3.2.2 \(\int \frac {-5000 x^2+1250 e^{5+x} x^2+(15000 x^2-3700 e^{5+x} x^2) \log (x)+(10000 x-2450 e^{5+x} x) \log ^2(x)+\log (\frac {1}{4} (4-e^{5+x})) (200 x-50 e^{5+x} x+(-400 x+100 e^{5+x} x) \log (x)+(-200+50 e^{5+x}) \log ^2(x))}{-2500 x^6+625 e^{5+x} x^6+(-5000 x^5+1250 e^{5+x} x^5) \log (x)+(-2500 x^4+625 e^{5+x} x^4) \log ^2(x)+\log ^2(\frac {1}{4} (4-e^{5+x})) (-4 x^4+e^{5+x} x^4+(-8 x^3+2 e^{5+x} x^3) \log (x)+(-4 x^2+e^{5+x} x^2) \log ^2(x))+\log (\frac {1}{4} (4-e^{5+x})) (200 x^5-50 e^{5+x} x^5+(400 x^4-100 e^{5+x} x^4) \log (x)+(200 x^3-50 e^{5+x} x^3) \log ^2(x))} \, dx\) [102]

3.2.2.1 Optimal result
3.2.2.2 Mathematica [A] (verified)
3.2.2.3 Rubi [F]
3.2.2.4 Maple [A] (verified)
3.2.2.5 Fricas [A] (verification not implemented)
3.2.2.6 Sympy [A] (verification not implemented)
3.2.2.7 Maxima [A] (verification not implemented)
3.2.2.8 Giac [B] (verification not implemented)
3.2.2.9 Mupad [B] (verification not implemented)

3.2.2.1 Optimal result

Integrand size = 309, antiderivative size = 35 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=\frac {2 \log (x)}{x \left (x-\frac {1}{25} \log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right )\right ) (x+\log (x))} \]

output
ln(x)/x/(x+ln(x))/(1/2*x-1/50*ln(-1/4*exp(5+x)+1))
 
3.2.2.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=-\frac {50 \log (x)}{x \left (-25 x+\log \left (1-\frac {e^{5+x}}{4}\right )\right ) (x+\log (x))} \]

input
Integrate[(-5000*x^2 + 1250*E^(5 + x)*x^2 + (15000*x^2 - 3700*E^(5 + x)*x^ 
2)*Log[x] + (10000*x - 2450*E^(5 + x)*x)*Log[x]^2 + Log[(4 - E^(5 + x))/4] 
*(200*x - 50*E^(5 + x)*x + (-400*x + 100*E^(5 + x)*x)*Log[x] + (-200 + 50* 
E^(5 + x))*Log[x]^2))/(-2500*x^6 + 625*E^(5 + x)*x^6 + (-5000*x^5 + 1250*E 
^(5 + x)*x^5)*Log[x] + (-2500*x^4 + 625*E^(5 + x)*x^4)*Log[x]^2 + Log[(4 - 
 E^(5 + x))/4]^2*(-4*x^4 + E^(5 + x)*x^4 + (-8*x^3 + 2*E^(5 + x)*x^3)*Log[ 
x] + (-4*x^2 + E^(5 + x)*x^2)*Log[x]^2) + Log[(4 - E^(5 + x))/4]*(200*x^5 
- 50*E^(5 + x)*x^5 + (400*x^4 - 100*E^(5 + x)*x^4)*Log[x] + (200*x^3 - 50* 
E^(5 + x)*x^3)*Log[x]^2)),x]
 
output
(-50*Log[x])/(x*(-25*x + Log[1 - E^(5 + x)/4])*(x + Log[x]))
 
3.2.2.3 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.

\(\displaystyle \int \frac {1250 e^{x+5} x^2-5000 x^2+\left (15000 x^2-3700 e^{x+5} x^2\right ) \log (x)+\left (10000 x-2450 e^{x+5} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right ) \left (-50 e^{x+5} x+200 x+\left (50 e^{x+5}-200\right ) \log ^2(x)+\left (100 e^{x+5} x-400 x\right ) \log (x)\right )}{625 e^{x+5} x^6-2500 x^6+\left (1250 e^{x+5} x^5-5000 x^5\right ) \log (x)+\left (625 e^{x+5} x^4-2500 x^4\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right ) \left (-50 e^{x+5} x^5+200 x^5+\left (400 x^4-100 e^{x+5} x^4\right ) \log (x)+\left (200 x^3-50 e^{x+5} x^3\right ) \log ^2(x)\right )+\log ^2\left (\frac {1}{4} \left (4-e^{x+5}\right )\right ) \left (e^{x+5} x^4-4 x^4+\left (2 e^{x+5} x^3-8 x^3\right ) \log (x)+\left (e^{x+5} x^2-4 x^2\right ) \log ^2(x)\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {50 x \left (-25 \left (e^{x+5}-4\right ) x+\left (49 e^{x+5}-200\right ) \log ^2(x)+2 \left (37 e^{x+5}-150\right ) x \log (x)\right )-50 \left (e^{x+5}-4\right ) \log \left (1-\frac {e^{x+5}}{4}\right ) \left (-x+\log ^2(x)+2 x \log (x)\right )}{\left (4-e^{x+5}\right ) x^2 \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {50 \log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right ) \left (-x+\log ^2(x)+2 x \log (x)\right )}{x^2 \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}-\frac {2450 \log ^2(x)}{x \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}+\frac {200 \log (x)}{\left (e^{x+5}-4\right ) x \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))}-\frac {50 (74 \log (x)-25)}{\left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 50 \int \frac {\log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right )}{x^2 \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2}dx-2450 \int \frac {1}{x \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2}dx+1250 \int \frac {1}{\left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}dx+1250 \int \frac {x}{\left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}dx-50 \int \frac {\log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right )}{\left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}dx-50 \int \frac {\log \left (\frac {1}{4} \left (4-e^{x+5}\right )\right )}{x \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))^2}dx+1200 \int \frac {1}{\left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))}dx+200 \int \frac {\log (x)}{\left (-4+e^{x+5}\right ) x \left (25 x-\log \left (1-\frac {e^{x+5}}{4}\right )\right )^2 (x+\log (x))}dx\)

input
Int[(-5000*x^2 + 1250*E^(5 + x)*x^2 + (15000*x^2 - 3700*E^(5 + x)*x^2)*Log 
[x] + (10000*x - 2450*E^(5 + x)*x)*Log[x]^2 + Log[(4 - E^(5 + x))/4]*(200* 
x - 50*E^(5 + x)*x + (-400*x + 100*E^(5 + x)*x)*Log[x] + (-200 + 50*E^(5 + 
 x))*Log[x]^2))/(-2500*x^6 + 625*E^(5 + x)*x^6 + (-5000*x^5 + 1250*E^(5 + 
x)*x^5)*Log[x] + (-2500*x^4 + 625*E^(5 + x)*x^4)*Log[x]^2 + Log[(4 - E^(5 
+ x))/4]^2*(-4*x^4 + E^(5 + x)*x^4 + (-8*x^3 + 2*E^(5 + x)*x^3)*Log[x] + ( 
-4*x^2 + E^(5 + x)*x^2)*Log[x]^2) + Log[(4 - E^(5 + x))/4]*(200*x^5 - 50*E 
^(5 + x)*x^5 + (400*x^4 - 100*E^(5 + x)*x^4)*Log[x] + (200*x^3 - 50*E^(5 + 
 x)*x^3)*Log[x]^2)),x]
 
output
$Aborted
 

3.2.2.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.2.2.4 Maple [A] (verified)

Time = 28.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89

method result size
risch \(\frac {50 \ln \left (x \right )}{\left (x +\ln \left (x \right )\right ) x \left (25 x -\ln \left (-\frac {{\mathrm e}^{5+x}}{4}+1\right )\right )}\) \(31\)
parallelrisch \(\frac {50 \ln \left (x \right )}{x \left (25 x^{2}+25 x \ln \left (x \right )-x \ln \left (-\frac {{\mathrm e}^{5+x}}{4}+1\right )-\ln \left (-\frac {{\mathrm e}^{5+x}}{4}+1\right ) \ln \left (x \right )\right )}\) \(46\)

input
int((((50*exp(5+x)-200)*ln(x)^2+(100*x*exp(5+x)-400*x)*ln(x)-50*x*exp(5+x) 
+200*x)*ln(-1/4*exp(5+x)+1)+(-2450*x*exp(5+x)+10000*x)*ln(x)^2+(-3700*x^2* 
exp(5+x)+15000*x^2)*ln(x)+1250*x^2*exp(5+x)-5000*x^2)/(((x^2*exp(5+x)-4*x^ 
2)*ln(x)^2+(2*x^3*exp(5+x)-8*x^3)*ln(x)+x^4*exp(5+x)-4*x^4)*ln(-1/4*exp(5+ 
x)+1)^2+((-50*x^3*exp(5+x)+200*x^3)*ln(x)^2+(-100*x^4*exp(5+x)+400*x^4)*ln 
(x)-50*x^5*exp(5+x)+200*x^5)*ln(-1/4*exp(5+x)+1)+(625*x^4*exp(5+x)-2500*x^ 
4)*ln(x)^2+(1250*x^5*exp(5+x)-5000*x^5)*ln(x)+625*x^6*exp(5+x)-2500*x^6),x 
,method=_RETURNVERBOSE)
 
output
50*ln(x)/(x+ln(x))/x/(25*x-ln(-1/4*exp(5+x)+1))
 
3.2.2.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=\frac {50 \, \log \left (x\right )}{25 \, x^{3} + 25 \, x^{2} \log \left (x\right ) - {\left (x^{2} + x \log \left (x\right )\right )} \log \left (-\frac {1}{4} \, e^{\left (x + 5\right )} + 1\right )} \]

input
integrate((((50*exp(5+x)-200)*log(x)^2+(100*x*exp(5+x)-400*x)*log(x)-50*x* 
exp(5+x)+200*x)*log(-1/4*exp(5+x)+1)+(-2450*x*exp(5+x)+10000*x)*log(x)^2+( 
-3700*x^2*exp(5+x)+15000*x^2)*log(x)+1250*x^2*exp(5+x)-5000*x^2)/(((x^2*ex 
p(5+x)-4*x^2)*log(x)^2+(2*x^3*exp(5+x)-8*x^3)*log(x)+x^4*exp(5+x)-4*x^4)*l 
og(-1/4*exp(5+x)+1)^2+((-50*x^3*exp(5+x)+200*x^3)*log(x)^2+(-100*x^4*exp(5 
+x)+400*x^4)*log(x)-50*x^5*exp(5+x)+200*x^5)*log(-1/4*exp(5+x)+1)+(625*x^4 
*exp(5+x)-2500*x^4)*log(x)^2+(1250*x^5*exp(5+x)-5000*x^5)*log(x)+625*x^6*e 
xp(5+x)-2500*x^6),x, algorithm=\
 
output
50*log(x)/(25*x^3 + 25*x^2*log(x) - (x^2 + x*log(x))*log(-1/4*e^(x + 5) + 
1))
 
3.2.2.6 Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=- \frac {50 \log {\left (x \right )}}{- 25 x^{3} - 25 x^{2} \log {\left (x \right )} + \left (x^{2} + x \log {\left (x \right )}\right ) \log {\left (1 - \frac {e^{x + 5}}{4} \right )}} \]

input
integrate((((50*exp(5+x)-200)*ln(x)**2+(100*x*exp(5+x)-400*x)*ln(x)-50*x*e 
xp(5+x)+200*x)*ln(-1/4*exp(5+x)+1)+(-2450*x*exp(5+x)+10000*x)*ln(x)**2+(-3 
700*x**2*exp(5+x)+15000*x**2)*ln(x)+1250*x**2*exp(5+x)-5000*x**2)/(((x**2* 
exp(5+x)-4*x**2)*ln(x)**2+(2*x**3*exp(5+x)-8*x**3)*ln(x)+x**4*exp(5+x)-4*x 
**4)*ln(-1/4*exp(5+x)+1)**2+((-50*x**3*exp(5+x)+200*x**3)*ln(x)**2+(-100*x 
**4*exp(5+x)+400*x**4)*ln(x)-50*x**5*exp(5+x)+200*x**5)*ln(-1/4*exp(5+x)+1 
)+(625*x**4*exp(5+x)-2500*x**4)*ln(x)**2+(1250*x**5*exp(5+x)-5000*x**5)*ln 
(x)+625*x**6*exp(5+x)-2500*x**6),x)
 
output
-50*log(x)/(-25*x**3 - 25*x**2*log(x) + (x**2 + x*log(x))*log(1 - exp(x + 
5)/4))
 
3.2.2.7 Maxima [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=\frac {50 \, \log \left (x\right )}{25 \, x^{3} + 2 \, x^{2} \log \left (2\right ) + {\left (25 \, x^{2} + 2 \, x \log \left (2\right )\right )} \log \left (x\right ) - {\left (x^{2} + x \log \left (x\right )\right )} \log \left (-e^{\left (x + 5\right )} + 4\right )} \]

input
integrate((((50*exp(5+x)-200)*log(x)^2+(100*x*exp(5+x)-400*x)*log(x)-50*x* 
exp(5+x)+200*x)*log(-1/4*exp(5+x)+1)+(-2450*x*exp(5+x)+10000*x)*log(x)^2+( 
-3700*x^2*exp(5+x)+15000*x^2)*log(x)+1250*x^2*exp(5+x)-5000*x^2)/(((x^2*ex 
p(5+x)-4*x^2)*log(x)^2+(2*x^3*exp(5+x)-8*x^3)*log(x)+x^4*exp(5+x)-4*x^4)*l 
og(-1/4*exp(5+x)+1)^2+((-50*x^3*exp(5+x)+200*x^3)*log(x)^2+(-100*x^4*exp(5 
+x)+400*x^4)*log(x)-50*x^5*exp(5+x)+200*x^5)*log(-1/4*exp(5+x)+1)+(625*x^4 
*exp(5+x)-2500*x^4)*log(x)^2+(1250*x^5*exp(5+x)-5000*x^5)*log(x)+625*x^6*e 
xp(5+x)-2500*x^6),x, algorithm=\
 
output
50*log(x)/(25*x^3 + 2*x^2*log(2) + (25*x^2 + 2*x*log(2))*log(x) - (x^2 + x 
*log(x))*log(-e^(x + 5) + 4))
 
3.2.2.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).

Time = 0.67 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.74 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx=\frac {50 \, \log \left (x\right )}{25 \, x^{3} + 2 \, x^{2} \log \left (2\right ) + 25 \, x^{2} \log \left (x\right ) + 2 \, x \log \left (2\right ) \log \left (x\right ) - x^{2} \log \left (-e^{\left (x + 5\right )} + 4\right ) - x \log \left (x\right ) \log \left (-e^{\left (x + 5\right )} + 4\right )} \]

input
integrate((((50*exp(5+x)-200)*log(x)^2+(100*x*exp(5+x)-400*x)*log(x)-50*x* 
exp(5+x)+200*x)*log(-1/4*exp(5+x)+1)+(-2450*x*exp(5+x)+10000*x)*log(x)^2+( 
-3700*x^2*exp(5+x)+15000*x^2)*log(x)+1250*x^2*exp(5+x)-5000*x^2)/(((x^2*ex 
p(5+x)-4*x^2)*log(x)^2+(2*x^3*exp(5+x)-8*x^3)*log(x)+x^4*exp(5+x)-4*x^4)*l 
og(-1/4*exp(5+x)+1)^2+((-50*x^3*exp(5+x)+200*x^3)*log(x)^2+(-100*x^4*exp(5 
+x)+400*x^4)*log(x)-50*x^5*exp(5+x)+200*x^5)*log(-1/4*exp(5+x)+1)+(625*x^4 
*exp(5+x)-2500*x^4)*log(x)^2+(1250*x^5*exp(5+x)-5000*x^5)*log(x)+625*x^6*e 
xp(5+x)-2500*x^6),x, algorithm=\
 
output
50*log(x)/(25*x^3 + 2*x^2*log(2) + 25*x^2*log(x) + 2*x*log(2)*log(x) - x^2 
*log(-e^(x + 5) + 4) - x*log(x)*log(-e^(x + 5) + 4))
 
3.2.2.9 Mupad [B] (verification not implemented)

Time = 14.59 (sec) , antiderivative size = 785, normalized size of antiderivative = 22.43 \[ \int \frac {-5000 x^2+1250 e^{5+x} x^2+\left (15000 x^2-3700 e^{5+x} x^2\right ) \log (x)+\left (10000 x-2450 e^{5+x} x\right ) \log ^2(x)+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x-50 e^{5+x} x+\left (-400 x+100 e^{5+x} x\right ) \log (x)+\left (-200+50 e^{5+x}\right ) \log ^2(x)\right )}{-2500 x^6+625 e^{5+x} x^6+\left (-5000 x^5+1250 e^{5+x} x^5\right ) \log (x)+\left (-2500 x^4+625 e^{5+x} x^4\right ) \log ^2(x)+\log ^2\left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (-4 x^4+e^{5+x} x^4+\left (-8 x^3+2 e^{5+x} x^3\right ) \log (x)+\left (-4 x^2+e^{5+x} x^2\right ) \log ^2(x)\right )+\log \left (\frac {1}{4} \left (4-e^{5+x}\right )\right ) \left (200 x^5-50 e^{5+x} x^5+\left (400 x^4-100 e^{5+x} x^4\right ) \log (x)+\left (200 x^3-50 e^{5+x} x^3\right ) \log ^2(x)\right )} \, dx =\text {Too large to display} \]

input
int((log(1 - exp(x + 5)/4)*(200*x - 50*x*exp(x + 5) - log(x)*(400*x - 100* 
x*exp(x + 5)) + log(x)^2*(50*exp(x + 5) - 200)) - log(x)*(3700*x^2*exp(x + 
 5) - 15000*x^2) + 1250*x^2*exp(x + 5) - 5000*x^2 + log(x)^2*(10000*x - 24 
50*x*exp(x + 5)))/(log(1 - exp(x + 5)/4)^2*(log(x)*(2*x^3*exp(x + 5) - 8*x 
^3) + x^4*exp(x + 5) + log(x)^2*(x^2*exp(x + 5) - 4*x^2) - 4*x^4) + log(x) 
*(1250*x^5*exp(x + 5) - 5000*x^5) + 625*x^6*exp(x + 5) + log(x)^2*(625*x^4 
*exp(x + 5) - 2500*x^4) - log(1 - exp(x + 5)/4)*(log(x)*(100*x^4*exp(x + 5 
) - 400*x^4) + 50*x^5*exp(x + 5) + log(x)^2*(50*x^3*exp(x + 5) - 200*x^3) 
- 200*x^5) - 2500*x^6),x)
 
output
((25*(100*x - 25*x*exp(x + 5) - 200*log(x)^2 - 300*x*log(x) + 49*exp(x + 5 
)*log(x)^2 + 74*x*exp(x + 5)*log(x)))/(2*x*(x + log(x))^2*(6*exp(x + 5) - 
25)) - (25*log(1 - (exp(5)*exp(x))/4)*(exp(x + 5) - 4)*(log(x)^2 - x + 2*x 
*log(x)))/(2*x^2*(x + log(x))^2*(6*exp(x + 5) - 25)))/(25*x - log(1 - (exp 
(5)*exp(x))/4)) - ((25*(1825*exp(x + 5) - 7500*x - 444*exp(2*x + 10) + 36* 
exp(3*x + 15) + 5450*x*exp(x + 5) - 1326*x*exp(2*x + 10) + 108*x*exp(3*x + 
 15) + 5375*x^2*exp(x + 5) - 75*x^3*exp(x + 5) + 25*x^4*exp(x + 5) + 25*x^ 
5*exp(x + 5) - 1308*x^2*exp(2*x + 10) + 30*x^3*exp(2*x + 10) + 18*x^4*exp( 
2*x + 10) + 6*x^5*exp(2*x + 10) + 108*x^2*exp(3*x + 15) - 7500*x^2 - 2500) 
)/(4*x*(x + 1)^3*(6*exp(x + 5) - 25)^3) + (25*log(x)*(1825*exp(x + 5) - 50 
00*x - 444*exp(2*x + 10) + 36*exp(3*x + 15) + 3625*x*exp(x + 5) - 882*x*ex 
p(2*x + 10) + 72*x*exp(3*x + 15) + 50*x^3*exp(x + 5) + 25*x^4*exp(x + 5) + 
 12*x^2*exp(2*x + 10) + 12*x^3*exp(2*x + 10) + 6*x^4*exp(2*x + 10) - 2500) 
)/(4*x*(x + 1)^3*(6*exp(x + 5) - 25)^3))/(x + log(x)) + ((25*(500*x - 98*e 
xp(x + 5) + 12*exp(2*x + 10) - 245*x*exp(x + 5) + 30*x*exp(2*x + 10) - 97* 
x^2*exp(x + 5) + x^3*exp(x + 5) + 12*x^2*exp(2*x + 10) + 200*x^2 + 200))/( 
4*x*(x + 1)*(6*exp(x + 5) - 25)^2) + (25*log(x)*(6*exp(2*x + 10) - 49*exp( 
x + 5) + x*exp(x + 5) + x^2*exp(x + 5) + 100))/(4*x*(x + 1)*(6*exp(x + 5) 
- 25)^2))/(log(x)^2 + 2*x*log(x) + x^2) - ((125*x)/24 + (25*x^2)/6 + (25*x 
^3)/12 + 25/12)/(x^2 + 3*x^3 + 3*x^4 + x^5) + (15625*(x^4 + 3*x^5 + 3*x...