\(\int \frac {(12 x+18 x^2+e^x (12+12 x+6 x^2)+(6 x+6 e^x x) \log (x)) \log (25 e^x x^2+25 x^3+(50 e^x x+50 x^2) \log (x)+(25 e^x+25 x) \log ^2(x))+(3 e^x x+3 x^2+(3 e^x+3 x) \log (x)) \log ^2(25 e^x x^2+25 x^3+(50 e^x x+50 x^2) \log (x)+(25 e^x+25 x) \log ^2(x))}{2 e^x x+2 x^2+(2 e^x+2 x) \log (x)} \, dx\) [596]

Optimal result
Mathematica [F]
Rubi [F]
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-2)]
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 180, antiderivative size = 21 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\frac {3}{2} x \log ^2\left (25 \left (e^x+x\right ) (x+\log (x))^2\right ) \] Output:

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

Mathematica [F]

\[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx \] Input:

Integrate[((12*x + 18*x^2 + E^x*(12 + 12*x + 6*x^2) + (6*x + 6*E^x*x)*Log[ 
x])*Log[25*E^x*x^2 + 25*x^3 + (50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x) 
*Log[x]^2] + (3*E^x*x + 3*x^2 + (3*E^x + 3*x)*Log[x])*Log[25*E^x*x^2 + 25* 
x^3 + (50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x)*Log[x]^2]^2)/(2*E^x*x + 
 2*x^2 + (2*E^x + 2*x)*Log[x]),x]
 

Output:

Integrate[((12*x + 18*x^2 + E^x*(12 + 12*x + 6*x^2) + (6*x + 6*E^x*x)*Log[ 
x])*Log[25*E^x*x^2 + 25*x^3 + (50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x) 
*Log[x]^2] + (3*E^x*x + 3*x^2 + (3*E^x + 3*x)*Log[x])*Log[25*E^x*x^2 + 25* 
x^3 + (50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x)*Log[x]^2]^2)/(2*E^x*x + 
 2*x^2 + (2*E^x + 2*x)*Log[x]), 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.

\(\displaystyle \int \frac {\left (3 x^2+3 e^x x+\left (3 x+3 e^x\right ) \log (x)\right ) \log ^2\left (25 x^3+25 e^x x^2+\left (50 x^2+50 e^x x\right ) \log (x)+\left (25 x+25 e^x\right ) \log ^2(x)\right )+\left (18 x^2+e^x \left (6 x^2+12 x+12\right )+12 x+\left (6 e^x x+6 x\right ) \log (x)\right ) \log \left (25 x^3+25 e^x x^2+\left (50 x^2+50 e^x x\right ) \log (x)+\left (25 x+25 e^x\right ) \log ^2(x)\right )}{2 x^2+2 e^x x+\left (2 x+2 e^x\right ) \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {3 \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) \left (2 e^x x^2+6 x^2+x^2 \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )+4 e^x x+4 x+4 e^x+2 e^x x \log (x)+2 x \log (x)+e^x x \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )+x \log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )+e^x \log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )\right )}{2 \left (x+e^x\right ) (x+\log (x))}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int \frac {\log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) \left (2 e^x x^2+\log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) x^2+6 x^2+4 e^x x+2 e^x \log (x) x+2 \log (x) x+e^x \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) x+\log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) x+4 x+4 e^x+e^x \log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )\right )}{\left (x+e^x\right ) (x+\log (x))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {3}{2} \int \left (\frac {\log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) \left (2 x^2+2 \log (x) x+\log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) x+4 x+\log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )+4\right )}{x+\log (x)}-\frac {2 (x-1) x \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )}{x+e^x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3}{2} \left (2 \int \frac {\int \frac {x^2}{x+e^x}dx}{x+e^x}dx-2 \int \frac {x \int \frac {x^2}{x+e^x}dx}{x+e^x}dx-2 \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) \int \frac {x^2}{x+e^x}dx+2 \int \frac {x^2 \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )}{x+\log (x)}dx+4 \int \frac {\int \frac {x^2}{x+e^x}dx}{x+\log (x)}dx+4 \int \frac {\int \frac {x^2}{x+e^x}dx}{x (x+\log (x))}dx+2 \int \frac {x \int \frac {x^2}{x+e^x}dx}{x+\log (x)}dx+2 \int \frac {\log (x) \int \frac {x^2}{x+e^x}dx}{x+\log (x)}dx-2 \int \frac {\int \frac {x}{x+e^x}dx}{x+e^x}dx+2 \int \frac {x \int \frac {x}{x+e^x}dx}{x+e^x}dx+\int \log ^2\left (25 \left (x+e^x\right ) (x+\log (x))^2\right )dx+2 \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right ) \int \frac {x}{x+e^x}dx+4 \int \frac {\log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )}{x+\log (x)}dx+4 \int \frac {x \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )}{x+\log (x)}dx+2 \int \frac {x \log (x) \log \left (25 \left (x+e^x\right ) (x+\log (x))^2\right )}{x+\log (x)}dx-4 \int \frac {\int \frac {x}{x+e^x}dx}{x+\log (x)}dx-4 \int \frac {\int \frac {x}{x+e^x}dx}{x (x+\log (x))}dx-2 \int \frac {x \int \frac {x}{x+e^x}dx}{x+\log (x)}dx-2 \int \frac {\log (x) \int \frac {x}{x+e^x}dx}{x+\log (x)}dx\right )\)

Input:

Int[((12*x + 18*x^2 + E^x*(12 + 12*x + 6*x^2) + (6*x + 6*E^x*x)*Log[x])*Lo 
g[25*E^x*x^2 + 25*x^3 + (50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x)*Log[x 
]^2] + (3*E^x*x + 3*x^2 + (3*E^x + 3*x)*Log[x])*Log[25*E^x*x^2 + 25*x^3 + 
(50*E^x*x + 50*x^2)*Log[x] + (25*E^x + 25*x)*Log[x]^2]^2)/(2*E^x*x + 2*x^2 
 + (2*E^x + 2*x)*Log[x]),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 3.94 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.24

method result size
parallelrisch \(\frac {3 x {\ln \left (\left (25 \,{\mathrm e}^{x}+25 x \right ) \ln \left (x \right )^{2}+\left (50 \,{\mathrm e}^{x} x +50 x^{2}\right ) \ln \left (x \right )+25 \,{\mathrm e}^{x} x^{2}+25 x^{3}\right )}^{2}}{2}\) \(47\)
risch \(\text {Expression too large to display}\) \(1627\)

Input:

int((((3*exp(x)+3*x)*ln(x)+3*exp(x)*x+3*x^2)*ln((25*exp(x)+25*x)*ln(x)^2+( 
50*exp(x)*x+50*x^2)*ln(x)+25*exp(x)*x^2+25*x^3)^2+((6*exp(x)*x+6*x)*ln(x)+ 
(6*x^2+12*x+12)*exp(x)+18*x^2+12*x)*ln((25*exp(x)+25*x)*ln(x)^2+(50*exp(x) 
*x+50*x^2)*ln(x)+25*exp(x)*x^2+25*x^3))/((2*exp(x)+2*x)*ln(x)+2*exp(x)*x+2 
*x^2),x,method=_RETURNVERBOSE)
 

Output:

3/2*x*ln((25*exp(x)+25*x)*ln(x)^2+(50*exp(x)*x+50*x^2)*ln(x)+25*exp(x)*x^2 
+25*x^3)^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (18) = 36\).

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.95 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\frac {3}{2} \, x \log \left (25 \, x^{3} + 25 \, x^{2} e^{x} + 25 \, {\left (x + e^{x}\right )} \log \left (x\right )^{2} + 50 \, {\left (x^{2} + x e^{x}\right )} \log \left (x\right )\right )^{2} \] Input:

integrate((((3*exp(x)+3*x)*log(x)+3*exp(x)*x+3*x^2)*log((25*exp(x)+25*x)*l 
og(x)^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3)^2+((6*exp(x)*x+6 
*x)*log(x)+(6*x^2+12*x+12)*exp(x)+18*x^2+12*x)*log((25*exp(x)+25*x)*log(x) 
^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3))/((2*exp(x)+2*x)*log( 
x)+2*exp(x)*x+2*x^2),x, algorithm="fricas")
 

Output:

3/2*x*log(25*x^3 + 25*x^2*e^x + 25*(x + e^x)*log(x)^2 + 50*(x^2 + x*e^x)*l 
og(x))^2
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((3*exp(x)+3*x)*ln(x)+3*exp(x)*x+3*x**2)*ln((25*exp(x)+25*x)*ln 
(x)**2+(50*exp(x)*x+50*x**2)*ln(x)+25*exp(x)*x**2+25*x**3)**2+((6*exp(x)*x 
+6*x)*ln(x)+(6*x**2+12*x+12)*exp(x)+18*x**2+12*x)*ln((25*exp(x)+25*x)*ln(x 
)**2+(50*exp(x)*x+50*x**2)*ln(x)+25*exp(x)*x**2+25*x**3))/((2*exp(x)+2*x)* 
ln(x)+2*exp(x)*x+2*x**2),x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (18) = 36\).

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.71 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=6 \, x \log \left (5\right )^{2} + \frac {3}{2} \, x \log \left (x + e^{x}\right )^{2} + 12 \, x \log \left (5\right ) \log \left (x + \log \left (x\right )\right ) + 6 \, x \log \left (x + \log \left (x\right )\right )^{2} + 6 \, {\left (x \log \left (5\right ) + x \log \left (x + \log \left (x\right )\right )\right )} \log \left (x + e^{x}\right ) \] Input:

integrate((((3*exp(x)+3*x)*log(x)+3*exp(x)*x+3*x^2)*log((25*exp(x)+25*x)*l 
og(x)^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3)^2+((6*exp(x)*x+6 
*x)*log(x)+(6*x^2+12*x+12)*exp(x)+18*x^2+12*x)*log((25*exp(x)+25*x)*log(x) 
^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3))/((2*exp(x)+2*x)*log( 
x)+2*exp(x)*x+2*x^2),x, algorithm="maxima")
 

Output:

6*x*log(5)^2 + 3/2*x*log(x + e^x)^2 + 12*x*log(5)*log(x + log(x)) + 6*x*lo 
g(x + log(x))^2 + 6*(x*log(5) + x*log(x + log(x)))*log(x + e^x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (18) = 36\).

Time = 1.40 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\frac {3}{2} \, x \log \left (25 \, x^{3} + 25 \, x^{2} e^{x} + 50 \, x^{2} \log \left (x\right ) + 50 \, x e^{x} \log \left (x\right ) + 25 \, x \log \left (x\right )^{2} + 25 \, e^{x} \log \left (x\right )^{2}\right )^{2} \] Input:

integrate((((3*exp(x)+3*x)*log(x)+3*exp(x)*x+3*x^2)*log((25*exp(x)+25*x)*l 
og(x)^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3)^2+((6*exp(x)*x+6 
*x)*log(x)+(6*x^2+12*x+12)*exp(x)+18*x^2+12*x)*log((25*exp(x)+25*x)*log(x) 
^2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3))/((2*exp(x)+2*x)*log( 
x)+2*exp(x)*x+2*x^2),x, algorithm="giac")
 

Output:

3/2*x*log(25*x^3 + 25*x^2*e^x + 50*x^2*log(x) + 50*x*e^x*log(x) + 25*x*log 
(x)^2 + 25*e^x*log(x)^2)^2
 

Mupad [B] (verification not implemented)

Time = 3.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\frac {3\,x\,{\ln \left (25\,x^2\,{\mathrm {e}}^x+{\ln \left (x\right )}^2\,\left (25\,x+25\,{\mathrm {e}}^x\right )+\ln \left (x\right )\,\left (50\,x\,{\mathrm {e}}^x+50\,x^2\right )+25\,x^3\right )}^2}{2} \] Input:

int((log(25*x^2*exp(x) + log(x)^2*(25*x + 25*exp(x)) + log(x)*(50*x*exp(x) 
 + 50*x^2) + 25*x^3)*(12*x + log(x)*(6*x + 6*x*exp(x)) + exp(x)*(12*x + 6* 
x^2 + 12) + 18*x^2) + log(25*x^2*exp(x) + log(x)^2*(25*x + 25*exp(x)) + lo 
g(x)*(50*x*exp(x) + 50*x^2) + 25*x^3)^2*(3*x*exp(x) + log(x)*(3*x + 3*exp( 
x)) + 3*x^2))/(2*x*exp(x) + log(x)*(2*x + 2*exp(x)) + 2*x^2),x)
 

Output:

(3*x*log(25*x^2*exp(x) + log(x)^2*(25*x + 25*exp(x)) + log(x)*(50*x*exp(x) 
 + 50*x^2) + 25*x^3)^2)/2
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.43 \[ \int \frac {\left (12 x+18 x^2+e^x \left (12+12 x+6 x^2\right )+\left (6 x+6 e^x x\right ) \log (x)\right ) \log \left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )+\left (3 e^x x+3 x^2+\left (3 e^x+3 x\right ) \log (x)\right ) \log ^2\left (25 e^x x^2+25 x^3+\left (50 e^x x+50 x^2\right ) \log (x)+\left (25 e^x+25 x\right ) \log ^2(x)\right )}{2 e^x x+2 x^2+\left (2 e^x+2 x\right ) \log (x)} \, dx=\frac {3 \mathrm {log}\left (25 e^{x} \mathrm {log}\left (x \right )^{2}+50 e^{x} \mathrm {log}\left (x \right ) x +25 e^{x} x^{2}+25 \mathrm {log}\left (x \right )^{2} x +50 \,\mathrm {log}\left (x \right ) x^{2}+25 x^{3}\right )^{2} x}{2} \] Input:

int((((3*exp(x)+3*x)*log(x)+3*exp(x)*x+3*x^2)*log((25*exp(x)+25*x)*log(x)^ 
2+(50*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3)^2+((6*exp(x)*x+6*x)*lo 
g(x)+(6*x^2+12*x+12)*exp(x)+18*x^2+12*x)*log((25*exp(x)+25*x)*log(x)^2+(50 
*exp(x)*x+50*x^2)*log(x)+25*exp(x)*x^2+25*x^3))/((2*exp(x)+2*x)*log(x)+2*e 
xp(x)*x+2*x^2),x)
 

Output:

(3*log(25*e**x*log(x)**2 + 50*e**x*log(x)*x + 25*e**x*x**2 + 25*log(x)**2* 
x + 50*log(x)*x**2 + 25*x**3)**2*x)/2