\(\int \frac {e (2+2 x-5 x^2+9 x^3)+e (-4 x^2+6 x^3) \log (x)+e (-x^2+x^3) \log ^2(x)+(e (-4 x+10 x^3)+8 e x^3 \log (x)+2 e x^3 \log ^2(x)) \log (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x})}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx\) [2047]

Optimal result

Integrand size = 132, antiderivative size = 29 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=e \left (-x+x^2 \log \left (\frac {2}{x}-x-x (2+\log (x))^2\right )\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=e \left (-x+x^2 \log \left (\frac {2}{x}-5 x-4 x \log (x)-x \log ^2(x)\right )\right ) \] Input:

Integrate[(E*(2 + 2*x - 5*x^2 + 9*x^3) + E*(-4*x^2 + 6*x^3)*Log[x] + E*(-x 
^2 + x^3)*Log[x]^2 + (E*(-4*x + 10*x^3) + 8*E*x^3*Log[x] + 2*E*x^3*Log[x]^ 
2)*Log[(2 - 5*x^2 - 4*x^2*Log[x] - x^2*Log[x]^2)/x])/(-2 + 5*x^2 + 4*x^2*L 
og[x] + x^2*Log[x]^2),x]
 

Output:

E*(-x + x^2*Log[2/x - 5*x - 4*x*Log[x] - x*Log[x]^2])
 

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.

\(\displaystyle \int \frac {e \left (9 x^3-5 x^2+2 x+2\right )+e \left (x^3-x^2\right ) \log ^2(x)+\left (e \left (10 x^3-4 x\right )+2 e x^3 \log ^2(x)+8 e x^3 \log (x)\right ) \log \left (\frac {-5 x^2+x^2 \left (-\log ^2(x)\right )-4 x^2 \log (x)+2}{x}\right )+e \left (6 x^3-4 x^2\right ) \log (x)}{5 x^2+x^2 \log ^2(x)+4 x^2 \log (x)-2} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e \left (9 x^3+x^3 \log ^2(x)+6 x^3 \log (x)-5 x^2-x^2 \log ^2(x)-4 x^2 \log (x)+2 x+2\right )}{5 x^2+x^2 \log ^2(x)+4 x^2 \log (x)-2}+2 e x \log \left (-5 x+\frac {2}{x}-x \log ^2(x)-4 x \log (x)\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 e \int \frac {x}{\log ^2(x) x^2+4 \log (x) x^2+5 x^2-2}dx+4 e \int \frac {x^3}{\log ^2(x) x^2+4 \log (x) x^2+5 x^2-2}dx+2 e \int \frac {x^3 \log (x)}{\log ^2(x) x^2+4 \log (x) x^2+5 x^2-2}dx+2 e \int x \log \left (-x \log ^2(x)-4 x \log (x)-5 x+\frac {2}{x}\right )dx+\frac {e x^2}{2}-e x\)

Input:

Int[(E*(2 + 2*x - 5*x^2 + 9*x^3) + E*(-4*x^2 + 6*x^3)*Log[x] + E*(-x^2 + x 
^3)*Log[x]^2 + (E*(-4*x + 10*x^3) + 8*E*x^3*Log[x] + 2*E*x^3*Log[x]^2)*Log 
[(2 - 5*x^2 - 4*x^2*Log[x] - x^2*Log[x]^2)/x])/(-2 + 5*x^2 + 4*x^2*Log[x] 
+ x^2*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 8.97 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41

method	result	size
parallelrisch	\(x^{2} \ln \left (-\frac {x^{2} \ln \left (x \right )^{2}+4 x^{2} \ln \left (x \right )+5 x^{2}-2}{x}\right ) {\mathrm e}-x \,{\mathrm e}\)	\(41\)
risch	\(x^{2} {\mathrm e} \ln \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )-x^{2} {\mathrm e} \ln \left (x \right )-\frac {i \pi \,{\mathrm e} x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )}{x}\right )}{2}+\frac {i \pi \,{\mathrm e} x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )}{x}\right )}^{2}}{2}+\frac {i \pi \,{\mathrm e} x^{2} \operatorname {csgn}\left (i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )}{x}\right )}^{2}}{2}+\frac {i \pi \,{\mathrm e} x^{2} {\operatorname {csgn}\left (\frac {i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )}{x}\right )}^{3}}{2}-i \pi \,{\mathrm e} x^{2} {\operatorname {csgn}\left (\frac {i \left (-2+\left (\ln \left (x \right )^{2}+4 \ln \left (x \right )+5\right ) x^{2}\right )}{x}\right )}^{2}+i {\mathrm e} \pi \,x^{2}-x \,{\mathrm e}\)	\(270\)

Input:

int(((2*x^3*exp(1)*ln(x)^2+8*x^3*exp(1)*ln(x)+(10*x^3-4*x)*exp(1))*ln((-x^ 
2*ln(x)^2-4*x^2*ln(x)-5*x^2+2)/x)+(x^3-x^2)*exp(1)*ln(x)^2+(6*x^3-4*x^2)*e 
xp(1)*ln(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*ln(x)^2+4*x^2*ln(x)+5*x^2-2), 
x,method=_RETURNVERBOSE)
 

Output:

x^2*ln(-(x^2*ln(x)^2+4*x^2*ln(x)+5*x^2-2)/x)*exp(1)-x*exp(1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x^{2} e \log \left (-\frac {x^{2} \log \left (x\right )^{2} + 4 \, x^{2} \log \left (x\right ) + 5 \, x^{2} - 2}{x}\right ) - x e \] Input:

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1)) 
*log((-x^2*log(x)^2-4*x^2*log(x)-5*x^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6* 
x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2*l 
og(x)+5*x^2-2),x, algorithm="fricas")
 

Output:

x^2*e*log(-(x^2*log(x)^2 + 4*x^2*log(x) + 5*x^2 - 2)/x) - x*e
 

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=e x^{2} \log {\left (\frac {- x^{2} \log {\left (x \right )}^{2} - 4 x^{2} \log {\left (x \right )} - 5 x^{2} + 2}{x} \right )} - e x \] Input:

integrate(((2*x**3*exp(1)*ln(x)**2+8*x**3*exp(1)*ln(x)+(10*x**3-4*x)*exp(1 
))*ln((-x**2*ln(x)**2-4*x**2*ln(x)-5*x**2+2)/x)+(x**3-x**2)*exp(1)*ln(x)** 
2+(6*x**3-4*x**2)*exp(1)*ln(x)+(9*x**3-5*x**2+2*x+2)*exp(1))/(x**2*ln(x)** 
2+4*x**2*ln(x)+5*x**2-2),x)
 

Output:

E*x**2*log((-x**2*log(x)**2 - 4*x**2*log(x) - 5*x**2 + 2)/x) - E*x
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x^{2} e \log \left (-x^{2} \log \left (x\right )^{2} - 4 \, x^{2} \log \left (x\right ) - 5 \, x^{2} + 2\right ) - x^{2} e \log \left (x\right ) - x e \] Input:

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1)) 
*log((-x^2*log(x)^2-4*x^2*log(x)-5*x^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6* 
x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2*l 
og(x)+5*x^2-2),x, algorithm="maxima")
 

Output:

x^2*e*log(-x^2*log(x)^2 - 4*x^2*log(x) - 5*x^2 + 2) - x^2*e*log(x) - x*e
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x^{2} e \log \left (-x^{2} \log \left (x\right )^{2} - 4 \, x^{2} \log \left (x\right ) - 5 \, x^{2} + 2\right ) - x^{2} e \log \left (x\right ) - x e \] Input:

integrate(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1)) 
*log((-x^2*log(x)^2-4*x^2*log(x)-5*x^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6* 
x^3-4*x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2*l 
og(x)+5*x^2-2),x, algorithm="giac")
 

Output:

x^2*e*log(-x^2*log(x)^2 - 4*x^2*log(x) - 5*x^2 + 2) - x^2*e*log(x) - x*e
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x^2\,\mathrm {e}\,\ln \left (-\frac {x^2\,{\ln \left (x\right )}^2+4\,x^2\,\ln \left (x\right )+5\,x^2-2}{x}\right )-x\,\mathrm {e} \] Input:

int((log(-(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2)/x)*(8*x^3*exp(1)*log(x 
) - exp(1)*(4*x - 10*x^3) + 2*x^3*exp(1)*log(x)^2) + exp(1)*(2*x - 5*x^2 + 
 9*x^3 + 2) - exp(1)*log(x)*(4*x^2 - 6*x^3) - exp(1)*log(x)^2*(x^2 - x^3)) 
/(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2),x)
 

Output:

x^2*exp(1)*log(-(4*x^2*log(x) + x^2*log(x)^2 + 5*x^2 - 2)/x) - x*exp(1)
 

Reduce [F]

\[ \int \frac {e \left (2+2 x-5 x^2+9 x^3\right )+e \left (-4 x^2+6 x^3\right ) \log (x)+e \left (-x^2+x^3\right ) \log ^2(x)+\left (e \left (-4 x+10 x^3\right )+8 e x^3 \log (x)+2 e x^3 \log ^2(x)\right ) \log \left (\frac {2-5 x^2-4 x^2 \log (x)-x^2 \log ^2(x)}{x}\right )}{-2+5 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx =\text {Too large to display} \] Input:

int(((2*x^3*exp(1)*log(x)^2+8*x^3*exp(1)*log(x)+(10*x^3-4*x)*exp(1))*log(( 
-x^2*log(x)^2-4*x^2*log(x)-5*x^2+2)/x)+(x^3-x^2)*exp(1)*log(x)^2+(6*x^3-4* 
x^2)*exp(1)*log(x)+(9*x^3-5*x^2+2*x+2)*exp(1))/(x^2*log(x)^2+4*x^2*log(x)+ 
5*x^2-2),x)
 

Output:

e*(9*int(x**3/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x) - 5*int(x** 
2/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x) + int((log(x)**2*x**3)/ 
(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x) - int((log(x)**2*x**2)/(l 
og(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x) + 2*int((log(( - log(x)**2* 
x**2 - 4*log(x)*x**2 - 5*x**2 + 2)/x)*log(x)**2*x**3)/(log(x)**2*x**2 + 4* 
log(x)*x**2 + 5*x**2 - 2),x) + 10*int((log(( - log(x)**2*x**2 - 4*log(x)*x 
**2 - 5*x**2 + 2)/x)*x**3)/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x 
) + 8*int((log(( - log(x)**2*x**2 - 4*log(x)*x**2 - 5*x**2 + 2)/x)*log(x)* 
x**3)/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x) - 4*int((log(( - lo 
g(x)**2*x**2 - 4*log(x)*x**2 - 5*x**2 + 2)/x)*x)/(log(x)**2*x**2 + 4*log(x 
)*x**2 + 5*x**2 - 2),x) + 6*int((log(x)*x**3)/(log(x)**2*x**2 + 4*log(x)*x 
**2 + 5*x**2 - 2),x) - 4*int((log(x)*x**2)/(log(x)**2*x**2 + 4*log(x)*x**2 
 + 5*x**2 - 2),x) + 2*int(x/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2), 
x) + 2*int(1/(log(x)**2*x**2 + 4*log(x)*x**2 + 5*x**2 - 2),x))