Integrand size = 162, antiderivative size = 27 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=16-x+\frac {\log (x)}{-4+x}+\frac {\log ^2(3)}{2-\log (\log (x))} \]
Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=-x+\frac {\log (x)}{-4+x}-\frac {\log ^2(3)}{-2+\log (\log (x))} \]
Integrate[((16 - 8*x + x^2)*Log[3]^2 + (-16 - 60*x + 32*x^2 - 4*x^3)*Log[x ] - 4*x*Log[x]^2 + ((16 + 60*x - 32*x^2 + 4*x^3)*Log[x] + 4*x*Log[x]^2)*Lo g[Log[x]] + ((-4 - 15*x + 8*x^2 - x^3)*Log[x] - x*Log[x]^2)*Log[Log[x]]^2) /((64*x - 32*x^2 + 4*x^3)*Log[x] + (-64*x + 32*x^2 - 4*x^3)*Log[x]*Log[Log [x]] + (16*x - 8*x^2 + x^3)*Log[x]*Log[Log[x]]^2),x]
Time = 0.44 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {7239, 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.
\(\displaystyle \int \frac {\left (x^2-8 x+16\right ) \log ^2(3)+\left (\left (-x^3+8 x^2-15 x-4\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))+\left (\left (4 x^3-32 x^2+60 x+16\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (-4 x^3+32 x^2-60 x-16\right ) \log (x)-4 x \log ^2(x)}{\left (x^3-8 x^2+16 x\right ) \log (x) \log ^2(\log (x))+\left (-4 x^3+32 x^2-64 x\right ) \log (x) \log (\log (x))+\left (4 x^3-32 x^2+64 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {-x^2+4 x+1}{(x-4) x}+\frac {\log ^2(3)}{x \log (x) (\log (\log (x))-2)^2}-\frac {\log (x)}{(x-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x+\frac {\log ^2(3)}{2-\log (\log (x))}-\frac {x \log (x)}{4 (4-x)}-\frac {\log (x)}{4}\) |
Int[((16 - 8*x + x^2)*Log[3]^2 + (-16 - 60*x + 32*x^2 - 4*x^3)*Log[x] - 4* x*Log[x]^2 + ((16 + 60*x - 32*x^2 + 4*x^3)*Log[x] + 4*x*Log[x]^2)*Log[Log[ x]] + ((-4 - 15*x + 8*x^2 - x^3)*Log[x] - x*Log[x]^2)*Log[Log[x]]^2)/((64* x - 32*x^2 + 4*x^3)*Log[x] + (-64*x + 32*x^2 - 4*x^3)*Log[x]*Log[Log[x]] + (16*x - 8*x^2 + x^3)*Log[x]*Log[Log[x]]^2),x]
3.7.49.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.61 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19
method | result | size |
default | \(-\frac {\ln \left (3\right )^{2}}{\ln \left (\ln \left (x \right )\right )-2}+\frac {x \ln \left (x \right )}{4 x -16}-x -\frac {\ln \left (x \right )}{4}\) | \(32\) |
parts | \(-\frac {\ln \left (3\right )^{2}}{\ln \left (\ln \left (x \right )\right )-2}+\frac {x \ln \left (x \right )}{4 x -16}-x -\frac {\ln \left (x \right )}{4}\) | \(32\) |
risch | \(-\frac {x^{2}-4 x -\ln \left (x \right )}{x -4}-\frac {\ln \left (3\right )^{2}}{\ln \left (\ln \left (x \right )\right )-2}\) | \(33\) |
parallelrisch | \(-\frac {32-\ln \left (x \right ) \ln \left (\ln \left (x \right )\right )+x \ln \left (3\right )^{2}+x^{2} \ln \left (\ln \left (x \right )\right )-4 \ln \left (3\right )^{2}-16 \ln \left (\ln \left (x \right )\right )-2 x^{2}+2 \ln \left (x \right )}{\left (\ln \left (\ln \left (x \right )\right )-2\right ) \left (x -4\right )}\) | \(57\) |
int(((-x*ln(x)^2+(-x^3+8*x^2-15*x-4)*ln(x))*ln(ln(x))^2+(4*x*ln(x)^2+(4*x^ 3-32*x^2+60*x+16)*ln(x))*ln(ln(x))-4*x*ln(x)^2+(-4*x^3+32*x^2-60*x-16)*ln( x)+(x^2-8*x+16)*ln(3)^2)/((x^3-8*x^2+16*x)*ln(x)*ln(ln(x))^2+(-4*x^3+32*x^ 2-64*x)*ln(x)*ln(ln(x))+(4*x^3-32*x^2+64*x)*ln(x)),x,method=_RETURNVERBOSE )
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=-\frac {{\left (x - 4\right )} \log \left (3\right )^{2} - 2 \, x^{2} + {\left (x^{2} - 4 \, x - \log \left (x\right )\right )} \log \left (\log \left (x\right )\right ) + 8 \, x + 2 \, \log \left (x\right )}{{\left (x - 4\right )} \log \left (\log \left (x\right )\right ) - 2 \, x + 8} \]
integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log (x)^2+(4*x^3-32*x^2+60*x+16)*log(x))*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x ^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(log (x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)) ,x, algorithm=\
-((x - 4)*log(3)^2 - 2*x^2 + (x^2 - 4*x - log(x))*log(log(x)) + 8*x + 2*lo g(x))/((x - 4)*log(log(x)) - 2*x + 8)
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=- x - \frac {\log {\left (3 \right )}^{2}}{\log {\left (\log {\left (x \right )} \right )} - 2} + \frac {\log {\left (x \right )}}{x - 4} \]
integrate(((-x*ln(x)**2+(-x**3+8*x**2-15*x-4)*ln(x))*ln(ln(x))**2+(4*x*ln( x)**2+(4*x**3-32*x**2+60*x+16)*ln(x))*ln(ln(x))-4*x*ln(x)**2+(-4*x**3+32*x **2-60*x-16)*ln(x)+(x**2-8*x+16)*ln(3)**2)/((x**3-8*x**2+16*x)*ln(x)*ln(ln (x))**2+(-4*x**3+32*x**2-64*x)*ln(x)*ln(ln(x))+(4*x**3-32*x**2+64*x)*ln(x) ),x)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=-\frac {{\left (\log \left (3\right )^{2} + 8\right )} x - 2 \, x^{2} - 4 \, \log \left (3\right )^{2} + {\left (x^{2} - 4 \, x - \log \left (x\right )\right )} \log \left (\log \left (x\right )\right ) + 2 \, \log \left (x\right )}{{\left (x - 4\right )} \log \left (\log \left (x\right )\right ) - 2 \, x + 8} \]
integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log (x)^2+(4*x^3-32*x^2+60*x+16)*log(x))*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x ^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(log (x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)) ,x, algorithm=\
-((log(3)^2 + 8)*x - 2*x^2 - 4*log(3)^2 + (x^2 - 4*x - log(x))*log(log(x)) + 2*log(x))/((x - 4)*log(log(x)) - 2*x + 8)
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=-x - \frac {\log \left (3\right )^{2}}{\log \left (\log \left (x\right )\right ) - 2} + \frac {\log \left (x\right )}{x - 4} \]
integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log (x)^2+(4*x^3-32*x^2+60*x+16)*log(x))*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x ^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(log (x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)) ,x, algorithm=\
Time = 9.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (16-8 x+x^2\right ) \log ^2(3)+\left (-16-60 x+32 x^2-4 x^3\right ) \log (x)-4 x \log ^2(x)+\left (\left (16+60 x-32 x^2+4 x^3\right ) \log (x)+4 x \log ^2(x)\right ) \log (\log (x))+\left (\left (-4-15 x+8 x^2-x^3\right ) \log (x)-x \log ^2(x)\right ) \log ^2(\log (x))}{\left (64 x-32 x^2+4 x^3\right ) \log (x)+\left (-64 x+32 x^2-4 x^3\right ) \log (x) \log (\log (x))+\left (16 x-8 x^2+x^3\right ) \log (x) \log ^2(\log (x))} \, dx=\frac {\ln \left (x\right )}{x-4}-\frac {{\ln \left (3\right )}^2}{\ln \left (\ln \left (x\right )\right )-2}-x \]
int(-(log(log(x))^2*(x*log(x)^2 + log(x)*(15*x - 8*x^2 + x^3 + 4)) - log(l og(x))*(4*x*log(x)^2 + log(x)*(60*x - 32*x^2 + 4*x^3 + 16)) + 4*x*log(x)^2 - log(3)^2*(x^2 - 8*x + 16) + log(x)*(60*x - 32*x^2 + 4*x^3 + 16))/(log(x )*(64*x - 32*x^2 + 4*x^3) - log(log(x))*log(x)*(64*x - 32*x^2 + 4*x^3) + l og(log(x))^2*log(x)*(16*x - 8*x^2 + x^3)),x)