}\) | \(37\) |
| norman | \(\frac {4+4 x \ln \left (\frac {-2 \ln \left (\ln \left (5\right )\right )+2 x +1}{\ln \left (\ln \left (5\right )\right )-x}\right )+4 x^{2}+4 x^{3}+4 x^{2} \ln \left (\frac {-2 \ln \left (\ln \left (5\right )\right )+2 x +1}{\ln \left (\ln \left (5\right )\right )-x}\right )}{x +\ln \left (\frac {-2 \ln \left (\ln \left (5\right )\right )+2 x +1}{\ln \left (\ln \left (5\right )\right )-x}\right )}\) | \(89\) |
| parallelrisch | \(\frac {16-96 x \ln \left (\ln \left (5\right )\right )^{2}-96 \ln \left (\ln \left (5\right )\right )^{2} \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right )+16 x^{3}+16 \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right ) x^{2}+112 x \ln \left (\ln \left (5\right )\right )+112 \ln \left (\ln \left (5\right )\right ) \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right )+16 x^{2}+16 \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right ) x -20 x -20 \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right )}{4 \ln \left (-\frac {2 \ln \left (\ln \left (5\right )\right )-2 x -1}{\ln \left (\ln \left (5\right )\right )-x}\right )+4 x}\) | \(190\) |
| derivativedivides | \(\frac {4 \left (\left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{3}+\left (-8 \ln \left (\ln \left (5\right )\right )-2\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-8 \ln \left (\ln \left (5\right )\right )^{2}-\ln \left (\ln \left (5\right )\right )+13\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2 \ln \left (\ln \left (5\right )\right )^{2}-\ln \left (\ln \left (5\right )\right )+6\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{2}+\left (-3-8 \ln \left (\ln \left (5\right )\right )\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2 \ln \left (\ln \left (5\right )\right )-1\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{2}-8 \ln \left (\ln \left (5\right )\right )^{2}+2 \ln \left (\ln \left (5\right )\right )+9\right ) \left (-\ln \left (\ln \left (5\right )\right )+x \right )^{2}}{\ln \left (\ln \left (5\right )\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+2 \ln \left (\ln \left (5\right )\right )+2 \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )-1}\) | \(263\) |
| default | \(\frac {4 \left (\left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{3}+\left (-8 \ln \left (\ln \left (5\right )\right )-2\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-8 \ln \left (\ln \left (5\right )\right )^{2}-\ln \left (\ln \left (5\right )\right )+13\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2 \ln \left (\ln \left (5\right )\right )^{2}-\ln \left (\ln \left (5\right )\right )+6\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{2}+\left (-3-8 \ln \left (\ln \left (5\right )\right )\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2 \ln \left (\ln \left (5\right )\right )-1\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )^{2}-8 \ln \left (\ln \left (5\right )\right )^{2}+2 \ln \left (\ln \left (5\right )\right )+9\right ) \left (-\ln \left (\ln \left (5\right )\right )+x \right )^{2}}{\ln \left (\ln \left (5\right )\right ) \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+\left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right ) \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )+2 \ln \left (\ln \left (5\right )\right )+2 \ln \left (-2-\frac {1}{-\ln \left (\ln \left (5\right )\right )+x}\right )-1}\) | \(263\) |
int((((16*x+8)*ln(ln(5))^2+(-32*x^2-24*x-4)*ln(ln(5))+16*x^3+16*x^2+4*x)*l n((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))^2+((32*x^2+16*x)*ln(ln(5))^2+(-64*x^ 3-48*x^2-8*x)*ln(ln(5))+32*x^4+32*x^3+8*x^2)*ln((-2*ln(ln(5))+2*x+1)/(ln(l n(5))-x))+(16*x^3+8*x^2-8)*ln(ln(5))^2+(-32*x^4-24*x^3-4*x^2+16*x+4)*ln(ln (5))+16*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*ln(ln(5))^2+(-4*x-1)*ln(ln(5))+2 *x^2+x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))^2+(4*x*ln(ln(5))^2+(-8*x^2- 2*x)*ln(ln(5))+4*x^3+2*x^2)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))+2*x^2*l n(ln(5))^2+(-4*x^3-x^2)*ln(ln(5))+2*x^4+x^3),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (29) = 58\).
Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.78 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=\frac {4 \, {\left (x^{3} + x^{2} + {\left (x^{2} + x\right )} \log \left (-\frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )}\right ) + 1\right )}}{x + \log \left (-\frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )}\right )} \]
integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16* x^2+4*x)*log((-2*log(log(5))+2*x+1)/(log(log(5))-x))^2+((32*x^2+16*x)*log( log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*lo g(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4- 24*x^3-4*x^2+16*x+4)*log(log(5))+16*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log( log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(log(lo g(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log(( -2*log(log(5))+2*x+1)/(log(log(5))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*lo g(log(5))+2*x^4+x^3),x, algorithm=\
4*(x^3 + x^2 + (x^2 + x)*log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))) + 1)/(x + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))))
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=4 x^{2} + 4 x + \frac {4}{x + \log {\left (\frac {2 x - 2 \log {\left (\log {\left (5 \right )} \right )} + 1}{- x + \log {\left (\log {\left (5 \right )} \right )}} \right )}} \]
integrate((((16*x+8)*ln(ln(5))**2+(-32*x**2-24*x-4)*ln(ln(5))+16*x**3+16*x **2+4*x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))**2+((32*x**2+16*x)*ln(ln(5 ))**2+(-64*x**3-48*x**2-8*x)*ln(ln(5))+32*x**4+32*x**3+8*x**2)*ln((-2*ln(l n(5))+2*x+1)/(ln(ln(5))-x))+(16*x**3+8*x**2-8)*ln(ln(5))**2+(-32*x**4-24*x **3-4*x**2+16*x+4)*ln(ln(5))+16*x**5+16*x**4+4*x**3-8*x**2-4*x+4)/((2*ln(l n(5))**2+(-4*x-1)*ln(ln(5))+2*x**2+x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x ))**2+(4*x*ln(ln(5))**2+(-8*x**2-2*x)*ln(ln(5))+4*x**3+2*x**2)*ln((-2*ln(l n(5))+2*x+1)/(ln(ln(5))-x))+2*x**2*ln(ln(5))**2+(-4*x**3-x**2)*ln(ln(5))+2 *x**4+x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (29) = 58\).
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.91 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=\frac {4 \, {\left (x^{3} + x^{2} - {\left (x^{2} + x\right )} \log \left (x - \log \left (\log \left (5\right )\right )\right ) + {\left (x^{2} + x\right )} \log \left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) - 1\right ) + 1\right )}}{x - \log \left (x - \log \left (\log \left (5\right )\right )\right ) + \log \left (-2 \, x + 2 \, \log \left (\log \left (5\right )\right ) - 1\right )} \]
integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16* x^2+4*x)*log((-2*log(log(5))+2*x+1)/(log(log(5))-x))^2+((32*x^2+16*x)*log( log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*lo g(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4- 24*x^3-4*x^2+16*x+4)*log(log(5))+16*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log( log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(log(lo g(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log(( -2*log(log(5))+2*x+1)/(log(log(5))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*lo g(log(5))+2*x^4+x^3),x, algorithm=\
4*(x^3 + x^2 - (x^2 + x)*log(x - log(log(5))) + (x^2 + x)*log(-2*x + 2*log (log(5)) - 1) + 1)/(x - log(x - log(log(5))) + log(-2*x + 2*log(log(5)) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (29) = 58\).
Time = 0.57 (sec) , antiderivative size = 223, normalized size of antiderivative = 9.70 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=\frac {4 \, {\left (\frac {2 \, {\left (2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1\right )} \log \left (\log \left (5\right )\right )}{x - \log \left (\log \left (5\right )\right )} + \frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )} - 4 \, \log \left (\log \left (5\right )\right ) - 1\right )}}{\frac {{\left (2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1\right )}^{2}}{{\left (x - \log \left (\log \left (5\right )\right )\right )}^{2}} - \frac {4 \, {\left (2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1\right )}}{x - \log \left (\log \left (5\right )\right )} + 4} + \frac {4 \, {\left (\frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )} - 2\right )}}{\frac {{\left (2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1\right )} \log \left (-\frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )}\right )}{x - \log \left (\log \left (5\right )\right )} + \frac {{\left (2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1\right )} \log \left (\log \left (5\right )\right )}{x - \log \left (\log \left (5\right )\right )} - 2 \, \log \left (-\frac {2 \, x - 2 \, \log \left (\log \left (5\right )\right ) + 1}{x - \log \left (\log \left (5\right )\right )}\right ) - 2 \, \log \left (\log \left (5\right )\right ) + 1} \]
integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16* x^2+4*x)*log((-2*log(log(5))+2*x+1)/(log(log(5))-x))^2+((32*x^2+16*x)*log( log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*lo g(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4- 24*x^3-4*x^2+16*x+4)*log(log(5))+16*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log( log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(log(lo g(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log(( -2*log(log(5))+2*x+1)/(log(log(5))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*lo g(log(5))+2*x^4+x^3),x, algorithm=\
4*(2*(2*x - 2*log(log(5)) + 1)*log(log(5))/(x - log(log(5))) + (2*x - 2*lo g(log(5)) + 1)/(x - log(log(5))) - 4*log(log(5)) - 1)/((2*x - 2*log(log(5) ) + 1)^2/(x - log(log(5)))^2 - 4*(2*x - 2*log(log(5)) + 1)/(x - log(log(5) )) + 4) + 4*((2*x - 2*log(log(5)) + 1)/(x - log(log(5))) - 2)/((2*x - 2*lo g(log(5)) + 1)*log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5))))/(x - log( log(5))) + (2*x - 2*log(log(5)) + 1)*log(log(5))/(x - log(log(5))) - 2*log (-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))) - 2*log(log(5)) + 1)
Time = 94.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+\left (4+16 x-4 x^2-24 x^3-32 x^4\right ) \log (\log (5))+\left (-8+8 x^2+16 x^3\right ) \log ^2(\log (5))+\left (8 x^2+32 x^3+32 x^4+\left (-8 x-48 x^2-64 x^3\right ) \log (\log (5))+\left (16 x+32 x^2\right ) \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (4 x+16 x^2+16 x^3+\left (-4-24 x-32 x^2\right ) \log (\log (5))+(8+16 x) \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{x^3+2 x^4+\left (-x^2-4 x^3\right ) \log (\log (5))+2 x^2 \log ^2(\log (5))+\left (2 x^2+4 x^3+\left (-2 x-8 x^2\right ) \log (\log (5))+4 x \log ^2(\log (5))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+\left (x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )} \, dx=4\,x+4\,x^2+\frac {4}{x+\ln \left (-\frac {2\,x-2\,\ln \left (\ln \left (5\right )\right )+1}{x-\ln \left (\ln \left (5\right )\right )}\right )} \]
int((log(log(5))^2*(8*x^2 + 16*x^3 - 8) - log(log(5))*(4*x^2 - 16*x + 24*x ^3 + 32*x^4 - 4) - 4*x + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))) ^2*(4*x - log(log(5))*(24*x + 32*x^2 + 4) + log(log(5))^2*(16*x + 8) + 16* x^2 + 16*x^3) - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + log(-(2*x - 2*log(log(5) ) + 1)/(x - log(log(5))))*(log(log(5))^2*(16*x + 32*x^2) + 8*x^2 + 32*x^3 + 32*x^4 - log(log(5))*(8*x + 48*x^2 + 64*x^3)) + 4)/(2*x^2*log(log(5))^2 + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5))))^2*(x + 2*log(log(5))^2 - log(log(5))*(4*x + 1) + 2*x^2) + log(-(2*x - 2*log(log(5)) + 1)/(x - lo g(log(5))))*(4*x*log(log(5))^2 - log(log(5))*(2*x + 8*x^2) + 2*x^2 + 4*x^3 ) - log(log(5))*(x^2 + 4*x^3) + x^3 + 2*x^4),x)