Integrand size = 102, antiderivative size = 19 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=-1+\frac {x}{4+x \log \left (\log \left (4+\frac {2}{x}\right )\right )} \]
Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{4+x \log \left (\log \left (4+\frac {2}{x}\right )\right )} \]
Integrate[(x + (4 + 8*x)*Log[(2 + 4*x)/x])/((16 + 32*x)*Log[(2 + 4*x)/x] + (8*x + 16*x^2)*Log[(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]] + (x^2 + 2*x^3)*Log [(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]]^2),x]
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.
| \(\displaystyle \int \frac {x+(8 x+4) \log \left (\frac {4 x+2}{x}\right )}{\left (16 x^2+8 x\right ) \log \left (\frac {4 x+2}{x}\right ) \log \left (\log \left (\frac {4 x+2}{x}\right )\right )+\left (2 x^3+x^2\right ) \log \left (\frac {4 x+2}{x}\right ) \log ^2\left (\log \left (\frac {4 x+2}{x}\right )\right )+(32 x+16) \log \left (\frac {4 x+2}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x+(8 x+4) \log \left (\frac {2}{x}+4\right )}{(2 x+1) \log \left (\frac {2}{x}+4\right ) \left (x \log \left (\log \left (\frac {2}{x}+4\right )\right )+4\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {8 x}{(2 x+1) \left (x \log \left (\log \left (\frac {2}{x}+4\right )\right )+4\right )^2}+\frac {x}{(2 x+1) \log \left (\frac {2}{x}+4\right ) \left (x \log \left (\log \left (\frac {2}{x}+4\right )\right )+4\right )^2}+\frac {4}{(2 x+1) \left (x \log \left (\log \left (\frac {2}{x}+4\right )\right )+4\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 \int \frac {1}{\left (x \log \left (\log \left (4+\frac {2}{x}\right )\right )+4\right )^2}dx+\frac {1}{2} \int \frac {1}{\log \left (4+\frac {2}{x}\right ) \left (x \log \left (\log \left (4+\frac {2}{x}\right )\right )+4\right )^2}dx-\frac {1}{2} \int \frac {1}{(2 x+1) \log \left (4+\frac {2}{x}\right ) \left (x \log \left (\log \left (4+\frac {2}{x}\right )\right )+4\right )^2}dx\) |
Int[(x + (4 + 8*x)*Log[(2 + 4*x)/x])/((16 + 32*x)*Log[(2 + 4*x)/x] + (8*x + 16*x^2)*Log[(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]] + (x^2 + 2*x^3)*Log[(2 + 4*x)/x]*Log[Log[(2 + 4*x)/x]]^2),x]
3.8.58.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11
| method | result | size |
| parallelrisch | \(\frac {x}{x \ln \left (\ln \left (\frac {4 x +2}{x}\right )\right )+4}\) | \(21\) |
| default | \(\frac {1}{-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )}\) | \(29\) |
| parts | \(\frac {4 \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right ) \left (1+2 x \right )}{\left (\frac {4 \left (1+2 x \right ) \ln \left (2\right )}{x}+\frac {4 \left (1+2 x \right ) \ln \left (\frac {1+2 x}{x}\right )}{x}+1\right ) x \left (-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )\right )}+\frac {1}{\left (\frac {4 \left (1+2 x \right ) \ln \left (2\right )}{x}+\frac {4 \left (1+2 x \right ) \ln \left (\frac {1+2 x}{x}\right )}{x}+1\right ) \left (-8+\frac {8 x +4}{x}+\ln \left (\ln \left (2\right )+\ln \left (\frac {1+2 x}{x}\right )\right )\right )}\) | \(154\) |
int(((8*x+4)*ln((4*x+2)/x)+x)/((2*x^3+x^2)*ln((4*x+2)/x)*ln(ln((4*x+2)/x)) ^2+(16*x^2+8*x)*ln((4*x+2)/x)*ln(ln((4*x+2)/x))+(32*x+16)*ln((4*x+2)/x)),x ,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 4} \]
integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log(( 4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2)/x)*log(log((4*x+2)/x))+(32*x+16)*log ((4*x+2)/x)),x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log {\left (\log {\left (\frac {4 x + 2}{x} \right )} \right )} + 4} \]
integrate(((8*x+4)*ln((4*x+2)/x)+x)/((2*x**3+x**2)*ln((4*x+2)/x)*ln(ln((4* x+2)/x))**2+(16*x**2+8*x)*ln((4*x+2)/x)*ln(ln((4*x+2)/x))+(32*x+16)*ln((4* x+2)/x)),x)
Time = 0.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x}{x \log \left (\log \left (2\right ) + \log \left (2 \, x + 1\right ) - \log \left (x\right )\right ) + 4} \]
integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log(( 4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2)/x)*log(log((4*x+2)/x))+(32*x+16)*log ((4*x+2)/x)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (19) = 38\).
Time = 0.43 (sec) , antiderivative size = 345, normalized size of antiderivative = 18.16 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {8 \, x^{2} \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 8 \, x^{2} \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + x^{2} \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + 4 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 4 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )}{8 \, x^{2} \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - 8 \, x^{2} \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + x^{2} \log \left (4 \, x + 2\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - x^{2} \log \left (x\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 4 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) - 4 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) \log \left (\log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )\right ) + 32 \, x \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 32 \, x \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) + 4 \, x \log \left (4 \, x + 2\right ) - 4 \, x \log \left (x\right ) + 16 \, \log \left (4 \, x + 2\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right ) - 16 \, \log \left (x\right ) \log \left (\frac {2 \, {\left (2 \, x + 1\right )}}{x}\right )} \]
integrate(((8*x+4)*log((4*x+2)/x)+x)/((2*x^3+x^2)*log((4*x+2)/x)*log(log(( 4*x+2)/x))^2+(16*x^2+8*x)*log((4*x+2)/x)*log(log((4*x+2)/x))+(32*x+16)*log ((4*x+2)/x)),x, algorithm=\
(8*x^2*log(4*x + 2)*log(2*(2*x + 1)/x) - 8*x^2*log(x)*log(2*(2*x + 1)/x) + x^2*log(2*(2*x + 1)/x) + 4*x*log(4*x + 2)*log(2*(2*x + 1)/x) - 4*x*log(x) *log(2*(2*x + 1)/x))/(8*x^2*log(4*x + 2)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) - 8*x^2*log(x)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) + x^2* log(4*x + 2)*log(log(2*(2*x + 1)/x)) - x^2*log(x)*log(log(2*(2*x + 1)/x)) + 4*x*log(4*x + 2)*log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) - 4*x*log(x) *log(2*(2*x + 1)/x)*log(log(2*(2*x + 1)/x)) + 32*x*log(4*x + 2)*log(2*(2*x + 1)/x) - 32*x*log(x)*log(2*(2*x + 1)/x) + 4*x*log(4*x + 2) - 4*x*log(x) + 16*log(4*x + 2)*log(2*(2*x + 1)/x) - 16*log(x)*log(2*(2*x + 1)/x))
Time = 13.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 8.79 \[ \int \frac {x+(4+8 x) \log \left (\frac {2+4 x}{x}\right )}{(16+32 x) \log \left (\frac {2+4 x}{x}\right )+\left (8 x+16 x^2\right ) \log \left (\frac {2+4 x}{x}\right ) \log \left (\log \left (\frac {2+4 x}{x}\right )\right )+\left (x^2+2 x^3\right ) \log \left (\frac {2+4 x}{x}\right ) \log ^2\left (\log \left (\frac {2+4 x}{x}\right )\right )} \, dx=\frac {x\,{\left (\ln \left (\frac {4\,x+2}{x}\right )+2\,x\,\ln \left (\frac {4\,x+2}{x}\right )\right )}^2\,\left (x+4\,\ln \left (\frac {4\,x+2}{x}\right )+8\,x\,\ln \left (\frac {4\,x+2}{x}\right )\right )}{\ln \left (\frac {4\,x+2}{x}\right )\,\left (2\,x+1\right )\,\left (x\,\ln \left (\ln \left (\frac {4\,x+2}{x}\right )\right )+4\right )\,\left (16\,x^2\,{\ln \left (\frac {4\,x+2}{x}\right )}^2+2\,x^2\,\ln \left (\frac {4\,x+2}{x}\right )+16\,x\,{\ln \left (\frac {4\,x+2}{x}\right )}^2+x\,\ln \left (\frac {4\,x+2}{x}\right )+4\,{\ln \left (\frac {4\,x+2}{x}\right )}^2\right )} \]
int((x + log((4*x + 2)/x)*(8*x + 4))/(log((4*x + 2)/x)*(32*x + 16) + log(l og((4*x + 2)/x))^2*log((4*x + 2)/x)*(x^2 + 2*x^3) + log(log((4*x + 2)/x))* log((4*x + 2)/x)*(8*x + 16*x^2)),x)