\(\int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+(25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} (20+20 x-10 x^2)) \log (x)+(10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} (x+x^2-x^3)) \log ^2(x)+(e^{4 x} x^2+e^{5 x} x^2) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx\) [1851]

Optimal result

Integrand size = 178, antiderivative size = 32 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=\left (x+\frac {e^x x}{x+\frac {5 e^{-2 x}}{\log (x)}}\right ) \left (2-e^x+\log (x)\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=-\frac {x \left (-2+e^x-\log (x)\right ) \left (5+e^{2 x} \left (e^x+x\right ) \log (x)\right )}{5+e^{2 x} x \log (x)} \] Input:

Integrate[(75 + 10*E^(3*x) - 5*E^(4*x) + E^x*(-25 - 25*x) + (25 + E^(4*x)* 
(-5 - 20*x) + 30*E^(2*x)*x + E^(3*x)*(20 + 20*x - 10*x^2))*Log[x] + (10*E^ 
(2*x)*x + 3*E^(4*x)*x^2 - 2*E^(6*x)*x^2 + E^(3*x)*(5 + 15*x) + E^(5*x)*(x 
+ x^2 - x^3))*Log[x]^2 + (E^(4*x)*x^2 + E^(5*x)*x^2)*Log[x]^3)/(25 + 10*E^ 
(2*x)*x*Log[x] + E^(4*x)*x^2*Log[x]^2),x]
 

Output:

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

\(\displaystyle \int \frac {\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)+\left (e^{3 x} \left (-10 x^2+20 x+20\right )+e^{4 x} (-20 x-5)+30 e^{2 x} x+25\right ) \log (x)+\left (3 e^{4 x} x^2-2 e^{6 x} x^2+e^{5 x} \left (-x^3+x^2+x\right )+10 e^{2 x} x+e^{3 x} (15 x+5)\right ) \log ^2(x)+10 e^{3 x}-5 e^{4 x}+e^x (-25 x-25)+75}{e^{4 x} x^2 \log ^2(x)+10 e^{2 x} x \log (x)+25} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)+\left (e^{3 x} \left (-10 x^2+20 x+20\right )+e^{4 x} (-20 x-5)+30 e^{2 x} x+25\right ) \log (x)+\left (3 e^{4 x} x^2-2 e^{6 x} x^2+e^{5 x} \left (-x^3+x^2+x\right )+10 e^{2 x} x+e^{3 x} (15 x+5)\right ) \log ^2(x)+10 e^{3 x}-5 e^{4 x}+e^x (-25 x-25)+75}{\left (e^{2 x} x \log (x)+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {25 (2 x \log (x)+\log (x)+1) \left (e^x x \log ^2(x)+2 e^x x \log (x)+5\right )}{x^2 \log ^2(x) \left (e^{2 x} x \log (x)+5\right )^2}+\frac {x^2 \log ^3(x)+3 x^2 \log ^2(x)-5 \log (x)-5}{x^2 \log ^2(x)}+\frac {5 \left (e^x x^2 \log ^3(x)+2 e^x x^2 \log ^2(x)+e^x x \log ^3(x)+2 e^x x \log ^2(x)+2 e^x x \log (x)+10 x \log (x)+10 \log (x)+10\right )}{x^2 \log ^2(x) \left (e^{2 x} x \log (x)+5\right )}-\frac {e^x \left (x^2-x-x \log (x)-1\right )}{x}-2 e^{2 x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -125 \int \frac {1}{x^2 \log ^2(x) \left (e^{2 x} x \log (x)+5\right )^2}dx+50 \int \frac {1}{x^2 \log ^2(x) \left (e^{2 x} x \log (x)+5\right )}dx-125 \int \frac {1}{x^2 \log (x) \left (e^{2 x} x \log (x)+5\right )^2}dx+50 \int \frac {1}{x^2 \log (x) \left (e^{2 x} x \log (x)+5\right )}dx-100 \int \frac {e^x}{\left (e^{2 x} x \log (x)+5\right )^2}dx-75 \int \frac {e^x}{x \left (e^{2 x} x \log (x)+5\right )^2}dx-250 \int \frac {1}{x \log (x) \left (e^{2 x} x \log (x)+5\right )^2}dx-50 \int \frac {e^x}{x \log (x) \left (e^{2 x} x \log (x)+5\right )^2}dx-50 \int \frac {e^x \log (x)}{\left (e^{2 x} x \log (x)+5\right )^2}dx-25 \int \frac {e^x \log (x)}{x \left (e^{2 x} x \log (x)+5\right )^2}dx+10 \int \frac {e^x}{e^{2 x} x \log (x)+5}dx+10 \int \frac {e^x}{x \left (e^{2 x} x \log (x)+5\right )}dx+50 \int \frac {1}{x \log (x) \left (e^{2 x} x \log (x)+5\right )}dx+10 \int \frac {e^x}{x \log (x) \left (e^{2 x} x \log (x)+5\right )}dx+5 \int \frac {e^x \log (x)}{e^{2 x} x \log (x)+5}dx+5 \int \frac {e^x \log (x)}{x \left (e^{2 x} x \log (x)+5\right )}dx-e^x x+2 x+2 e^x-e^{2 x}+x \log (x)+e^x \log (x)+\frac {5}{x \log (x)}\)

Input:

Int[(75 + 10*E^(3*x) - 5*E^(4*x) + E^x*(-25 - 25*x) + (25 + E^(4*x)*(-5 - 
20*x) + 30*E^(2*x)*x + E^(3*x)*(20 + 20*x - 10*x^2))*Log[x] + (10*E^(2*x)* 
x + 3*E^(4*x)*x^2 - 2*E^(6*x)*x^2 + E^(3*x)*(5 + 15*x) + E^(5*x)*(x + x^2 
- x^3))*Log[x]^2 + (E^(4*x)*x^2 + E^(5*x)*x^2)*Log[x]^3)/(25 + 10*E^(2*x)* 
x*Log[x] + E^(4*x)*x^2*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(83\) vs. \(2(29)=58\).

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.62

\[\left ({\mathrm e}^{x}+x \right ) \ln \left (x \right )-\frac {\left ({\mathrm e}^{2 x} x^{2}+x \,{\mathrm e}^{3 x}-2 \,{\mathrm e}^{x} x^{2}-2 x \,{\mathrm e}^{2 x}+5\right ) {\mathrm e}^{-x}}{x}+\frac {5 \left (x \,{\mathrm e}^{3 x}-2 x \,{\mathrm e}^{2 x}+5\right ) {\mathrm e}^{-x}}{x \left (\ln \left (x \right ) {\mathrm e}^{2 x} x +5\right )}\]

Input:

int(((x^2*exp(x)^5+x^2*exp(x)^4)*ln(x)^3+(-2*x^2*exp(x)^6+(-x^3+x^2+x)*exp 
(x)^5+3*x^2*exp(x)^4+(15*x+5)*exp(x)^3+10*x*exp(x)^2)*ln(x)^2+((-20*x-5)*e 
xp(x)^4+(-10*x^2+20*x+20)*exp(x)^3+30*x*exp(x)^2+25)*ln(x)-5*exp(x)^4+10*e 
xp(x)^3+(-25*x-25)*exp(x)+75)/(x^2*exp(x)^4*ln(x)^2+10*x*exp(x)^2*ln(x)+25 
),x)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (30) = 60\).

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.44 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=\frac {{\left (x^{2} e^{\left (2 \, x\right )} + x e^{\left (3 \, x\right )}\right )} \log \left (x\right )^{2} - 5 \, x e^{x} + {\left (2 \, x^{2} e^{\left (2 \, x\right )} - x e^{\left (4 \, x\right )} - {\left (x^{2} - 2 \, x\right )} e^{\left (3 \, x\right )} + 5 \, x\right )} \log \left (x\right ) + 10 \, x}{x e^{\left (2 \, x\right )} \log \left (x\right ) + 5} \] Input:

integrate(((x^2*exp(x)^5+x^2*exp(x)^4)*log(x)^3+(-2*x^2*exp(x)^6+(-x^3+x^2 
+x)*exp(x)^5+3*x^2*exp(x)^4+(15*x+5)*exp(x)^3+10*x*exp(x)^2)*log(x)^2+((-2 
0*x-5)*exp(x)^4+(-10*x^2+20*x+20)*exp(x)^3+30*x*exp(x)^2+25)*log(x)-5*exp( 
x)^4+10*exp(x)^3+(-25*x-25)*exp(x)+75)/(x^2*exp(x)^4*log(x)^2+10*x*exp(x)^ 
2*log(x)+25),x, algorithm="fricas")
 

Output:

((x^2*e^(2*x) + x*e^(3*x))*log(x)^2 - 5*x*e^x + (2*x^2*e^(2*x) - x*e^(4*x) 
 - (x^2 - 2*x)*e^(3*x) + 5*x)*log(x) + 10*x)/(x*e^(2*x)*log(x) + 5)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=x \log {\left (x \right )} + 2 x + \frac {\left (- 5 x \log {\left (x \right )}^{2} - 10 x \log {\left (x \right )}\right ) e^{x} - 25}{x^{2} e^{2 x} \log {\left (x \right )}^{2} + 5 x \log {\left (x \right )}} + \left (- x + \log {\left (x \right )} + 2\right ) e^{x} - e^{2 x} + \frac {5}{x \log {\left (x \right )}} \] Input:

integrate(((x**2*exp(x)**5+x**2*exp(x)**4)*ln(x)**3+(-2*x**2*exp(x)**6+(-x 
**3+x**2+x)*exp(x)**5+3*x**2*exp(x)**4+(15*x+5)*exp(x)**3+10*x*exp(x)**2)* 
ln(x)**2+((-20*x-5)*exp(x)**4+(-10*x**2+20*x+20)*exp(x)**3+30*x*exp(x)**2+ 
25)*ln(x)-5*exp(x)**4+10*exp(x)**3+(-25*x-25)*exp(x)+75)/(x**2*exp(x)**4*l 
n(x)**2+10*x*exp(x)**2*ln(x)+25),x)
 

Output:

x*log(x) + 2*x + ((-5*x*log(x)**2 - 10*x*log(x))*exp(x) - 25)/(x**2*exp(2* 
x)*log(x)**2 + 5*x*log(x)) + (-x + log(x) + 2)*exp(x) - exp(2*x) + 5/(x*lo 
g(x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.56 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=-\frac {x e^{\left (4 \, x\right )} \log \left (x\right ) - {\left (x \log \left (x\right )^{2} - {\left (x^{2} - 2 \, x\right )} \log \left (x\right )\right )} e^{\left (3 \, x\right )} - {\left (x^{2} \log \left (x\right )^{2} + 2 \, x^{2} \log \left (x\right )\right )} e^{\left (2 \, x\right )} + 5 \, x e^{x} - 5 \, x \log \left (x\right ) - 10 \, x}{x e^{\left (2 \, x\right )} \log \left (x\right ) + 5} \] Input:

integrate(((x^2*exp(x)^5+x^2*exp(x)^4)*log(x)^3+(-2*x^2*exp(x)^6+(-x^3+x^2 
+x)*exp(x)^5+3*x^2*exp(x)^4+(15*x+5)*exp(x)^3+10*x*exp(x)^2)*log(x)^2+((-2 
0*x-5)*exp(x)^4+(-10*x^2+20*x+20)*exp(x)^3+30*x*exp(x)^2+25)*log(x)-5*exp( 
x)^4+10*exp(x)^3+(-25*x-25)*exp(x)+75)/(x^2*exp(x)^4*log(x)^2+10*x*exp(x)^ 
2*log(x)+25),x, algorithm="maxima")
 

Output:

-(x*e^(4*x)*log(x) - (x*log(x)^2 - (x^2 - 2*x)*log(x))*e^(3*x) - (x^2*log( 
x)^2 + 2*x^2*log(x))*e^(2*x) + 5*x*e^x - 5*x*log(x) - 10*x)/(x*e^(2*x)*log 
(x) + 5)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (30) = 60\).

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.78 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} \log \left (x\right )^{2} - x^{2} e^{\left (3 \, x\right )} \log \left (x\right ) + 2 \, x^{2} e^{\left (2 \, x\right )} \log \left (x\right ) + x e^{\left (3 \, x\right )} \log \left (x\right )^{2} - x e^{\left (4 \, x\right )} \log \left (x\right ) + 2 \, x e^{\left (3 \, x\right )} \log \left (x\right ) - 5 \, x e^{x} + 5 \, x \log \left (x\right ) + 10 \, x}{x e^{\left (2 \, x\right )} \log \left (x\right ) + 5} \] Input:

integrate(((x^2*exp(x)^5+x^2*exp(x)^4)*log(x)^3+(-2*x^2*exp(x)^6+(-x^3+x^2 
+x)*exp(x)^5+3*x^2*exp(x)^4+(15*x+5)*exp(x)^3+10*x*exp(x)^2)*log(x)^2+((-2 
0*x-5)*exp(x)^4+(-10*x^2+20*x+20)*exp(x)^3+30*x*exp(x)^2+25)*log(x)-5*exp( 
x)^4+10*exp(x)^3+(-25*x-25)*exp(x)+75)/(x^2*exp(x)^4*log(x)^2+10*x*exp(x)^ 
2*log(x)+25),x, algorithm="giac")
 

Output:

(x^2*e^(2*x)*log(x)^2 - x^2*e^(3*x)*log(x) + 2*x^2*e^(2*x)*log(x) + x*e^(3 
*x)*log(x)^2 - x*e^(4*x)*log(x) + 2*x*e^(3*x)*log(x) - 5*x*e^x + 5*x*log(x 
) + 10*x)/(x*e^(2*x)*log(x) + 5)
 

Mupad [B] (verification not implemented)

Time = 3.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.16 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=2\,x-{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (x-2\right )-\frac {5\,{\mathrm {e}}^{-x}}{x}+\ln \left (x\right )\,\left (x+{\mathrm {e}}^x\right )+\frac {5\,\left (25\,{\mathrm {e}}^x-15\,x\,{\mathrm {e}}^{3\,x}+5\,x\,{\mathrm {e}}^{4\,x}-20\,x^2\,{\mathrm {e}}^{3\,x}+10\,x^2\,{\mathrm {e}}^{4\,x}+2\,x^2\,{\mathrm {e}}^{5\,x}-x^2\,{\mathrm {e}}^{6\,x}+50\,x\,{\mathrm {e}}^x\right )}{x\,\left (x\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right )+5\right )\,\left (5\,{\mathrm {e}}^{2\,x}+10\,x\,{\mathrm {e}}^{2\,x}-x\,{\mathrm {e}}^{4\,x}\right )} \] Input:

int((10*exp(3*x) - 5*exp(4*x) + log(x)^2*(10*x*exp(2*x) + 3*x^2*exp(4*x) - 
 2*x^2*exp(6*x) + exp(5*x)*(x + x^2 - x^3) + exp(3*x)*(15*x + 5)) - exp(x) 
*(25*x + 25) + log(x)*(exp(3*x)*(20*x - 10*x^2 + 20) + 30*x*exp(2*x) - exp 
(4*x)*(20*x + 5) + 25) + log(x)^3*(x^2*exp(4*x) + x^2*exp(5*x)) + 75)/(x^2 
*exp(4*x)*log(x)^2 + 10*x*exp(2*x)*log(x) + 25),x)
 

Output:

2*x - exp(2*x) - exp(x)*(x - 2) - (5*exp(-x))/x + log(x)*(x + exp(x)) + (5 
*(25*exp(x) - 15*x*exp(3*x) + 5*x*exp(4*x) - 20*x^2*exp(3*x) + 10*x^2*exp( 
4*x) + 2*x^2*exp(5*x) - x^2*exp(6*x) + 50*x*exp(x)))/(x*(x*exp(2*x)*log(x) 
 + 5)*(5*exp(2*x) + 10*x*exp(2*x) - x*exp(4*x)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.66 \[ \int \frac {75+10 e^{3 x}-5 e^{4 x}+e^x (-25-25 x)+\left (25+e^{4 x} (-5-20 x)+30 e^{2 x} x+e^{3 x} \left (20+20 x-10 x^2\right )\right ) \log (x)+\left (10 e^{2 x} x+3 e^{4 x} x^2-2 e^{6 x} x^2+e^{3 x} (5+15 x)+e^{5 x} \left (x+x^2-x^3\right )\right ) \log ^2(x)+\left (e^{4 x} x^2+e^{5 x} x^2\right ) \log ^3(x)}{25+10 e^{2 x} x \log (x)+e^{4 x} x^2 \log ^2(x)} \, dx=\frac {x \left (-e^{4 x} \mathrm {log}\left (x \right )+e^{3 x} \mathrm {log}\left (x \right )^{2}-e^{3 x} \mathrm {log}\left (x \right ) x +2 e^{3 x} \mathrm {log}\left (x \right )+e^{2 x} \mathrm {log}\left (x \right )^{2} x +2 e^{2 x} \mathrm {log}\left (x \right ) x -5 e^{x}+5 \,\mathrm {log}\left (x \right )+10\right )}{e^{2 x} \mathrm {log}\left (x \right ) x +5} \] Input:

int(((x^2*exp(x)^5+x^2*exp(x)^4)*log(x)^3+(-2*x^2*exp(x)^6+(-x^3+x^2+x)*ex 
p(x)^5+3*x^2*exp(x)^4+(15*x+5)*exp(x)^3+10*x*exp(x)^2)*log(x)^2+((-20*x-5) 
*exp(x)^4+(-10*x^2+20*x+20)*exp(x)^3+30*x*exp(x)^2+25)*log(x)-5*exp(x)^4+1 
0*exp(x)^3+(-25*x-25)*exp(x)+75)/(x^2*exp(x)^4*log(x)^2+10*x*exp(x)^2*log( 
x)+25),x)
 

Output:

(x*( - e**(4*x)*log(x) + e**(3*x)*log(x)**2 - e**(3*x)*log(x)*x + 2*e**(3* 
x)*log(x) + e**(2*x)*log(x)**2*x + 2*e**(2*x)*log(x)*x - 5*e**x + 5*log(x) 
 + 10))/(e**(2*x)*log(x)*x + 5)