\(\int \frac {3 e^{x+x^2}+e^{x+x^2} (-3+15 x-6 x^2) \log (x)+e^x (-1+5 x) \log ^2(x)+(-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)) \log (\frac {e^{-1/e} (3 e^{x^2} x+x \log (x))}{\log (x)})}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx\) [293]

Optimal result

Integrand size = 113, antiderivative size = 31 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=e^x \left (5-\log \left (e^{-1/e} \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )\right ) \] Output:

exp(x)*(5-ln((3*exp(x^2)*x/ln(x)+x)/exp(exp(-1))))
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=e^{-1+x} \left (1+5 e-e \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right ) \] Input:

Integrate[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(- 
1 + 5*x)*Log[x]^2 + (-3*E^(x + x^2)*x*Log[x] - E^x*x*Log[x]^2)*Log[(3*E^x^ 
2*x + x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]
 

Output:

E^(-1 + x)*(1 + 5*E - E*Log[x + (3*E^x^2*x)/Log[x]])
 

Rubi [A] (verified)

Time = 4.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7292, 7293, 2009}

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.

\(\displaystyle \int \frac {3 e^{x^2+x}+\left (-3 e^{x^2+x} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )+e^{x^2+x} \left (-6 x^2+15 x-3\right ) \log (x)+e^x (5 x-1) \log ^2(x)}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {3 e^{x^2+x}+\left (-3 e^{x^2+x} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )+e^{x^2+x} \left (-6 x^2+15 x-3\right ) \log (x)+e^x (5 x-1) \log ^2(x)}{x \log (x) \left (3 e^{x^2}+\log (x)\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^x \left (2 x^2 \log (x)-1\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^{x-1} \left (-2 e x^2 \log (x)-e x \log (x) \log \left (\frac {3 e^{x^2} x}{\log (x)}+x\right )+(1+5 e) x \log (x)-e \log (x)+e\right )}{x \log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle (1+5 e) e^{x-1}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\)

Input:

Int[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(-1 + 5* 
x)*Log[x]^2 + (-3*E^(x + x^2)*x*Log[x] - E^x*x*Log[x]^2)*Log[(3*E^x^2*x + 
x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]
 

Output:

E^(-1 + x)*(1 + 5*E) - E^x*Log[(x*(3*E^x^2 + Log[x]))/Log[x]]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 6.36 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06

method	result	size
parallelrisch	\(-\ln \left (\frac {x \left (\ln \left (x \right )+3 \,{\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-{\mathrm e}^{-1}}}{\ln \left (x \right )}\right ) {\mathrm e}^{x}+5 \,{\mathrm e}^{x}\)	\(33\)
risch	\(-{\mathrm e}^{x} \ln \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )+\frac {\left (i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{3} {\mathrm e}+i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x \right ) {\mathrm e}+i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}-i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\mathrm e}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}+i \pi {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{3} {\mathrm e}-i \pi {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} \operatorname {csgn}\left (i x \right ) {\mathrm e}+2-2 \ln \left (3\right ) {\mathrm e}-2 \,{\mathrm e} \ln \left (x \right )+2 \ln \left (\ln \left (x \right )\right ) {\mathrm e}+10 \,{\mathrm e}\right ) {\mathrm e}^{-1+x}}{2}\)	\(333\)

Input:

int(((-x*exp(x)*ln(x)^2-3*x*exp(x)*exp(x^2)*ln(x))*ln((x*ln(x)+3*exp(x^2)* 
x)/exp(1/exp(1))/ln(x))+(5*x-1)*exp(x)*ln(x)^2+(-6*x^2+15*x-3)*exp(x)*exp( 
x^2)*ln(x)+3*exp(x)*exp(x^2))/(x*ln(x)^2+3*x*exp(x^2)*ln(x)),x,method=_RET 
URNVERBOSE)
 

Output:

-ln(x*(ln(x)+3*exp(x^2))/exp(1/exp(1))/ln(x))*exp(x)+5*exp(x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left (e^{\left (x^{2} + x\right )} \log \left (\frac {{\left (3 \, x e^{\left (x^{2}\right )} + x \log \left (x\right )\right )} e^{\left (-e^{\left (-1\right )}\right )}}{\log \left (x\right )}\right ) - 5 \, e^{\left (x^{2} + x\right )}\right )} e^{\left (-x^{2}\right )} \] Input:

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 
*exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) 
*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) 
),x, algorithm="fricas")
 

Output:

-(e^(x^2 + x)*log((3*x*e^(x^2) + x*log(x))*e^(-e^(-1))/log(x)) - 5*e^(x^2 
+ x))*e^(-x^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=\text {Timed out} \] Input:

integrate(((-x*exp(x)*ln(x)**2-3*x*exp(x)*exp(x**2)*ln(x))*ln((x*ln(x)+3*e 
xp(x**2)*x)/exp(1/exp(1))/ln(x))+(5*x-1)*exp(x)*ln(x)**2+(-6*x**2+15*x-3)* 
exp(x)*exp(x**2)*ln(x)+3*exp(x)*exp(x**2))/(x*ln(x)**2+3*x*exp(x**2)*ln(x) 
),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left ({\left (e \log \left (x\right ) - 5 \, e - 1\right )} e^{x} + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \left (x\right )\right ) - e^{\left (x + 1\right )} \log \left (\log \left (x\right )\right )\right )} e^{\left (-1\right )} \] Input:

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 
*exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) 
*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) 
),x, algorithm="maxima")
 

Output:

-((e*log(x) - 5*e - 1)*e^x + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1) 
*log(log(x)))*e^(-1)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left (e^{\left (x + 1\right )} \log \left (x\right ) + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \left (x\right )\right ) - e^{\left (x + 1\right )} \log \left (\log \left (x\right )\right ) - 5 \, e^{\left (x + 1\right )} - e^{x}\right )} e^{\left (-1\right )} \] Input:

integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 
*exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) 
*exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) 
),x, algorithm="giac")
 

Output:

-(e^(x + 1)*log(x) + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1)*log(log 
(x)) - 5*e^(x + 1) - e^x)*e^(-1)
 

Mupad [B] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\mathrm {e}}^x\,\left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^{-1}}\,\left (3\,x\,{\mathrm {e}}^{x^2}+x\,\ln \left (x\right )\right )}{\ln \left (x\right )}\right )-5\right ) \] Input:

int((3*exp(x^2)*exp(x) - log((exp(-exp(-1))*(3*x*exp(x^2) + x*log(x)))/log 
(x))*(x*exp(x)*log(x)^2 + 3*x*exp(x^2)*exp(x)*log(x)) + exp(x)*log(x)^2*(5 
*x - 1) - exp(x^2)*exp(x)*log(x)*(6*x^2 - 15*x + 3))/(x*log(x)^2 + 3*x*exp 
(x^2)*log(x)),x)
 

Output:

-exp(x)*(log((exp(-exp(-1))*(3*x*exp(x^2) + x*log(x)))/log(x)) - 5)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=e^{x} \left (-\mathrm {log}\left (\frac {3 e^{x^{2}} x +\mathrm {log}\left (x \right ) x}{e^{\frac {1}{e}} \mathrm {log}\left (x \right )}\right )+5\right ) \] Input:

int(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3*exp(x 
^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3)*exp(x 
)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x)),x)
 

Output:

e**x*( - log((3*e**(x**2)*x + log(x)*x)/(e**(1/e)*log(x))) + 5)