\(\int \frac {-18 x^2-6 x^3+e^2 (-18 x-6 x^2)+e^2 (-18 x-6 x^2) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} (-3 x \log ^2(x)+(18+6 x+e^x (-9 x-3 x^2)) \log ^2(x) \log (3+x))}{(3 x^3+x^4) \log ^2(x)} \, dx\) [958]

Optimal result

Integrand size = 117, antiderivative size = 33 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {3 \left (\frac {2 \left (e^2+x\right )}{\log (x)}-\frac {\left (5+e^{e^x}\right ) \log (3+x)}{x}\right )}{x} \] Output:

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

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {3 \left (-2 x \left (e^2+x\right )+\left (5+e^{e^x}\right ) \log (x) \log (3+x)\right )}{x^2 \log (x)} \] Input:

Integrate[(-18*x^2 - 6*x^3 + E^2*(-18*x - 6*x^2) + E^2*(-18*x - 6*x^2)*Log 
[x] - 15*x*Log[x]^2 + (90 + 30*x)*Log[x]^2*Log[3 + x] + E^E^x*(-3*x*Log[x] 
^2 + (18 + 6*x + E^x*(-9*x - 3*x^2))*Log[x]^2*Log[3 + x]))/((3*x^3 + x^4)* 
Log[x]^2),x]
 

Output:

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

\(\displaystyle \int \frac {-6 x^3-18 x^2+e^2 \left (-6 x^2-18 x\right )+e^{e^x} \left (\left (e^x \left (-3 x^2-9 x\right )+6 x+18\right ) \log ^2(x) \log (x+3)-3 x \log ^2(x)\right )+e^2 \left (-6 x^2-18 x\right ) \log (x)-15 x \log ^2(x)+(30 x+90) \log ^2(x) \log (x+3)}{\left (x^4+3 x^3\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {-6 x^3-18 x^2+e^2 \left (-6 x^2-18 x\right )+e^{e^x} \left (\left (e^x \left (-3 x^2-9 x\right )+6 x+18\right ) \log ^2(x) \log (x+3)-3 x \log ^2(x)\right )+e^2 \left (-6 x^2-18 x\right ) \log (x)-15 x \log ^2(x)+(30 x+90) \log ^2(x) \log (x+3)}{x^3 (x+3) \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {18 e^{e^x} \log (x+3)}{x^3 (x+3)}+\frac {30 \log (x+3)}{x^3}-\frac {3 e^{e^x}}{x^2 (x+3)}-\frac {15}{x^2 (x+3)}-\frac {6 e^2}{x^2 \log ^2(x)}+\frac {6 e^{e^x} \log (x+3)}{x^2 (x+3)}-\frac {3 e^{x+e^x} \log (x+3)}{x^2}-\frac {6 e^2}{x^2 \log (x)}-\frac {18}{x (x+3) \log ^2(x)}-\frac {6}{(x+3) \log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -6 \int \frac {\int \frac {e^{e^x}}{x^3}dx}{x+3}dx+6 \log (x+3) \int \frac {e^{e^x}}{x^3}dx-\int \frac {e^{e^x}}{x^2}dx+3 \int \frac {\int \frac {e^{x+e^x}}{x^2}dx}{x+3}dx-3 \log (x+3) \int \frac {e^{x+e^x}}{x^2}dx+\frac {1}{3} \int \frac {e^{e^x}}{x}dx-\frac {1}{3} \int \frac {e^{e^x}}{x+3}dx-6 \int \frac {1}{(x+3) \log ^2(x)}dx-18 \int \frac {1}{x (x+3) \log ^2(x)}dx-\frac {15 \log (x+3)}{x^2}+\frac {6 e^2}{x \log (x)}\)

Input:

Int[(-18*x^2 - 6*x^3 + E^2*(-18*x - 6*x^2) + E^2*(-18*x - 6*x^2)*Log[x] - 
15*x*Log[x]^2 + (90 + 30*x)*Log[x]^2*Log[3 + x] + E^E^x*(-3*x*Log[x]^2 + ( 
18 + 6*x + E^x*(-9*x - 3*x^2))*Log[x]^2*Log[3 + x]))/((3*x^3 + x^4)*Log[x] 
^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

\[-\frac {15 \ln \left (3+x \right )}{x^{2}}+\frac {6 x +6 \,{\mathrm e}^{2}}{x \ln \left (x \right )}-\frac {3 \ln \left (3+x \right ) {\mathrm e}^{{\mathrm e}^{x}}}{x^{2}}\]

Input:

int(((((-3*x^2-9*x)*exp(x)+18+6*x)*ln(x)^2*ln(3+x)-3*x*ln(x)^2)*exp(exp(x) 
)+(30*x+90)*ln(x)^2*ln(3+x)-15*x*ln(x)^2+(-6*x^2-18*x)*exp(2)*ln(x)+(-6*x^ 
2-18*x)*exp(2)-6*x^3-18*x^2)/(x^4+3*x^3)/ln(x)^2,x)
 

Output:

-15/x^2*ln(3+x)+6/x*(x+exp(2))/ln(x)-3/x^2*ln(3+x)*exp(exp(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {3 \, {\left (e^{\left (e^{x}\right )} \log \left (x + 3\right ) \log \left (x\right ) - 2 \, x^{2} - 2 \, x e^{2} + 5 \, \log \left (x + 3\right ) \log \left (x\right )\right )}}{x^{2} \log \left (x\right )} \] Input:

integrate(((((-3*x^2-9*x)*exp(x)+18+6*x)*log(x)^2*log(3+x)-3*x*log(x)^2)*e 
xp(exp(x))+(30*x+90)*log(x)^2*log(3+x)-15*x*log(x)^2+(-6*x^2-18*x)*exp(2)* 
log(x)+(-6*x^2-18*x)*exp(2)-6*x^3-18*x^2)/(x^4+3*x^3)/log(x)^2,x, algorith 
m="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {6 x + 6 e^{2}}{x \log {\left (x \right )}} - \frac {3 e^{e^{x}} \log {\left (x + 3 \right )}}{x^{2}} - \frac {15 \log {\left (x + 3 \right )}}{x^{2}} \] Input:

integrate(((((-3*x**2-9*x)*exp(x)+18+6*x)*ln(x)**2*ln(3+x)-3*x*ln(x)**2)*e 
xp(exp(x))+(30*x+90)*ln(x)**2*ln(3+x)-15*x*ln(x)**2+(-6*x**2-18*x)*exp(2)* 
ln(x)+(-6*x**2-18*x)*exp(2)-6*x**3-18*x**2)/(x**4+3*x**3)/ln(x)**2,x)
 

Output:

(6*x + 6*exp(2))/(x*log(x)) - 3*exp(exp(x))*log(x + 3)/x**2 - 15*log(x + 3 
)/x**2
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=-\frac {3 \, {\left (e^{\left (e^{x}\right )} \log \left (x + 3\right ) \log \left (x\right ) - 2 \, x^{2} - 2 \, x e^{2} + 5 \, \log \left (x + 3\right ) \log \left (x\right )\right )}}{x^{2} \log \left (x\right )} \] Input:

integrate(((((-3*x^2-9*x)*exp(x)+18+6*x)*log(x)^2*log(3+x)-3*x*log(x)^2)*e 
xp(exp(x))+(30*x+90)*log(x)^2*log(3+x)-15*x*log(x)^2+(-6*x^2-18*x)*exp(2)* 
log(x)+(-6*x^2-18*x)*exp(2)-6*x^3-18*x^2)/(x^4+3*x^3)/log(x)^2,x, algorith 
m="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=\int { \frac {3 \, {\left (10 \, {\left (x + 3\right )} \log \left (x + 3\right ) \log \left (x\right )^{2} - 2 \, x^{3} - 2 \, {\left (x^{2} + 3 \, x\right )} e^{2} \log \left (x\right ) - 5 \, x \log \left (x\right )^{2} - 6 \, x^{2} - 2 \, {\left (x^{2} + 3 \, x\right )} e^{2} - {\left ({\left ({\left (x^{2} + 3 \, x\right )} e^{x} - 2 \, x - 6\right )} \log \left (x + 3\right ) \log \left (x\right )^{2} + x \log \left (x\right )^{2}\right )} e^{\left (e^{x}\right )}\right )}}{{\left (x^{4} + 3 \, x^{3}\right )} \log \left (x\right )^{2}} \,d x } \] Input:

integrate(((((-3*x^2-9*x)*exp(x)+18+6*x)*log(x)^2*log(3+x)-3*x*log(x)^2)*e 
xp(exp(x))+(30*x+90)*log(x)^2*log(3+x)-15*x*log(x)^2+(-6*x^2-18*x)*exp(2)* 
log(x)+(-6*x^2-18*x)*exp(2)-6*x^3-18*x^2)/(x^4+3*x^3)/log(x)^2,x, algorith 
m="giac")
 

Output:

integrate(3*(10*(x + 3)*log(x + 3)*log(x)^2 - 2*x^3 - 2*(x^2 + 3*x)*e^2*lo 
g(x) - 5*x*log(x)^2 - 6*x^2 - 2*(x^2 + 3*x)*e^2 - (((x^2 + 3*x)*e^x - 2*x 
- 6)*log(x + 3)*log(x)^2 + x*log(x)^2)*e^(e^x))/((x^4 + 3*x^3)*log(x)^2), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=\int -\frac {15\,x\,{\ln \left (x\right )}^2+{\mathrm {e}}^2\,\left (6\,x^2+18\,x\right )+18\,x^2+6\,x^3+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x\,{\ln \left (x\right )}^2-\ln \left (x+3\right )\,{\ln \left (x\right )}^2\,\left (6\,x-{\mathrm {e}}^x\,\left (3\,x^2+9\,x\right )+18\right )\right )+{\mathrm {e}}^2\,\ln \left (x\right )\,\left (6\,x^2+18\,x\right )-\ln \left (x+3\right )\,{\ln \left (x\right )}^2\,\left (30\,x+90\right )}{{\ln \left (x\right )}^2\,\left (x^4+3\,x^3\right )} \,d x \] Input:

int(-(15*x*log(x)^2 + exp(2)*(18*x + 6*x^2) + 18*x^2 + 6*x^3 + exp(exp(x)) 
*(3*x*log(x)^2 - log(x + 3)*log(x)^2*(6*x - exp(x)*(9*x + 3*x^2) + 18)) + 
exp(2)*log(x)*(18*x + 6*x^2) - log(x + 3)*log(x)^2*(30*x + 90))/(log(x)^2* 
(3*x^3 + x^4)),x)
 

Output:

int(-(15*x*log(x)^2 + exp(2)*(18*x + 6*x^2) + 18*x^2 + 6*x^3 + exp(exp(x)) 
*(3*x*log(x)^2 - log(x + 3)*log(x)^2*(6*x - exp(x)*(9*x + 3*x^2) + 18)) + 
exp(2)*log(x)*(18*x + 6*x^2) - log(x + 3)*log(x)^2*(30*x + 90))/(log(x)^2* 
(3*x^3 + x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {-18 x^2-6 x^3+e^2 \left (-18 x-6 x^2\right )+e^2 \left (-18 x-6 x^2\right ) \log (x)-15 x \log ^2(x)+(90+30 x) \log ^2(x) \log (3+x)+e^{e^x} \left (-3 x \log ^2(x)+\left (18+6 x+e^x \left (-9 x-3 x^2\right )\right ) \log ^2(x) \log (3+x)\right )}{\left (3 x^3+x^4\right ) \log ^2(x)} \, dx=\frac {-3 e^{e^{x}} \mathrm {log}\left (x +3\right ) \mathrm {log}\left (x \right )-15 \,\mathrm {log}\left (x +3\right ) \mathrm {log}\left (x \right )+6 e^{2} x +6 x^{2}}{\mathrm {log}\left (x \right ) x^{2}} \] Input:

int(((((-3*x^2-9*x)*exp(x)+18+6*x)*log(x)^2*log(3+x)-3*x*log(x)^2)*exp(exp 
(x))+(30*x+90)*log(x)^2*log(3+x)-15*x*log(x)^2+(-6*x^2-18*x)*exp(2)*log(x) 
+(-6*x^2-18*x)*exp(2)-6*x^3-18*x^2)/(x^4+3*x^3)/log(x)^2,x)
 

Output:

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