\(\int \frac {-2 x^2+e^6 (-x+x^2)+e^3 (-2 x^2+x^3)+(x+e^3 x+x^2-3 x^3+x^4) \log (x)+2 x^2 \log ^2(x)+(e^6+x+e^3 x+(-e^6-x-e^3 x+x^2) \log (x)-x^2 \log ^2(x)) \log (e^6+x+e^3 x+x^2 \log (x))}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx\) [1019]

Optimal result

Integrand size = 145, antiderivative size = 30 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=\frac {(-x+\log (x)) \left (-x+\log \left (e^6+x+x \left (e^3+x \log (x)\right )\right )\right )}{x} \] Output:

(ln(exp(3)^2+(exp(3)+x*ln(x))*x+x)-x)/x*(ln(x)-x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=x-\log (x)-\log \left (e^6+x+e^3 x+x^2 \log (x)\right )+\frac {\log (x) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{x} \] Input:

Integrate[(-2*x^2 + E^6*(-x + x^2) + E^3*(-2*x^2 + x^3) + (x + E^3*x + x^2 
 - 3*x^3 + x^4)*Log[x] + 2*x^2*Log[x]^2 + (E^6 + x + E^3*x + (-E^6 - x - E 
^3*x + x^2)*Log[x] - x^2*Log[x]^2)*Log[E^6 + x + E^3*x + x^2*Log[x]])/(E^6 
*x^2 + x^3 + E^3*x^3 + x^4*Log[x]),x]
 

Output:

x - Log[x] - Log[E^6 + x + E^3*x + x^2*Log[x]] + (Log[x]*Log[E^6 + x + E^3 
*x + x^2*Log[x]])/x
 

Rubi [F]

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.

\(\displaystyle \int \frac {-2 x^2+e^6 \left (x^2-x\right )+2 x^2 \log ^2(x)+\left (-x^2 \log ^2(x)+\left (x^2-e^3 x-x-e^6\right ) \log (x)+e^3 x+x+e^6\right ) \log \left (x^2 \log (x)+e^3 x+x+e^6\right )+e^3 \left (x^3-2 x^2\right )+\left (x^4-3 x^3+x^2+e^3 x+x\right ) \log (x)}{x^4 \log (x)+e^3 x^3+x^3+e^6 x^2} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-2 x^2+e^6 \left (x^2-x\right )+2 x^2 \log ^2(x)+\left (-x^2 \log ^2(x)+\left (x^2-e^3 x-x-e^6\right ) \log (x)+e^3 x+x+e^6\right ) \log \left (x^2 \log (x)+e^3 x+x+e^6\right )+e^3 \left (x^3-2 x^2\right )+\left (x^4-3 x^3+x^2+e^3 x+x\right ) \log (x)}{x^4 \log (x)+\left (1+e^3\right ) x^3+e^6 x^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {2 \log ^2(x)}{x^2 \log (x)+\left (1+e^3\right ) x+e^6}+\frac {(1-\log (x)) \log \left (x^2 \log (x)+\left (1+e^3\right ) x+e^6\right )}{x^2}+\frac {2}{x^2 (-\log (x))-\left (1+e^3\right ) x-e^6}+\frac {e^3 (x-2)}{x^2 \log (x)+\left (1+e^3\right ) x+e^6}+\frac {e^6 (x-1)}{x \left (x^2 \log (x)+\left (1+e^3\right ) x+e^6\right )}+\frac {\left (x^3-3 x^2+x+e^3+1\right ) \log (x)}{x \left (x^2 \log (x)+\left (1+e^3\right ) x+e^6\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\left (3+3 e^3-e^6\right ) \int \frac {1}{-\log (x) x^2-\left (1+e^3\right ) x-e^6}dx+2 e^3 \int \frac {1}{-\log (x) x^2-\left (1+e^3\right ) x-e^6}dx+2 \int \frac {1}{-\log (x) x^2-\left (1+e^3\right ) x-e^6}dx+e^6 \int \frac {1}{x \left (-\log (x) x^2-\left (1+e^3\right ) x-e^6\right )}dx+e^6 \int \frac {1}{\log (x) x^2+\left (1+e^3\right ) x+e^6}dx-\left (1+2 e^3+2 e^6\right ) \int \frac {1}{x^2 \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}dx+2 \left (1+e^3\right )^2 \int \frac {1}{x^2 \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}dx-\left (1+e^3-3 e^6\right ) \int \frac {1}{x \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}dx-\left (1+e^3\right ) \int \frac {x}{\log (x) x^2+\left (1+e^3\right ) x+e^6}dx+e^3 \int \frac {x}{\log (x) x^2+\left (1+e^3\right ) x+e^6}dx+\int \frac {\log \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}{x^2}dx-\int \frac {\log (x) \log \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}{x^2}dx+2 e^{12} \int \frac {1}{x^4 \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}dx+3 e^6 \left (1+e^3\right ) \int \frac {1}{x^3 \left (\log (x) x^2+\left (1+e^3\right ) x+e^6\right )}dx+\frac {2 e^6}{3 x^3}+\frac {1+e^3}{2 x^2}+x-\frac {3}{x}-3 \log (x)-\frac {2 \log (x)}{x}\)

Input:

Int[(-2*x^2 + E^6*(-x + x^2) + E^3*(-2*x^2 + x^3) + (x + E^3*x + x^2 - 3*x 
^3 + x^4)*Log[x] + 2*x^2*Log[x]^2 + (E^6 + x + E^3*x + (-E^6 - x - E^3*x + 
 x^2)*Log[x] - x^2*Log[x]^2)*Log[E^6 + x + E^3*x + x^2*Log[x]])/(E^6*x^2 + 
 x^3 + E^3*x^3 + x^4*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53

method	result	size
risch	\(\frac {\ln \left (x \right ) \ln \left (x^{2} \ln \left (x \right )+{\mathrm e}^{6}+x \,{\mathrm e}^{3}+x \right )}{x}+x -3 \ln \left (x \right )-\ln \left (\ln \left (x \right )+\frac {x \,{\mathrm e}^{3}+{\mathrm e}^{6}+x}{x^{2}}\right )\)	\(46\)
parallelrisch	\(\frac {x^{2}-x \ln \left (x \right )-\ln \left (x^{2} \ln \left (x \right )+{\mathrm e}^{6}+x \,{\mathrm e}^{3}+x \right ) x +\ln \left (x \right ) \ln \left (x^{2} \ln \left (x \right )+{\mathrm e}^{6}+x \,{\mathrm e}^{3}+x \right )}{x}\)	\(54\)

Input:

int(((-x^2*ln(x)^2+(-exp(3)^2-x*exp(3)+x^2-x)*ln(x)+exp(3)^2+x*exp(3)+x)*l 
n(x^2*ln(x)+exp(3)^2+x*exp(3)+x)+2*x^2*ln(x)^2+(x*exp(3)+x^4-3*x^3+x^2+x)* 
ln(x)+(x^2-x)*exp(3)^2+(x^3-2*x^2)*exp(3)-2*x^2)/(x^4*ln(x)+x^2*exp(3)^2+x 
^3*exp(3)+x^3),x,method=_RETURNVERBOSE)
 

Output:

1/x*ln(x)*ln(x^2*ln(x)+exp(6)+x*exp(3)+x)+x-3*ln(x)-ln(ln(x)+(x*exp(3)+exp 
(6)+x)/x^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=\frac {x^{2} - {\left (x - \log \left (x\right )\right )} \log \left (x^{2} \log \left (x\right ) + x e^{3} + x + e^{6}\right ) - x \log \left (x\right )}{x} \] Input:

integrate(((-x^2*log(x)^2+(-exp(3)^2-x*exp(3)+x^2-x)*log(x)+exp(3)^2+x*exp 
(3)+x)*log(x^2*log(x)+exp(3)^2+x*exp(3)+x)+2*x^2*log(x)^2+(x*exp(3)+x^4-3* 
x^3+x^2+x)*log(x)+(x^2-x)*exp(3)^2+(x^3-2*x^2)*exp(3)-2*x^2)/(x^4*log(x)+x 
^2*exp(3)^2+x^3*exp(3)+x^3),x, algorithm="fricas")
 

Output:

(x^2 - (x - log(x))*log(x^2*log(x) + x*e^3 + x + e^6) - x*log(x))/x
 

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=x - 3 \log {\left (x \right )} - \log {\left (\log {\left (x \right )} + \frac {x + x e^{3} + e^{6}}{x^{2}} \right )} + \frac {\log {\left (x \right )} \log {\left (x^{2} \log {\left (x \right )} + x + x e^{3} + e^{6} \right )}}{x} \] Input:

integrate(((-x**2*ln(x)**2+(-exp(3)**2-x*exp(3)+x**2-x)*ln(x)+exp(3)**2+x* 
exp(3)+x)*ln(x**2*ln(x)+exp(3)**2+x*exp(3)+x)+2*x**2*ln(x)**2+(x*exp(3)+x* 
*4-3*x**3+x**2+x)*ln(x)+(x**2-x)*exp(3)**2+(x**3-2*x**2)*exp(3)-2*x**2)/(x 
**4*ln(x)+x**2*exp(3)**2+x**3*exp(3)+x**3),x)
 

Output:

x - 3*log(x) - log(log(x) + (x + x*exp(3) + exp(6))/x**2) + log(x)*log(x** 
2*log(x) + x + x*exp(3) + exp(6))/x
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=\frac {x^{2} + \log \left (x^{2} \log \left (x\right ) + x {\left (e^{3} + 1\right )} + e^{6}\right ) \log \left (x\right )}{x} - 3 \, \log \left (x\right ) - \log \left (\frac {x^{2} \log \left (x\right ) + x {\left (e^{3} + 1\right )} + e^{6}}{x^{2}}\right ) \] Input:

integrate(((-x^2*log(x)^2+(-exp(3)^2-x*exp(3)+x^2-x)*log(x)+exp(3)^2+x*exp 
(3)+x)*log(x^2*log(x)+exp(3)^2+x*exp(3)+x)+2*x^2*log(x)^2+(x*exp(3)+x^4-3* 
x^3+x^2+x)*log(x)+(x^2-x)*exp(3)^2+(x^3-2*x^2)*exp(3)-2*x^2)/(x^4*log(x)+x 
^2*exp(3)^2+x^3*exp(3)+x^3),x, algorithm="maxima")
 

Output:

(x^2 + log(x^2*log(x) + x*(e^3 + 1) + e^6)*log(x))/x - 3*log(x) - log((x^2 
*log(x) + x*(e^3 + 1) + e^6)/x^2)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=\frac {x^{2} - x \log \left (-x^{2} \log \left (x\right ) - x e^{3} - x - e^{6}\right ) - x \log \left (x\right ) + \log \left (x^{2} \log \left (x\right ) + x e^{3} + x + e^{6}\right ) \log \left (x\right )}{x} \] Input:

integrate(((-x^2*log(x)^2+(-exp(3)^2-x*exp(3)+x^2-x)*log(x)+exp(3)^2+x*exp 
(3)+x)*log(x^2*log(x)+exp(3)^2+x*exp(3)+x)+2*x^2*log(x)^2+(x*exp(3)+x^4-3* 
x^3+x^2+x)*log(x)+(x^2-x)*exp(3)^2+(x^3-2*x^2)*exp(3)-2*x^2)/(x^4*log(x)+x 
^2*exp(3)^2+x^3*exp(3)+x^3),x, algorithm="giac")
 

Output:

(x^2 - x*log(-x^2*log(x) - x*e^3 - x - e^6) - x*log(x) + log(x^2*log(x) + 
x*e^3 + x + e^6)*log(x))/x
 

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=x-\ln \left (x+{\mathrm {e}}^6+x^2\,\ln \left (x\right )+x\,{\mathrm {e}}^3\right )-\ln \left (\frac {1}{x^2}\right )-3\,\ln \left (x\right )+\frac {\ln \left (x+{\mathrm {e}}^6+x^2\,\ln \left (x\right )+x\,{\mathrm {e}}^3\right )\,\ln \left (x\right )}{x} \] Input:

int((log(x)*(x + x*exp(3) + x^2 - 3*x^3 + x^4) + log(x + exp(6) + x^2*log( 
x) + x*exp(3))*(x + exp(6) + x*exp(3) - x^2*log(x)^2 - log(x)*(x + exp(6) 
+ x*exp(3) - x^2)) + 2*x^2*log(x)^2 - exp(3)*(2*x^2 - x^3) - exp(6)*(x - x 
^2) - 2*x^2)/(x^4*log(x) + x^3*exp(3) + x^2*exp(6) + x^3),x)
 

Output:

x - log(x + exp(6) + x^2*log(x) + x*exp(3)) - log(1/x^2) - 3*log(x) + (log 
(x + exp(6) + x^2*log(x) + x*exp(3))*log(x))/x
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {-2 x^2+e^6 \left (-x+x^2\right )+e^3 \left (-2 x^2+x^3\right )+\left (x+e^3 x+x^2-3 x^3+x^4\right ) \log (x)+2 x^2 \log ^2(x)+\left (e^6+x+e^3 x+\left (-e^6-x-e^3 x+x^2\right ) \log (x)-x^2 \log ^2(x)\right ) \log \left (e^6+x+e^3 x+x^2 \log (x)\right )}{e^6 x^2+x^3+e^3 x^3+x^4 \log (x)} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (x \right ) x^{2}+e^{6}+e^{3} x +x \right ) \mathrm {log}\left (x \right )-\mathrm {log}\left (\mathrm {log}\left (x \right ) x^{2}+e^{6}+e^{3} x +x \right ) x -\mathrm {log}\left (x \right ) x +x^{2}}{x} \] Input:

int(((-x^2*log(x)^2+(-exp(3)^2-x*exp(3)+x^2-x)*log(x)+exp(3)^2+x*exp(3)+x) 
*log(x^2*log(x)+exp(3)^2+x*exp(3)+x)+2*x^2*log(x)^2+(x*exp(3)+x^4-3*x^3+x^ 
2+x)*log(x)+(x^2-x)*exp(3)^2+(x^3-2*x^2)*exp(3)-2*x^2)/(x^4*log(x)+x^2*exp 
(3)^2+x^3*exp(3)+x^3),x)
 

Output:

(log(log(x)*x**2 + e**6 + e**3*x + x)*log(x) - log(log(x)*x**2 + e**6 + e* 
*3*x + x)*x - log(x)*x + x**2)/x