\(\int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x))}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 (-44 x^2+8 x^3)+(44 x-8 e^2 x-8 x^2) \log (x)+4 \log ^2(x)} \, dx\) [152]

Optimal result

Integrand size = 125, antiderivative size = 33 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=-\frac {4}{e^2}+e^{\frac {2 x^2}{\frac {11}{2}-e^2-x+\frac {\log (x)}{x}}} \] Output:

exp(x^2/(ln(x)/x+11/2-x-exp(2)))^2-4/exp(2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \] Input:

Integrate[(E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))*(-8*x^2 + 88*x^ 
3 - 16*E^2*x^3 - 8*x^4 + 24*x^2*Log[x]))/(121*x^2 + 4*E^4*x^2 - 44*x^3 + 4 
*x^4 + E^2*(-44*x^2 + 8*x^3) + (44*x - 8*E^2*x - 8*x^2)*Log[x] + 4*Log[x]^ 
2),x]
 

Output:

E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))
 

Rubi [A] (verified)

Time = 1.83 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6, 6, 7239, 27, 25, 7257}

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.

\(\displaystyle \int \frac {e^{\frac {4 x^3}{-2 x^2-2 e^2 x+11 x+2 \log (x)}} \left (-8 x^4-16 e^2 x^3+88 x^3-8 x^2+24 x^2 \log (x)\right )}{4 x^4-44 x^3+4 e^4 x^2+121 x^2+\left (-8 x^2-8 e^2 x+44 x\right ) \log (x)+e^2 \left (8 x^3-44 x^2\right )+4 \log ^2(x)} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^{\frac {4 x^3}{-2 x^2-2 e^2 x+11 x+2 \log (x)}} \left (-8 x^4+\left (88-16 e^2\right ) x^3-8 x^2+24 x^2 \log (x)\right )}{4 x^4-44 x^3+4 e^4 x^2+121 x^2+\left (-8 x^2-8 e^2 x+44 x\right ) \log (x)+e^2 \left (8 x^3-44 x^2\right )+4 \log ^2(x)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^{\frac {4 x^3}{-2 x^2-2 e^2 x+11 x+2 \log (x)}} \left (-8 x^4+\left (88-16 e^2\right ) x^3-8 x^2+24 x^2 \log (x)\right )}{4 x^4-44 x^3+\left (121+4 e^4\right ) x^2+\left (-8 x^2-8 e^2 x+44 x\right ) \log (x)+e^2 \left (8 x^3-44 x^2\right )+4 \log ^2(x)}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {8 x^2 e^{-\frac {4 x^3}{x \left (2 x+2 e^2-11\right )-2 \log (x)}} \left (-x^2-\left (2 e^2-11\right ) x+3 \log (x)-1\right )}{\left (x \left (2 x+2 e^2-11\right )-2 \log (x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 8 \int -\frac {e^{\frac {4 x^3}{\left (-2 x-2 e^2+11\right ) x+2 \log (x)}} x^2 \left (x^2-\left (11-2 e^2\right ) x-3 \log (x)+1\right )}{\left (\left (-2 x-2 e^2+11\right ) x+2 \log (x)\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -8 \int \frac {e^{\frac {4 x^3}{\left (-2 x-2 e^2+11\right ) x+2 \log (x)}} x^2 \left (x^2-\left (11-2 e^2\right ) x-3 \log (x)+1\right )}{\left (\left (-2 x-2 e^2+11\right ) x+2 \log (x)\right )^2}dx\)

\(\Big \downarrow \) 7257

\(\displaystyle \exp \left (\frac {4 x^3}{\left (-2 x-2 e^2+11\right ) x+2 \log (x)}\right )\)

Input:

Int[(E^((4*x^3)/(11*x - 2*E^2*x - 2*x^2 + 2*Log[x]))*(-8*x^2 + 88*x^3 - 16 
*E^2*x^3 - 8*x^4 + 24*x^2*Log[x]))/(121*x^2 + 4*E^4*x^2 - 44*x^3 + 4*x^4 + 
 E^2*(-44*x^2 + 8*x^3) + (44*x - 8*E^2*x - 8*x^2)*Log[x] + 4*Log[x]^2),x]
 

Output:

E^((4*x^3)/((11 - 2*E^2 - 2*x)*x + 2*Log[x]))
 

Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7257

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim 
p[q*(F^v/Log[F]), x] /;  !FalseQ[q]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 7.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82

method	result	size
risch	\({\mathrm e}^{-\frac {4 x^{3}}{2 \,{\mathrm e}^{2} x +2 x^{2}-2 \ln \left (x \right )-11 x}}\)	\(27\)
parallelrisch	\({\mathrm e}^{-\frac {4 x^{3}}{2 \,{\mathrm e}^{2} x +2 x^{2}-2 \ln \left (x \right )-11 x}}\)	\(29\)

Input:

int((24*x^2*ln(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*ln(x)-2*e 
xp(2)*x-2*x^2+11*x))^2/(4*ln(x)^2+(-8*exp(2)*x-8*x^2+44*x)*ln(x)+4*x^2*exp 
(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x,method=_RETURNVERBOSE)
 

Output:

exp(-4*x^3/(2*exp(2)*x+2*x^2-2*ln(x)-11*x))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=e^{\left (-\frac {4 \, x^{3}}{2 \, x^{2} + 2 \, x e^{2} - 11 \, x - 2 \, \log \left (x\right )}\right )} \] Input:

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*lo 
g(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*log(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x) 
+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm=" 
fricas")
 

Output:

e^(-4*x^3/(2*x^2 + 2*x*e^2 - 11*x - 2*log(x)))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((24*x**2*ln(x)-16*x**3*exp(2)-8*x**4+88*x**3-8*x**2)*exp(2*x**3/ 
(2*ln(x)-2*exp(2)*x-2*x**2+11*x))**2/(4*ln(x)**2+(-8*exp(2)*x-8*x**2+44*x) 
*ln(x)+4*x**2*exp(2)**2+(8*x**3-44*x**2)*exp(2)+4*x**4-44*x**3+121*x**2),x 
)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*lo 
g(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*log(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x) 
+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm=" 
maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=e^{\left (-\frac {4 \, x^{3}}{2 \, x^{2} + 2 \, x e^{2} - 11 \, x - 2 \, \log \left (x\right )}\right )} \] Input:

integrate((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*lo 
g(x)-2*exp(2)*x-2*x^2+11*x))^2/(4*log(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x) 
+4*x^2*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x, algorithm=" 
giac")
 

Output:

e^(-4*x^3/(2*x^2 + 2*x*e^2 - 11*x - 2*log(x)))
 

Mupad [B] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx={\mathrm {e}}^{\frac {4\,x^3}{11\,x+2\,\ln \left (x\right )-2\,x\,{\mathrm {e}}^2-2\,x^2}} \] Input:

int(-(exp((4*x^3)/(11*x + 2*log(x) - 2*x*exp(2) - 2*x^2))*(16*x^3*exp(2) - 
 24*x^2*log(x) + 8*x^2 - 88*x^3 + 8*x^4))/(4*log(x)^2 - exp(2)*(44*x^2 - 8 
*x^3) + 4*x^2*exp(4) + 121*x^2 - 44*x^3 + 4*x^4 - log(x)*(8*x*exp(2) - 44* 
x + 8*x^2)),x)
 

Output:

exp((4*x^3)/(11*x + 2*log(x) - 2*x*exp(2) - 2*x^2))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {4 x^3}{11 x-2 e^2 x-2 x^2+2 \log (x)}} \left (-8 x^2+88 x^3-16 e^2 x^3-8 x^4+24 x^2 \log (x)\right )}{121 x^2+4 e^4 x^2-44 x^3+4 x^4+e^2 \left (-44 x^2+8 x^3\right )+\left (44 x-8 e^2 x-8 x^2\right ) \log (x)+4 \log ^2(x)} \, dx=e^{\frac {4 x^{3}}{2 \,\mathrm {log}\left (x \right )-2 e^{2} x -2 x^{2}+11 x}} \] Input:

int((24*x^2*log(x)-16*x^3*exp(2)-8*x^4+88*x^3-8*x^2)*exp(2*x^3/(2*log(x)-2 
*exp(2)*x-2*x^2+11*x))^2/(4*log(x)^2+(-8*exp(2)*x-8*x^2+44*x)*log(x)+4*x^2 
*exp(2)^2+(8*x^3-44*x^2)*exp(2)+4*x^4-44*x^3+121*x^2),x)
 

Output:

e**((4*x**3)/(2*log(x) - 2*e**2*x - 2*x**2 + 11*x))