\(\int \frac {e^{\frac {2}{5} (5 e^{\frac {-2 e^2+x \log ^2(\frac {2+3 x}{x})}{x^2}}-x)} (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2(\frac {2+3 x}{x})}{x^2}} (e^2 (80+120 x)-40 x \log (\frac {2+3 x}{x})+(-20 x-30 x^2) \log ^2(\frac {2+3 x}{x})))}{10 x^3+15 x^4} \, dx\) [1596]

Optimal result

Integrand size = 135, antiderivative size = 38 \[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=e^{2 e^{\frac {-\frac {2 e^2}{x}+\log ^2\left (4-\frac {-2+x}{x}\right )}{x}}-\frac {2 x}{5}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=e^{2 e^{-\frac {2 e^2}{x^2}+\frac {\log ^2\left (3+\frac {2}{x}\right )}{x}}-\frac {2 x}{5}} \] Input:

Integrate[(E^((2*(5*E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)*(-4*x 
^3 - 6*x^4 + E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2)*(E^2*(80 + 120*x) - 4 
0*x*Log[(2 + 3*x)/x] + (-20*x - 30*x^2)*Log[(2 + 3*x)/x]^2)))/(10*x^3 + 15 
*x^4),x]
 

Output:

E^(2*E^((-2*E^2)/x^2 + Log[3 + 2/x]^2/x) - (2*x)/5)
 

Rubi [A] (verified)

Time = 3.64 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2026, 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.

\(\displaystyle \int \frac {\left (-6 x^4-4 x^3+e^{\frac {x \log ^2\left (\frac {3 x+2}{x}\right )-2 e^2}{x^2}} \left (\left (-30 x^2-20 x\right ) \log ^2\left (\frac {3 x+2}{x}\right )+e^2 (120 x+80)-40 x \log \left (\frac {3 x+2}{x}\right )\right )\right ) \exp \left (\frac {2}{5} \left (5 e^{\frac {x \log ^2\left (\frac {3 x+2}{x}\right )-2 e^2}{x^2}}-x\right )\right )}{15 x^4+10 x^3} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (-6 x^4-4 x^3+e^{\frac {x \log ^2\left (\frac {3 x+2}{x}\right )-2 e^2}{x^2}} \left (\left (-30 x^2-20 x\right ) \log ^2\left (\frac {3 x+2}{x}\right )+e^2 (120 x+80)-40 x \log \left (\frac {3 x+2}{x}\right )\right )\right ) \exp \left (\frac {2}{5} \left (5 e^{\frac {x \log ^2\left (\frac {3 x+2}{x}\right )-2 e^2}{x^2}}-x\right )\right )}{x^3 (15 x+10)}dx\)

\(\Big \downarrow \) 7257

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

Input:

Int[(E^((2*(5*E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)*(-4*x^3 - 6 
*x^4 + E^((-2*E^2 + x*Log[(2 + 3*x)/x]^2)/x^2)*(E^2*(80 + 120*x) - 40*x*Lo 
g[(2 + 3*x)/x] + (-20*x - 30*x^2)*Log[(2 + 3*x)/x]^2)))/(10*x^3 + 15*x^4), 
x]
 

Output:

E^((2*(5/E^((2*E^2 - x*Log[(2 + 3*x)/x]^2)/x^2) - x))/5)
 

Defintions of rubi rules used

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[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 = 77.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
risch	\({\mathrm e}^{2 \,{\mathrm e}^{-\frac {-x \ln \left (\frac {2+3 x}{x}\right )^{2}+2 \,{\mathrm e}^{2}}{x^{2}}}-\frac {2 x}{5}}\)	\(34\)
parallelrisch	\({\mathrm e}^{2 \,{\mathrm e}^{-\frac {-x \ln \left (\frac {2+3 x}{x}\right )^{2}+2 \,{\mathrm e}^{2}}{x^{2}}}-\frac {2 x}{5}}\)	\(34\)

Input:

int((((-30*x^2-20*x)*ln((2+3*x)/x)^2-40*x*ln((2+3*x)/x)+(120*x+80)*exp(2)) 
*exp((x*ln((2+3*x)/x)^2-2*exp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*ln((2+3*x)/ 
x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x,method=_RETURNVERBOSE)
 

Output:

exp(2*exp(-(-x*ln((2+3*x)/x)^2+2*exp(2))/x^2)-2/5*x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )} \] Input:

integrate((((-30*x^2-20*x)*log((2+3*x)/x)^2-40*x*log((2+3*x)/x)+(120*x+80) 
*exp(2))*exp((x*log((2+3*x)/x)^2-2*exp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*lo 
g((2+3*x)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="frica 
s")
 

Output:

e^(-2/5*x + 2*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))
 

Sympy [A] (verification not implemented)

Time = 2.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=e^{- \frac {2 x}{5} + 2 e^{\frac {x \log {\left (\frac {3 x + 2}{x} \right )}^{2} - 2 e^{2}}{x^{2}}}} \] Input:

integrate((((-30*x**2-20*x)*ln((2+3*x)/x)**2-40*x*ln((2+3*x)/x)+(120*x+80) 
*exp(2))*exp((x*ln((2+3*x)/x)**2-2*exp(2))/x**2)-6*x**4-4*x**3)*exp(exp((x 
*ln((2+3*x)/x)**2-2*exp(2))/x**2)-1/5*x)**2/(15*x**4+10*x**3),x)
 

Output:

exp(-2*x/5 + 2*exp((x*log((3*x + 2)/x)**2 - 2*exp(2))/x**2))
 

Maxima [F]

\[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=\int { -\frac {2 \, {\left (3 \, x^{4} + 2 \, x^{3} + 5 \, {\left ({\left (3 \, x^{2} + 2 \, x\right )} \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 4 \, {\left (3 \, x + 2\right )} e^{2} + 4 \, x \log \left (\frac {3 \, x + 2}{x}\right )\right )} e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )} e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )}}{5 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )}} \,d x } \] Input:

integrate((((-30*x^2-20*x)*log((2+3*x)/x)^2-40*x*log((2+3*x)/x)+(120*x+80) 
*exp(2))*exp((x*log((2+3*x)/x)^2-2*exp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*lo 
g((2+3*x)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="maxim 
a")
 

Output:

-2/5*integrate((3*x^4 + 2*x^3 + 5*((3*x^2 + 2*x)*log((3*x + 2)/x)^2 - 4*(3 
*x + 2)*e^2 + 4*x*log((3*x + 2)/x))*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2) 
)*e^(-2/5*x + 2*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))/(3*x^4 + 2*x^3), x 
)
 

Giac [F]

\[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx=\int { -\frac {2 \, {\left (3 \, x^{4} + 2 \, x^{3} + 5 \, {\left ({\left (3 \, x^{2} + 2 \, x\right )} \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 4 \, {\left (3 \, x + 2\right )} e^{2} + 4 \, x \log \left (\frac {3 \, x + 2}{x}\right )\right )} e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )} e^{\left (-\frac {2}{5} \, x + 2 \, e^{\left (\frac {x \log \left (\frac {3 \, x + 2}{x}\right )^{2} - 2 \, e^{2}}{x^{2}}\right )}\right )}}{5 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )}} \,d x } \] Input:

integrate((((-30*x^2-20*x)*log((2+3*x)/x)^2-40*x*log((2+3*x)/x)+(120*x+80) 
*exp(2))*exp((x*log((2+3*x)/x)^2-2*exp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*lo 
g((2+3*x)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x, algorithm="giac" 
)
 

Output:

integrate(-2/5*(3*x^4 + 2*x^3 + 5*((3*x^2 + 2*x)*log((3*x + 2)/x)^2 - 4*(3 
*x + 2)*e^2 + 4*x*log((3*x + 2)/x))*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2) 
)*e^(-2/5*x + 2*e^((x*log((3*x + 2)/x)^2 - 2*e^2)/x^2))/(3*x^4 + 2*x^3), x 
)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 2.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx={\mathrm {e}}^{-\frac {2\,x}{5}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^2}{x^2}}\,{\mathrm {e}}^{\frac {{\ln \left (\frac {3\,x+2}{x}\right )}^2}{x}}} \] Input:

int(-(exp(2*exp(-(2*exp(2) - x*log((3*x + 2)/x)^2)/x^2) - (2*x)/5)*(exp(-( 
2*exp(2) - x*log((3*x + 2)/x)^2)/x^2)*(log((3*x + 2)/x)^2*(20*x + 30*x^2) 
+ 40*x*log((3*x + 2)/x) - exp(2)*(120*x + 80)) + 4*x^3 + 6*x^4))/(10*x^3 + 
 15*x^4),x)
 

Output:

exp(-(2*x)/5)*exp(2*exp(-(2*exp(2))/x^2)*exp(log((3*x + 2)/x)^2/x))
 

Reduce [F]

\[ \int \frac {e^{\frac {2}{5} \left (5 e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}}-x\right )} \left (-4 x^3-6 x^4+e^{\frac {-2 e^2+x \log ^2\left (\frac {2+3 x}{x}\right )}{x^2}} \left (e^2 (80+120 x)-40 x \log \left (\frac {2+3 x}{x}\right )+\left (-20 x-30 x^2\right ) \log ^2\left (\frac {2+3 x}{x}\right )\right )\right )}{10 x^3+15 x^4} \, dx =\text {Too large to display} \] Input:

int((((-30*x^2-20*x)*log((2+3*x)/x)^2-40*x*log((2+3*x)/x)+(120*x+80)*exp(2 
))*exp((x*log((2+3*x)/x)^2-2*exp(2))/x^2)-6*x^4-4*x^3)*exp(exp((x*log((2+3 
*x)/x)^2-2*exp(2))/x^2)-1/5*x)^2/(15*x^4+10*x^3),x)
 

Output:

(2*(40*int(e**((2*e**(log((3*x + 2)/x)**2/x)*x + e**((2*e**2)/x**2)*log((3 
*x + 2)/x)**2)/(e**((2*e**2)/x**2)*x))/(3*e**((10*e**2 + 2*x**3)/(5*x**2)) 
*x**4 + 2*e**((10*e**2 + 2*x**3)/(5*x**2))*x**3),x)*e**2 + 60*int(e**((2*e 
**(log((3*x + 2)/x)**2/x)*x + e**((2*e**2)/x**2)*log((3*x + 2)/x)**2)/(e** 
((2*e**2)/x**2)*x))/(3*e**((10*e**2 + 2*x**3)/(5*x**2))*x**3 + 2*e**((10*e 
**2 + 2*x**3)/(5*x**2))*x**2),x)*e**2 - 2*int(e**((2*e**(log((3*x + 2)/x)* 
*2/x))/e**((2*e**2)/x**2))/(3*e**((2*x)/5)*x + 2*e**((2*x)/5)),x) - 10*int 
((e**((2*e**(log((3*x + 2)/x)**2/x)*x + e**((2*e**2)/x**2)*log((3*x + 2)/x 
)**2)/(e**((2*e**2)/x**2)*x))*log((3*x + 2)/x)**2)/(3*e**((10*e**2 + 2*x** 
3)/(5*x**2))*x**3 + 2*e**((10*e**2 + 2*x**3)/(5*x**2))*x**2),x) - 15*int(( 
e**((2*e**(log((3*x + 2)/x)**2/x)*x + e**((2*e**2)/x**2)*log((3*x + 2)/x)* 
*2)/(e**((2*e**2)/x**2)*x))*log((3*x + 2)/x)**2)/(3*e**((10*e**2 + 2*x**3) 
/(5*x**2))*x**2 + 2*e**((10*e**2 + 2*x**3)/(5*x**2))*x),x) - 20*int((e**(( 
2*e**(log((3*x + 2)/x)**2/x)*x + e**((2*e**2)/x**2)*log((3*x + 2)/x)**2)/( 
e**((2*e**2)/x**2)*x))*log((3*x + 2)/x))/(3*e**((10*e**2 + 2*x**3)/(5*x**2 
))*x**3 + 2*e**((10*e**2 + 2*x**3)/(5*x**2))*x**2),x) - 3*int((e**((2*e**( 
log((3*x + 2)/x)**2/x))/e**((2*e**2)/x**2))*x)/(3*e**((2*x)/5)*x + 2*e**(( 
2*x)/5)),x)))/5