Integrand size = 199, antiderivative size = 30 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log \left (\frac {4+x}{x}+x \left (4-x+\log \left (\frac {1}{x}-x^2+\log (x)\right )\right )\right )} \]
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log \left (1+\frac {4}{x}+4 x-x^2+x \log \left (\frac {1}{x}-x^2+\log (x)\right )\right )} \]
Integrate[(4 - 3*x^2 - 3*x^3 + 6*x^5 - 2*x^6 + (4*x - 4*x^3 + 2*x^4)*Log[x ] + (-x^2 + x^5 - x^3*Log[x])*Log[(1 - x^3 + x*Log[x])/x])/((4*x + x^2 + 4 *x^3 - 5*x^4 - x^5 - 4*x^6 + x^7 + (4*x^2 + x^3 + 4*x^4 - x^5)*Log[x] + (x ^3 - x^6 + x^4*Log[x])*Log[(1 - x^3 + x*Log[x])/x])*Log[(4 + x + 4*x^2 - x ^3 + x^2*Log[(1 - x^3 + x*Log[x])/x])/x]^2),x]
Time = 0.53 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {7237}
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.
\(\displaystyle \int \frac {-2 x^6+6 x^5-3 x^3-3 x^2+\left (2 x^4-4 x^3+4 x\right ) \log (x)+\left (x^5-x^3 \log (x)-x^2\right ) \log \left (\frac {-x^3+x \log (x)+1}{x}\right )+4}{\left (x^7-4 x^6-x^5-5 x^4+4 x^3+x^2+\left (-x^6+x^4 \log (x)+x^3\right ) \log \left (\frac {-x^3+x \log (x)+1}{x}\right )+\left (-x^5+4 x^4+x^3+4 x^2\right ) \log (x)+4 x\right ) \log ^2\left (\frac {-x^3+4 x^2+x^2 \log \left (\frac {-x^3+x \log (x)+1}{x}\right )+x+4}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {1}{\log \left (\frac {-x^3+4 x^2+x^2 \log \left (\frac {-x^3+x \log (x)+1}{x}\right )+x+4}{x}\right )}\) |
Int[(4 - 3*x^2 - 3*x^3 + 6*x^5 - 2*x^6 + (4*x - 4*x^3 + 2*x^4)*Log[x] + (- x^2 + x^5 - x^3*Log[x])*Log[(1 - x^3 + x*Log[x])/x])/((4*x + x^2 + 4*x^3 - 5*x^4 - x^5 - 4*x^6 + x^7 + (4*x^2 + x^3 + 4*x^4 - x^5)*Log[x] + (x^3 - x ^6 + x^4*Log[x])*Log[(1 - x^3 + x*Log[x])/x])*Log[(4 + x + 4*x^2 - x^3 + x ^2*Log[(1 - x^3 + x*Log[x])/x])/x]^2),x]
3.13.43.3.1 Defintions of rubi rules used
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.51 (sec) , antiderivative size = 3007, normalized size of antiderivative = 100.23
\[\text {output too large to display}\]
int(((-x^3*ln(x)+x^5-x^2)*ln((x*ln(x)-x^3+1)/x)+(2*x^4-4*x^3+4*x)*ln(x)-2* x^6+6*x^5-3*x^3-3*x^2+4)/((x^4*ln(x)-x^6+x^3)*ln((x*ln(x)-x^3+1)/x)+(-x^5+ 4*x^4+x^3+4*x^2)*ln(x)+x^7-4*x^6-x^5-5*x^4+4*x^3+x^2+4*x)/ln((x^2*ln((x*ln (x)-x^3+1)/x)-x^3+4*x^2+x+4)/x)^2,x)
-2*I/(-Pi*csgn(-8-2*x+2*x^3-8*x^2-I*x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1 )/x)^2+I*x^2*Pi*csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln(x)-x^3+1)/x)^2+2*I*x^ 2*Pi*csgn(I*(x*ln(x)-x^3+1)/x)^2+I*x^2*Pi*csgn(I*(x*ln(x)-x^3+1)/x)^3-2*x^ 2*ln(-x*ln(x)+x^3-1)+I*x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x* ln(x)-x^3+1)/x)+2*x^2*ln(x)-2*I*Pi*x^2)*csgn(I/x*(-8*I-2*I*x+2*I*x^3-8*I*x ^2+x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1)/x)^2-x^2*Pi*csgn(I*(x*ln(x)-x^3 +1))*csgn(I*(x*ln(x)-x^3+1)/x)^2-2*x^2*Pi*csgn(I*(x*ln(x)-x^3+1)/x)^2-x^2* Pi*csgn(I*(x*ln(x)-x^3+1)/x)^3-2*I*x^2*ln(-x*ln(x)+x^3-1)-x^2*Pi*csgn(I/x) *csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln(x)-x^3+1)/x)+2*I*x^2*ln(x)+2*Pi*x^2) )^2+Pi*csgn(-8-2*x+2*x^3-8*x^2-I*x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1)/x )^2+I*x^2*Pi*csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln(x)-x^3+1)/x)^2+2*I*x^2*P i*csgn(I*(x*ln(x)-x^3+1)/x)^2+I*x^2*Pi*csgn(I*(x*ln(x)-x^3+1)/x)^3-2*x^2*l n(-x*ln(x)+x^3-1)+I*x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln( x)-x^3+1)/x)+2*x^2*ln(x)-2*I*Pi*x^2)*csgn(I/x*(-8*I-2*I*x+2*I*x^3-8*I*x^2+ x^2*Pi*csgn(I/x)*csgn(I*(x*ln(x)-x^3+1)/x)^2-x^2*Pi*csgn(I*(x*ln(x)-x^3+1) )*csgn(I*(x*ln(x)-x^3+1)/x)^2-2*x^2*Pi*csgn(I*(x*ln(x)-x^3+1)/x)^2-x^2*Pi* csgn(I*(x*ln(x)-x^3+1)/x)^3-2*I*x^2*ln(-x*ln(x)+x^3-1)-x^2*Pi*csgn(I/x)*cs gn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln(x)-x^3+1)/x)+2*I*x^2*ln(x)+2*Pi*x^2))*c sgn(I/x)-Pi*csgn(I/x*(-8*I-2*I*x+2*I*x^3-8*I*x^2+x^2*Pi*csgn(I/x)*csgn(I*( x*ln(x)-x^3+1)/x)^2-x^2*Pi*csgn(I*(x*ln(x)-x^3+1))*csgn(I*(x*ln(x)-x^3+...
Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log \left (-\frac {x^{3} - x^{2} \log \left (-\frac {x^{3} - x \log \left (x\right ) - 1}{x}\right ) - 4 \, x^{2} - x - 4}{x}\right )} \]
integrate(((-x^3*log(x)+x^5-x^2)*log((x*log(x)-x^3+1)/x)+(2*x^4-4*x^3+4*x) *log(x)-2*x^6+6*x^5-3*x^3-3*x^2+4)/((x^4*log(x)-x^6+x^3)*log((x*log(x)-x^3 +1)/x)+(-x^5+4*x^4+x^3+4*x^2)*log(x)+x^7-4*x^6-x^5-5*x^4+4*x^3+x^2+4*x)/lo g((x^2*log((x*log(x)-x^3+1)/x)-x^3+4*x^2+x+4)/x)^2,x, algorithm=\
Time = 159.82 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log {\left (\frac {- x^{3} + x^{2} \log {\left (\frac {- x^{3} + x \log {\left (x \right )} + 1}{x} \right )} + 4 x^{2} + x + 4}{x} \right )}} \]
integrate(((-x**3*ln(x)+x**5-x**2)*ln((x*ln(x)-x**3+1)/x)+(2*x**4-4*x**3+4 *x)*ln(x)-2*x**6+6*x**5-3*x**3-3*x**2+4)/((x**4*ln(x)-x**6+x**3)*ln((x*ln( x)-x**3+1)/x)+(-x**5+4*x**4+x**3+4*x**2)*ln(x)+x**7-4*x**6-x**5-5*x**4+4*x **3+x**2+4*x)/ln((x**2*ln((x*ln(x)-x**3+1)/x)-x**3+4*x**2+x+4)/x)**2,x)
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log \left (-x^{3} + x^{2} {\left (\log \left (-x^{3} + x \log \left (x\right ) + 1\right ) + 4\right )} - x^{2} \log \left (x\right ) + x + 4\right ) - \log \left (x\right )} \]
integrate(((-x^3*log(x)+x^5-x^2)*log((x*log(x)-x^3+1)/x)+(2*x^4-4*x^3+4*x) *log(x)-2*x^6+6*x^5-3*x^3-3*x^2+4)/((x^4*log(x)-x^6+x^3)*log((x*log(x)-x^3 +1)/x)+(-x^5+4*x^4+x^3+4*x^2)*log(x)+x^7-4*x^6-x^5-5*x^4+4*x^3+x^2+4*x)/lo g((x^2*log((x*log(x)-x^3+1)/x)-x^3+4*x^2+x+4)/x)^2,x, algorithm=\
Time = 0.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\log \left (-x^{3} + x^{2} \log \left (-x^{3} + x \log \left (x\right ) + 1\right ) - x^{2} \log \left (x\right ) + 4 \, x^{2} + x + 4\right ) - \log \left (x\right )} \]
integrate(((-x^3*log(x)+x^5-x^2)*log((x*log(x)-x^3+1)/x)+(2*x^4-4*x^3+4*x) *log(x)-2*x^6+6*x^5-3*x^3-3*x^2+4)/((x^4*log(x)-x^6+x^3)*log((x*log(x)-x^3 +1)/x)+(-x^5+4*x^4+x^3+4*x^2)*log(x)+x^7-4*x^6-x^5-5*x^4+4*x^3+x^2+4*x)/lo g((x^2*log((x*log(x)-x^3+1)/x)-x^3+4*x^2+x+4)/x)^2,x, algorithm=\
Time = 10.83 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {4-3 x^2-3 x^3+6 x^5-2 x^6+\left (4 x-4 x^3+2 x^4\right ) \log (x)+\left (-x^2+x^5-x^3 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{\left (4 x+x^2+4 x^3-5 x^4-x^5-4 x^6+x^7+\left (4 x^2+x^3+4 x^4-x^5\right ) \log (x)+\left (x^3-x^6+x^4 \log (x)\right ) \log \left (\frac {1-x^3+x \log (x)}{x}\right )\right ) \log ^2\left (\frac {4+x+4 x^2-x^3+x^2 \log \left (\frac {1-x^3+x \log (x)}{x}\right )}{x}\right )} \, dx=\frac {1}{\ln \left (4\,x+x\,\ln \left (\frac {x\,\ln \left (x\right )-x^3+1}{x}\right )+\frac {4}{x}-x^2+1\right )} \]
int(-(3*x^2 + 3*x^3 - 6*x^5 + 2*x^6 + log((x*log(x) - x^3 + 1)/x)*(x^3*log (x) + x^2 - x^5) - log(x)*(4*x - 4*x^3 + 2*x^4) - 4)/(log((x + x^2*log((x* log(x) - x^3 + 1)/x) + 4*x^2 - x^3 + 4)/x)^2*(4*x + log(x)*(4*x^2 + x^3 + 4*x^4 - x^5) + x^2 + 4*x^3 - 5*x^4 - x^5 - 4*x^6 + x^7 + log((x*log(x) - x ^3 + 1)/x)*(x^4*log(x) + x^3 - x^6))),x)