Integrand size = 173, antiderivative size = 31 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=5-\frac {2}{5 x \left (x-\log \left (\frac {x^2}{(x+x (1+x) \log (x))^2}\right )\right )} \]
Time = 6.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=\frac {2}{5 x \left (-x+\log \left (\frac {1}{(1+\log (x)+x \log (x))^2}\right )\right )} \]
Integrate[(4 + 8*x + (8*x + 4*x^2)*Log[x] + (-2 + (-2 - 2*x)*Log[x])*Log[( 1 + (2 + 2*x)*Log[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)])/(5*x^4 + (5*x^4 + 5*x^5)*Log[x] + (-10*x^3 + (-10*x^3 - 10*x^4)*Log[x])*Log[(1 + (2 + 2*x)*L og[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)] + (5*x^2 + (5*x^2 + 5*x^3)*Log[x]) *Log[(1 + (2 + 2*x)*Log[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)]^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.
\(\displaystyle \int \frac {((-2 x-2) \log (x)-2) \log \left (\frac {1}{\left (x^2+2 x+1\right ) \log ^2(x)+(2 x+2) \log (x)+1}\right )+\left (4 x^2+8 x\right ) \log (x)+8 x+4}{5 x^4+\left (5 x^5+5 x^4\right ) \log (x)+\left (5 x^2+\left (5 x^3+5 x^2\right ) \log (x)\right ) \log ^2\left (\frac {1}{\left (x^2+2 x+1\right ) \log ^2(x)+(2 x+2) \log (x)+1}\right )+\left (\left (-10 x^4-10 x^3\right ) \log (x)-10 x^3\right ) \log \left (\frac {1}{\left (x^2+2 x+1\right ) \log ^2(x)+(2 x+2) \log (x)+1}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {8 x-2 \log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )+\log (x) \left (4 x (x+2)-2 (x+1) \log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )\right )+4}{5 x^2 ((x+1) \log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {2 \left (4 x-\log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )+\log (x) \left (2 x (x+2)-(x+1) \log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )\right )+2\right )}{x^2 ((x+1) \log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int \frac {4 x-\log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )+\log (x) \left (2 x (x+2)-(x+1) \log \left (\frac {1}{(x \log (x)+\log (x)+1)^2}\right )\right )+2}{x^2 ((x+1) \log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{5} \int \left (-\frac {\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right ) \log (x)}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}-\frac {\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right ) \log (x)}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}+\frac {4 \log (x)}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}+\frac {2 \log (x)}{(x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}-\frac {\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}+\frac {4}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}+\frac {2}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \left (2 \int \frac {1}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx+\int \frac {1}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )}dx+\int \frac {\log (x)}{x^2 (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )}dx+3 \int \frac {1}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx+\int \frac {\log (x)}{(x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx+3 \int \frac {\log (x)}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )^2}dx+\int \frac {\log (x)}{x (x \log (x)+\log (x)+1) \left (x-\log \left (\frac {1}{((x+1) \log (x)+1)^2}\right )\right )}dx\right )\) |
Int[(4 + 8*x + (8*x + 4*x^2)*Log[x] + (-2 + (-2 - 2*x)*Log[x])*Log[(1 + (2 + 2*x)*Log[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)])/(5*x^4 + (5*x^4 + 5*x^5) *Log[x] + (-10*x^3 + (-10*x^3 - 10*x^4)*Log[x])*Log[(1 + (2 + 2*x)*Log[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)] + (5*x^2 + (5*x^2 + 5*x^3)*Log[x])*Log[( 1 + (2 + 2*x)*Log[x] + (1 + 2*x + x^2)*Log[x]^2)^(-1)]^2),x]
3.21.100.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.84 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45
method | result | size |
parallelrisch | \(-\frac {2}{5 x \left (x -\ln \left (\frac {1}{x^{2} \ln \left (x \right )^{2}+2 x \ln \left (x \right )^{2}+\ln \left (x \right )^{2}+2 x \ln \left (x \right )+2 \ln \left (x \right )+1}\right )\right )}\) | \(45\) |
default | \(-\frac {4 i}{5 \left (\pi \operatorname {csgn}\left (i \left (1+\ln \left (x \right ) \left (1+x \right )\right )\right )^{2} \operatorname {csgn}\left (i \left (1+\ln \left (x \right ) \left (1+x \right )\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (1+\ln \left (x \right ) \left (1+x \right )\right )\right ) \operatorname {csgn}\left (i \left (1+\ln \left (x \right ) \left (1+x \right )\right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \left (1+\ln \left (x \right ) \left (1+x \right )\right )^{2}\right )^{3}+2 i x +4 i \ln \left (1+\ln \left (x \right ) \left (1+x \right )\right )\right ) x}\) | \(105\) |
risch | \(-\frac {4 i}{5 x \left (\pi \operatorname {csgn}\left (i \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )\right )^{2} \operatorname {csgn}\left (i \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )\right ) \operatorname {csgn}\left (i \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )^{2}\right )^{3}+2 i x +4 i \ln \left (x \ln \left (x \right )+\ln \left (x \right )+1\right )\right )}\) | \(105\) |
int((((-2-2*x)*ln(x)-2)*ln(1/((x^2+2*x+1)*ln(x)^2+(2+2*x)*ln(x)+1))+(4*x^2 +8*x)*ln(x)+8*x+4)/(((5*x^3+5*x^2)*ln(x)+5*x^2)*ln(1/((x^2+2*x+1)*ln(x)^2+ (2+2*x)*ln(x)+1))^2+((-10*x^4-10*x^3)*ln(x)-10*x^3)*ln(1/((x^2+2*x+1)*ln(x )^2+(2+2*x)*ln(x)+1))+(5*x^5+5*x^4)*ln(x)+5*x^4),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=-\frac {2}{5 \, {\left (x^{2} - x \log \left (\frac {1}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right )^{2} + 2 \, {\left (x + 1\right )} \log \left (x\right ) + 1}\right )\right )}} \]
integrate((((-2-2*x)*log(x)-2)*log(1/((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+ 1))+(4*x^2+8*x)*log(x)+8*x+4)/(((5*x^3+5*x^2)*log(x)+5*x^2)*log(1/((x^2+2* x+1)*log(x)^2+(2+2*x)*log(x)+1))^2+((-10*x^4-10*x^3)*log(x)-10*x^3)*log(1/ ((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+1))+(5*x^5+5*x^4)*log(x)+5*x^4),x, al gorithm=\
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=\frac {2}{- 5 x^{2} + 5 x \log {\left (\frac {1}{\left (2 x + 2\right ) \log {\left (x \right )} + \left (x^{2} + 2 x + 1\right ) \log {\left (x \right )}^{2} + 1} \right )}} \]
integrate((((-2-2*x)*ln(x)-2)*ln(1/((x**2+2*x+1)*ln(x)**2+(2+2*x)*ln(x)+1) )+(4*x**2+8*x)*ln(x)+8*x+4)/(((5*x**3+5*x**2)*ln(x)+5*x**2)*ln(1/((x**2+2* x+1)*ln(x)**2+(2+2*x)*ln(x)+1))**2+((-10*x**4-10*x**3)*ln(x)-10*x**3)*ln(1 /((x**2+2*x+1)*ln(x)**2+(2+2*x)*ln(x)+1))+(5*x**5+5*x**4)*ln(x)+5*x**4),x)
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=-\frac {2}{5 \, {\left (x^{2} + 2 \, x \log \left ({\left (x + 1\right )} \log \left (x\right ) + 1\right )\right )}} \]
integrate((((-2-2*x)*log(x)-2)*log(1/((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+ 1))+(4*x^2+8*x)*log(x)+8*x+4)/(((5*x^3+5*x^2)*log(x)+5*x^2)*log(1/((x^2+2* x+1)*log(x)^2+(2+2*x)*log(x)+1))^2+((-10*x^4-10*x^3)*log(x)-10*x^3)*log(1/ ((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+1))+(5*x^5+5*x^4)*log(x)+5*x^4),x, al gorithm=\
Time = 1.89 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=-\frac {2}{5 \, {\left (x^{2} + x \log \left (x^{2} \log \left (x\right )^{2} + 2 \, x \log \left (x\right )^{2} + 2 \, x \log \left (x\right ) + \log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 1\right )\right )}} \]
integrate((((-2-2*x)*log(x)-2)*log(1/((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+ 1))+(4*x^2+8*x)*log(x)+8*x+4)/(((5*x^3+5*x^2)*log(x)+5*x^2)*log(1/((x^2+2* x+1)*log(x)^2+(2+2*x)*log(x)+1))^2+((-10*x^4-10*x^3)*log(x)-10*x^3)*log(1/ ((x^2+2*x+1)*log(x)^2+(2+2*x)*log(x)+1))+(5*x^5+5*x^4)*log(x)+5*x^4),x, al gorithm=\
Timed out. \[ \int \frac {4+8 x+\left (8 x+4 x^2\right ) \log (x)+(-2+(-2-2 x) \log (x)) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )}{5 x^4+\left (5 x^4+5 x^5\right ) \log (x)+\left (-10 x^3+\left (-10 x^3-10 x^4\right ) \log (x)\right ) \log \left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )+\left (5 x^2+\left (5 x^2+5 x^3\right ) \log (x)\right ) \log ^2\left (\frac {1}{1+(2+2 x) \log (x)+\left (1+2 x+x^2\right ) \log ^2(x)}\right )} \, dx=\int \frac {8\,x-\ln \left (\frac {1}{\left (x^2+2\,x+1\right )\,{\ln \left (x\right )}^2+\left (2\,x+2\right )\,\ln \left (x\right )+1}\right )\,\left (\ln \left (x\right )\,\left (2\,x+2\right )+2\right )+\ln \left (x\right )\,\left (4\,x^2+8\,x\right )+4}{\ln \left (x\right )\,\left (5\,x^5+5\,x^4\right )-\ln \left (\frac {1}{\left (x^2+2\,x+1\right )\,{\ln \left (x\right )}^2+\left (2\,x+2\right )\,\ln \left (x\right )+1}\right )\,\left (\ln \left (x\right )\,\left (10\,x^4+10\,x^3\right )+10\,x^3\right )+{\ln \left (\frac {1}{\left (x^2+2\,x+1\right )\,{\ln \left (x\right )}^2+\left (2\,x+2\right )\,\ln \left (x\right )+1}\right )}^2\,\left (\ln \left (x\right )\,\left (5\,x^3+5\,x^2\right )+5\,x^2\right )+5\,x^4} \,d x \]
int((8*x - log(1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1))*(log(x )*(2*x + 2) + 2) + log(x)*(8*x + 4*x^2) + 4)/(log(x)*(5*x^4 + 5*x^5) - log (1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1))*(log(x)*(10*x^3 + 10 *x^4) + 10*x^3) + log(1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1)) ^2*(log(x)*(5*x^2 + 5*x^3) + 5*x^2) + 5*x^4),x)
int((8*x - log(1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1))*(log(x )*(2*x + 2) + 2) + log(x)*(8*x + 4*x^2) + 4)/(log(x)*(5*x^4 + 5*x^5) - log (1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1))*(log(x)*(10*x^3 + 10 *x^4) + 10*x^3) + log(1/(log(x)*(2*x + 2) + log(x)^2*(2*x + x^2 + 1) + 1)) ^2*(log(x)*(5*x^2 + 5*x^3) + 5*x^2) + 5*x^4), x)