\(\int \frac {e^x (125+220 x-90 x^2+100 x^3+e^2 (-10 x+20 x^2))+e^x (125+125 x-20 x^2+10 e^2 x^2+50 x^3) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 (400 x+96 x^2+160 x^3)+(2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 (400 x+96 x^2+160 x^3)) \log (x)+(625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 (100 x+24 x^2+40 x^3)) \log ^2(x)} \, dx\) [2211]

Optimal result

Integrand size = 211, antiderivative size = 33 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {e^x}{2 \left (\frac {1}{5} \left (3+e^2\right )+\frac {5}{2 x}+x\right ) (2+\log (x))} \] Output:

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

Mathematica [A] (verified)

Time = 3.96 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 e^x x}{\left (25+6 x+2 e^2 x+10 x^2\right ) (2+\log (x))} \] Input:

Integrate[(E^x*(125 + 220*x - 90*x^2 + 100*x^3 + E^2*(-10*x + 20*x^2)) + E 
^x*(125 + 125*x - 20*x^2 + 10*E^2*x^2 + 50*x^3)*Log[x])/(2500 + 1200*x + 2 
144*x^2 + 16*E^4*x^2 + 480*x^3 + 400*x^4 + E^2*(400*x + 96*x^2 + 160*x^3) 
+ (2500 + 1200*x + 2144*x^2 + 16*E^4*x^2 + 480*x^3 + 400*x^4 + E^2*(400*x 
+ 96*x^2 + 160*x^3))*Log[x] + (625 + 300*x + 536*x^2 + 4*E^4*x^2 + 120*x^3 
 + 100*x^4 + E^2*(100*x + 24*x^2 + 40*x^3))*Log[x]^2),x]
 

Output:

(5*E^x*x)/((25 + 6*x + 2*E^2*x + 10*x^2)*(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.

\(\displaystyle \int \frac {e^x \left (100 x^3-90 x^2+e^2 \left (20 x^2-10 x\right )+220 x+125\right )+e^x \left (50 x^3+10 e^2 x^2-20 x^2+125 x+125\right ) \log (x)}{400 x^4+480 x^3+16 e^4 x^2+2144 x^2+e^2 \left (160 x^3+96 x^2+400 x\right )+\left (100 x^4+120 x^3+4 e^4 x^2+536 x^2+e^2 \left (40 x^3+24 x^2+100 x\right )+300 x+625\right ) \log ^2(x)+\left (400 x^4+480 x^3+16 e^4 x^2+2144 x^2+e^2 \left (160 x^3+96 x^2+400 x\right )+1200 x+2500\right ) \log (x)+1200 x+2500} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^x \left (100 x^3-90 x^2+e^2 \left (20 x^2-10 x\right )+220 x+125\right )+e^x \left (50 x^3+10 e^2 x^2-20 x^2+125 x+125\right ) \log (x)}{400 x^4+480 x^3+\left (2144+16 e^4\right ) x^2+e^2 \left (160 x^3+96 x^2+400 x\right )+\left (100 x^4+120 x^3+4 e^4 x^2+536 x^2+e^2 \left (40 x^3+24 x^2+100 x\right )+300 x+625\right ) \log ^2(x)+\left (400 x^4+480 x^3+16 e^4 x^2+2144 x^2+e^2 \left (160 x^3+96 x^2+400 x\right )+1200 x+2500\right ) \log (x)+1200 x+2500}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {5 e^x \left (20 x^3+2 \left (2 e^2-9\right ) x^2+\left (10 x^3+2 \left (e^2-2\right ) x^2+25 x+25\right ) \log (x)-2 \left (e^2-22\right ) x+25\right )}{\left (10 x^2+2 \left (3+e^2\right ) x+25\right )^2 (\log (x)+2)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \int \frac {e^x \left (20 x^3-2 \left (9-2 e^2\right ) x^2+2 \left (22-e^2\right ) x+\left (10 x^3-2 \left (2-e^2\right ) x^2+25 x+25\right ) \log (x)+25\right )}{\left (10 x^2+2 \left (3+e^2\right ) x+25\right )^2 (\log (x)+2)^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle 5 \left (50 \int \frac {e^x}{\left (10 x^2+2 \left (3+e^2\right ) x+25\right )^2 (\log (x)+2)}dx+2 \left (3+e^2\right ) \int \frac {e^x x}{\left (10 x^2+2 \left (3+e^2\right ) x+25\right )^2 (\log (x)+2)}dx-\frac {10 i \int \frac {e^x}{\left (-20 x+2 i \sqrt {241-6 e^2-e^4}-2 e^2-6\right ) (\log (x)+2)^2}dx}{\sqrt {241-6 e^2-e^4}}-\frac {10 i \int \frac {e^x}{\left (20 x+2 i \sqrt {241-6 e^2-e^4}+2 e^2+6\right ) (\log (x)+2)^2}dx}{\sqrt {241-6 e^2-e^4}}-\frac {10 i \int \frac {e^x}{\left (-20 x+2 i \sqrt {241-6 e^2-e^4}-2 e^2-6\right ) (\log (x)+2)}dx}{\sqrt {241-6 e^2-e^4}}+\left (1+\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right ) \int \frac {e^x}{\left (20 x-2 i \sqrt {241-6 e^2-e^4}+2 \left (3+e^2\right )\right ) (\log (x)+2)}dx-\frac {10 i \int \frac {e^x}{\left (20 x+2 i \sqrt {241-6 e^2-e^4}+2 e^2+6\right ) (\log (x)+2)}dx}{\sqrt {241-6 e^2-e^4}}+\left (1-\frac {i \left (3+e^2\right )}{\sqrt {241-6 e^2-e^4}}\right ) \int \frac {e^x}{\left (20 x+2 i \sqrt {241-6 e^2-e^4}+2 \left (3+e^2\right )\right ) (\log (x)+2)}dx\right )\)

Input:

Int[(E^x*(125 + 220*x - 90*x^2 + 100*x^3 + E^2*(-10*x + 20*x^2)) + E^x*(12 
5 + 125*x - 20*x^2 + 10*E^2*x^2 + 50*x^3)*Log[x])/(2500 + 1200*x + 2144*x^ 
2 + 16*E^4*x^2 + 480*x^3 + 400*x^4 + E^2*(400*x + 96*x^2 + 160*x^3) + (250 
0 + 1200*x + 2144*x^2 + 16*E^4*x^2 + 480*x^3 + 400*x^4 + E^2*(400*x + 96*x 
^2 + 160*x^3))*Log[x] + (625 + 300*x + 536*x^2 + 4*E^4*x^2 + 120*x^3 + 100 
*x^4 + E^2*(100*x + 24*x^2 + 40*x^3))*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {5 x \,{\mathrm e}^{x}}{\left (2 \,{\mathrm e}^{2} x +10 x^{2}+6 x +25\right ) \left (\ln \left (x \right )+2\right )}\)	\(29\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{x}}{\left (2 \,{\mathrm e}^{2} x +10 x^{2}+6 x +25\right ) \left (\ln \left (x \right )+2\right )}\)	\(29\)

Input:

int(((10*x^2*exp(2)+50*x^3-20*x^2+125*x+125)*exp(x)*ln(x)+((20*x^2-10*x)*e 
xp(2)+100*x^3-90*x^2+220*x+125)*exp(x))/((4*x^2*exp(2)^2+(40*x^3+24*x^2+10 
0*x)*exp(2)+100*x^4+120*x^3+536*x^2+300*x+625)*ln(x)^2+(16*x^2*exp(2)^2+(1 
60*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200*x+2500)*ln(x)+16 
*x^2*exp(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200* 
x+2500),x,method=_RETURNVERBOSE)
 

Output:

5*x*exp(x)/(2*exp(2)*x+10*x^2+6*x+25)/(ln(x)+2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 \, x e^{x}}{20 \, x^{2} + 4 \, x e^{2} + {\left (10 \, x^{2} + 2 \, x e^{2} + 6 \, x + 25\right )} \log \left (x\right ) + 12 \, x + 50} \] Input:

integrate(((10*x^2*exp(2)+50*x^3-20*x^2+125*x+125)*exp(x)*log(x)+((20*x^2- 
10*x)*exp(2)+100*x^3-90*x^2+220*x+125)*exp(x))/((4*x^2*exp(2)^2+(40*x^3+24 
*x^2+100*x)*exp(2)+100*x^4+120*x^3+536*x^2+300*x+625)*log(x)^2+(16*x^2*exp 
(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200*x+2500)* 
log(x)+16*x^2*exp(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144* 
x^2+1200*x+2500),x, algorithm="fricas")
 

Output:

5*x*e^x/(20*x^2 + 4*x*e^2 + (10*x^2 + 2*x*e^2 + 6*x + 25)*log(x) + 12*x + 
50)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (24) = 48\).

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 x e^{x}}{10 x^{2} \log {\left (x \right )} + 20 x^{2} + 6 x \log {\left (x \right )} + 2 x e^{2} \log {\left (x \right )} + 12 x + 4 x e^{2} + 25 \log {\left (x \right )} + 50} \] Input:

integrate(((10*x**2*exp(2)+50*x**3-20*x**2+125*x+125)*exp(x)*ln(x)+((20*x* 
*2-10*x)*exp(2)+100*x**3-90*x**2+220*x+125)*exp(x))/((4*x**2*exp(2)**2+(40 
*x**3+24*x**2+100*x)*exp(2)+100*x**4+120*x**3+536*x**2+300*x+625)*ln(x)**2 
+(16*x**2*exp(2)**2+(160*x**3+96*x**2+400*x)*exp(2)+400*x**4+480*x**3+2144 
*x**2+1200*x+2500)*ln(x)+16*x**2*exp(2)**2+(160*x**3+96*x**2+400*x)*exp(2) 
+400*x**4+480*x**3+2144*x**2+1200*x+2500),x)
 

Output:

5*x*exp(x)/(10*x**2*log(x) + 20*x**2 + 6*x*log(x) + 2*x*exp(2)*log(x) + 12 
*x + 4*x*exp(2) + 25*log(x) + 50)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 \, x e^{x}}{20 \, x^{2} + 4 \, x {\left (e^{2} + 3\right )} + {\left (10 \, x^{2} + 2 \, x {\left (e^{2} + 3\right )} + 25\right )} \log \left (x\right ) + 50} \] Input:

integrate(((10*x^2*exp(2)+50*x^3-20*x^2+125*x+125)*exp(x)*log(x)+((20*x^2- 
10*x)*exp(2)+100*x^3-90*x^2+220*x+125)*exp(x))/((4*x^2*exp(2)^2+(40*x^3+24 
*x^2+100*x)*exp(2)+100*x^4+120*x^3+536*x^2+300*x+625)*log(x)^2+(16*x^2*exp 
(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200*x+2500)* 
log(x)+16*x^2*exp(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144* 
x^2+1200*x+2500),x, algorithm="maxima")
 

Output:

5*x*e^x/(20*x^2 + 4*x*(e^2 + 3) + (10*x^2 + 2*x*(e^2 + 3) + 25)*log(x) + 5 
0)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 \, x e^{x}}{10 \, x^{2} \log \left (x\right ) + 2 \, x e^{2} \log \left (x\right ) + 20 \, x^{2} + 4 \, x e^{2} + 6 \, x \log \left (x\right ) + 12 \, x + 25 \, \log \left (x\right ) + 50} \] Input:

integrate(((10*x^2*exp(2)+50*x^3-20*x^2+125*x+125)*exp(x)*log(x)+((20*x^2- 
10*x)*exp(2)+100*x^3-90*x^2+220*x+125)*exp(x))/((4*x^2*exp(2)^2+(40*x^3+24 
*x^2+100*x)*exp(2)+100*x^4+120*x^3+536*x^2+300*x+625)*log(x)^2+(16*x^2*exp 
(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200*x+2500)* 
log(x)+16*x^2*exp(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144* 
x^2+1200*x+2500),x, algorithm="giac")
 

Output:

5*x*e^x/(10*x^2*log(x) + 2*x*e^2*log(x) + 20*x^2 + 4*x*e^2 + 6*x*log(x) + 
12*x + 25*log(x) + 50)
 

Mupad [B] (verification not implemented)

Time = 2.99 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5\,x\,{\mathrm {e}}^x}{\left (\ln \left (x\right )+2\right )\,\left (6\,x+2\,x\,{\mathrm {e}}^2+10\,x^2+25\right )} \] Input:

int((exp(x)*(220*x - exp(2)*(10*x - 20*x^2) - 90*x^2 + 100*x^3 + 125) + ex 
p(x)*log(x)*(125*x + 10*x^2*exp(2) - 20*x^2 + 50*x^3 + 125))/(1200*x + log 
(x)*(1200*x + exp(2)*(400*x + 96*x^2 + 160*x^3) + 16*x^2*exp(4) + 2144*x^2 
 + 480*x^3 + 400*x^4 + 2500) + exp(2)*(400*x + 96*x^2 + 160*x^3) + log(x)^ 
2*(300*x + exp(2)*(100*x + 24*x^2 + 40*x^3) + 4*x^2*exp(4) + 536*x^2 + 120 
*x^3 + 100*x^4 + 625) + 16*x^2*exp(4) + 2144*x^2 + 480*x^3 + 400*x^4 + 250 
0),x)
 

Output:

(5*x*exp(x))/((log(x) + 2)*(6*x + 2*x*exp(2) + 10*x^2 + 25))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {e^x \left (125+220 x-90 x^2+100 x^3+e^2 \left (-10 x+20 x^2\right )\right )+e^x \left (125+125 x-20 x^2+10 e^2 x^2+50 x^3\right ) \log (x)}{2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )+\left (2500+1200 x+2144 x^2+16 e^4 x^2+480 x^3+400 x^4+e^2 \left (400 x+96 x^2+160 x^3\right )\right ) \log (x)+\left (625+300 x+536 x^2+4 e^4 x^2+120 x^3+100 x^4+e^2 \left (100 x+24 x^2+40 x^3\right )\right ) \log ^2(x)} \, dx=\frac {5 e^{x} x}{2 \,\mathrm {log}\left (x \right ) e^{2} x +10 \,\mathrm {log}\left (x \right ) x^{2}+6 \,\mathrm {log}\left (x \right ) x +25 \,\mathrm {log}\left (x \right )+4 e^{2} x +20 x^{2}+12 x +50} \] Input:

int(((10*x^2*exp(2)+50*x^3-20*x^2+125*x+125)*exp(x)*log(x)+((20*x^2-10*x)* 
exp(2)+100*x^3-90*x^2+220*x+125)*exp(x))/((4*x^2*exp(2)^2+(40*x^3+24*x^2+1 
00*x)*exp(2)+100*x^4+120*x^3+536*x^2+300*x+625)*log(x)^2+(16*x^2*exp(2)^2+ 
(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+1200*x+2500)*log(x) 
+16*x^2*exp(2)^2+(160*x^3+96*x^2+400*x)*exp(2)+400*x^4+480*x^3+2144*x^2+12 
00*x+2500),x)
 

Output:

(5*e**x*x)/(2*log(x)*e**2*x + 10*log(x)*x**2 + 6*log(x)*x + 25*log(x) + 4* 
e**2*x + 20*x**2 + 12*x + 50)