\(\int \frac {e^x (-6+11 x-3 e^3 x)+e^x (5 x+x^2) \log (x)}{1008 x-504 e^3 x+63 e^6 x+(1008 x+168 x^2+e^3 (-252 x-42 x^2)) \log (x)+(252 x+84 x^2+7 x^3) \log ^2(x)} \, dx\) [215]

Optimal result

Integrand size = 92, antiderivative size = 31 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 x \left (\log (x)+\frac {3 \left (4-e^3+2 \log (x)\right )}{x}\right )} \] Output:

1/7*exp(x)/(3*(2*ln(x)+4-exp(3))/x+ln(x))/x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^x}{7 \left (12-3 e^3+6 \log (x)+x \log (x)\right )} \] Input:

Integrate[(E^x*(-6 + 11*x - 3*E^3*x) + E^x*(5*x + x^2)*Log[x])/(1008*x - 5 
04*E^3*x + 63*E^6*x + (1008*x + 168*x^2 + E^3*(-252*x - 42*x^2))*Log[x] + 
(252*x + 84*x^2 + 7*x^3)*Log[x]^2),x]
 

Output:

E^x/(7*(12 - 3*E^3 + 6*Log[x] + x*Log[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.

\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+63 e^6 x-504 e^3 x+1008 x} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+\left (1008-504 e^3\right ) x+63 e^6 x}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^x \left (x^2+5 x\right ) \log (x)+e^x \left (-3 e^3 x+11 x-6\right )}{\left (168 x^2+e^3 \left (-42 x^2-252 x\right )+1008 x\right ) \log (x)+\left (7 x^3+84 x^2+252 x\right ) \log ^2(x)+\left (1008-504 e^3+63 e^6\right ) x}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {e^x \left (\left (11-3 e^3\right ) x+(x+5) x \log (x)-6\right )}{7 x \left (3 \left (e^3-4\right )-(x+6) \log (x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int -\frac {e^x \left (-((x+5) \log (x) x)-\left (11-3 e^3\right ) x+6\right )}{x \left ((x+6) \log (x)+3 \left (4-e^3\right )\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{7} \int \frac {e^x \left (-((x+5) \log (x) x)-\left (11-3 e^3\right ) x+6\right )}{x \left ((x+6) \log (x)+3 \left (4-e^3\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{7} \int \left (\frac {e^x (-x-5)}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )}+\frac {e^x \left (x^2+3 e^3 x+36\right )}{x (x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{7} \left (-\int \frac {e^x}{-x \log (x)-6 \log (x)-12 \left (1-\frac {e^3}{4}\right )}dx-\int \frac {e^x}{\left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx-6 \int \frac {e^x}{x \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx+3 \left (4-e^3\right ) \int \frac {e^x}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )^2}dx-\int \frac {e^x}{(x+6) \left (x \log (x)+6 \log (x)+12 \left (1-\frac {e^3}{4}\right )\right )}dx\right )\)

Input:

Int[(E^x*(-6 + 11*x - 3*E^3*x) + E^x*(5*x + x^2)*Log[x])/(1008*x - 504*E^3 
*x + 63*E^6*x + (1008*x + 168*x^2 + E^3*(-252*x - 42*x^2))*Log[x] + (252*x 
 + 84*x^2 + 7*x^3)*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\)	\(22\)
parallelrisch	\(-\frac {{\mathrm e}^{x}}{7 \left (-x \ln \left (x \right )+3 \,{\mathrm e}^{3}-6 \ln \left (x \right )-12\right )}\)	\(22\)

Input:

int(((x^2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84*x^2+25 
2*x)*ln(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*ln(x)+63*x*exp(3)^2-5 
04*x*exp(3)+1008*x),x,method=_RETURNVERBOSE)
 

Output:

-1/7*exp(x)/(-x*ln(x)+3*exp(3)-6*ln(x)-12)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \] Input:

integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 
*x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e 
xp(3)^2-504*x*exp(3)+1008*x),x, algorithm="fricas")
 

Output:

1/7*e^x/((x + 6)*log(x) - 3*e^3 + 12)
 

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 x \log {\left (x \right )} + 42 \log {\left (x \right )} - 21 e^{3} + 84} \] Input:

integrate(((x**2+5*x)*exp(x)*ln(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x**3+8 
4*x**2+252*x)*ln(x)**2+((-42*x**2-252*x)*exp(3)+168*x**2+1008*x)*ln(x)+63* 
x*exp(3)**2-504*x*exp(3)+1008*x),x)
 

Output:

exp(x)/(7*x*log(x) + 42*log(x) - 21*exp(3) + 84)
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.58 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left ({\left (x + 6\right )} \log \left (x\right ) - 3 \, e^{3} + 12\right )}} \] Input:

integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 
*x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e 
xp(3)^2-504*x*exp(3)+1008*x),x, algorithm="maxima")
 

Output:

1/7*e^x/((x + 6)*log(x) - 3*e^3 + 12)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \, {\left (x \log \left (x\right ) - 3 \, e^{3} + 6 \, \log \left (x\right ) + 12\right )}} \] Input:

integrate(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84 
*x^2+252*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*e 
xp(3)^2-504*x*exp(3)+1008*x),x, algorithm="giac")
 

Output:

1/7*e^x/(x*log(x) - 3*e^3 + 6*log(x) + 12)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (3\,x\,{\mathrm {e}}^3-11\,x+6\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x^2+5\,x\right )}{\left (7\,x^3+84\,x^2+252\,x\right )\,{\ln \left (x\right )}^2+\left (1008\,x-{\mathrm {e}}^3\,\left (42\,x^2+252\,x\right )+168\,x^2\right )\,\ln \left (x\right )+1008\,x-504\,x\,{\mathrm {e}}^3+63\,x\,{\mathrm {e}}^6} \,d x \] Input:

int(-(exp(x)*(3*x*exp(3) - 11*x + 6) - exp(x)*log(x)*(5*x + x^2))/(1008*x 
- 504*x*exp(3) + 63*x*exp(6) + log(x)^2*(252*x + 84*x^2 + 7*x^3) + log(x)* 
(1008*x - exp(3)*(252*x + 42*x^2) + 168*x^2)),x)
 

Output:

-int((exp(x)*(3*x*exp(3) - 11*x + 6) - exp(x)*log(x)*(5*x + x^2))/(1008*x 
- 504*x*exp(3) + 63*x*exp(6) + log(x)^2*(252*x + 84*x^2 + 7*x^3) + log(x)* 
(1008*x - exp(3)*(252*x + 42*x^2) + 168*x^2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^x \left (-6+11 x-3 e^3 x\right )+e^x \left (5 x+x^2\right ) \log (x)}{1008 x-504 e^3 x+63 e^6 x+\left (1008 x+168 x^2+e^3 \left (-252 x-42 x^2\right )\right ) \log (x)+\left (252 x+84 x^2+7 x^3\right ) \log ^2(x)} \, dx=\frac {e^{x}}{7 \,\mathrm {log}\left (x \right ) x +42 \,\mathrm {log}\left (x \right )-21 e^{3}+84} \] Input:

int(((x^2+5*x)*exp(x)*log(x)+(-3*x*exp(3)+11*x-6)*exp(x))/((7*x^3+84*x^2+2 
52*x)*log(x)^2+((-42*x^2-252*x)*exp(3)+168*x^2+1008*x)*log(x)+63*x*exp(3)^ 
2-504*x*exp(3)+1008*x),x)
 

Output:

e**x/(7*(log(x)*x + 6*log(x) - 3*e**3 + 12))