\(\int \frac {-30+54 x+e^{2-x} (15-12 x-12 x^2-3 x^3)+(-120 x-12 x^2+e^{2-x} (60 x+36 x^2+6 x^3)) \log (5+x)+(60 x+12 x^2+e^{2-x} (-30 x-21 x^2-3 x^3)) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx\) [419]

Optimal result

Integrand size = 134, antiderivative size = 28 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3 \left (x-(x-x \log (5+x))^2\right )}{-2+e^{2-x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3 e^x x \left (-1+x-2 x \log (5+x)+x \log ^2(5+x)\right )}{-e^2+2 e^x} \] Input:

Integrate[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 
 12*x^2 + E^(2 - x)*(60*x + 36*x^2 + 6*x^3))*Log[5 + x] + (60*x + 12*x^2 + 
 E^(2 - x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20 - 
4*x) + 4*x + E^(4 - 2*x)*(5 + x)),x]
 

Output:

(3*E^x*x*(-1 + x - 2*x*Log[5 + x] + x*Log[5 + x]^2))/(-E^2 + 2*E^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.

\(\displaystyle \int \frac {e^{2-x} \left (-3 x^3-12 x^2-12 x+15\right )+\left (12 x^2+e^{2-x} \left (-3 x^3-21 x^2-30 x\right )+60 x\right ) \log ^2(x+5)+\left (-12 x^2+e^{2-x} \left (6 x^3+36 x^2+60 x\right )-120 x\right ) \log (x+5)+54 x-30}{e^{2-x} (-4 x-20)+4 x+e^{4-2 x} (x+5)+20} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{2 x} \left (e^{2-x} \left (-3 x^3-12 x^2-12 x+15\right )+\left (12 x^2+e^{2-x} \left (-3 x^3-21 x^2-30 x\right )+60 x\right ) \log ^2(x+5)+\left (-12 x^2+e^{2-x} \left (6 x^3+36 x^2+60 x\right )-120 x\right ) \log (x+5)+54 x-30\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {3 e^{x-2} \left (x^3+x^3 \log ^2(x+5)-2 x^3 \log (x+5)+4 x^2+7 x^2 \log ^2(x+5)-12 x^2 \log (x+5)+4 x+10 x \log ^2(x+5)-20 x \log (x+5)-5\right )}{x+5}-\frac {6 e^{2 x-2} \left (x^3+x^3 \log ^2(x+5)-2 x^3 \log (x+5)+4 x^2+7 x^2 \log ^2(x+5)-12 x^2 \log (x+5)+4 x+10 x \log ^2(x+5)-20 x \log (x+5)-5\right )}{\left (e^2-2 e^x\right ) (x+5)}-\frac {6 e^{2 x} x \left (x+x \log ^2(x+5)-2 x \log (x+5)-1\right )}{\left (2 e^x-e^2\right )^2}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {3 e^x \left (2 x \left (e^2 \left (x^2+6 x+10\right )-2 e^x (x+10)\right ) \log (x+5)-e^2 \left (x^3+4 x^2+4 x-5\right )+2 e^x (9 x-5)-x (x+5) \left (e^2 (x+2)-4 e^x\right ) \log ^2(x+5)\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \int -\frac {e^x \left (-x (x+5) \left (4 e^x-e^2 (x+2)\right ) \log ^2(x+5)+2 x \left (2 e^x (x+10)-e^2 \left (x^2+6 x+10\right )\right ) \log (x+5)+2 e^x (5-9 x)-e^2 \left (-x^3-4 x^2-4 x+5\right )\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -3 \int \frac {e^x \left (-x (x+5) \left (4 e^x-e^2 (x+2)\right ) \log ^2(x+5)+2 x \left (2 e^x (x+10)-e^2 \left (x^2+6 x+10\right )\right ) \log (x+5)+2 e^x (5-9 x)-e^2 \left (-x^3-4 x^2-4 x+5\right )\right )}{\left (e^2-2 e^x\right )^2 (x+5)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -3 \int \left (\frac {e^{x+2} x \left (x \log ^2(x+5)-2 x \log (x+5)+x-1\right )}{\left (e^2-2 e^x\right )^2}-\frac {e^x \left (2 \log ^2(x+5) x^2-2 \log (x+5) x^2+10 \log ^2(x+5) x-20 \log (x+5) x+9 x-5\right )}{\left (-e^2+2 e^x\right ) (x+5)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \left (\int \frac {e^{x+2} x^2 \log ^2(x+5)}{\left (-e^2+2 e^x\right )^2}dx-2 \int \frac {e^{x+2} x^2 \log (x+5)}{\left (-e^2+2 e^x\right )^2}dx+50 \int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx+50 \int \frac {\int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx}{x+5}dx-2 \int \frac {e^x x \log ^2(x+5)}{-e^2+2 e^x}dx-50 \log (x+5) \int \frac {e^x}{\left (-e^2+2 e^x\right ) (x+5)}dx+10 \int \frac {e^x \log (x+5)}{-e^2+2 e^x}dx+2 \int \frac {e^x x \log (x+5)}{-e^2+2 e^x}dx+\operatorname {PolyLog}\left (2,2 e^{x-2}\right )+\frac {e^2 x^2}{2 \left (e^2-2 e^x\right )}-\frac {x^2}{2}-\frac {e^2 x}{2 \left (e^2-2 e^x\right )}+\frac {x}{2}+x \log \left (1-2 e^{x-2}\right )-5 \log \left (e^2-2 e^x\right )\right )\)

Input:

Int[(-30 + 54*x + E^(2 - x)*(15 - 12*x - 12*x^2 - 3*x^3) + (-120*x - 12*x^ 
2 + E^(2 - x)*(60*x + 36*x^2 + 6*x^3))*Log[5 + x] + (60*x + 12*x^2 + E^(2 
- x)*(-30*x - 21*x^2 - 3*x^3))*Log[5 + x]^2)/(20 + E^(2 - x)*(-20 - 4*x) + 
 4*x + E^(4 - 2*x)*(5 + x)),x]
 

Output:

$Aborted
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(57\) vs. \(2(27)=54\).

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.07

\[-\frac {3 x^{2} \ln \left (5+x \right )^{2}}{{\mathrm e}^{2-x}-2}+\frac {6 x^{2} \ln \left (5+x \right )}{{\mathrm e}^{2-x}-2}-\frac {3 x \left (-1+x \right )}{{\mathrm e}^{2-x}-2}\]

Input:

int((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*ln(5+x)^2+((6*x^3+36*x^2+ 
60*x)*exp(2-x)-12*x^2-120*x)*ln(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54*x 
-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x)
 

Output:

-3*x^2/(exp(2-x)-2)*ln(5+x)^2+6*x^2/(exp(2-x)-2)*ln(5+x)-3*x*(-1+x)/(exp(2 
-x)-2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} \log \left (x + 5\right )^{2} - 2 \, x^{2} \log \left (x + 5\right ) + x^{2} - x\right )}}{e^{\left (-x + 2\right )} - 2} \] Input:

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 
36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 
-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="fr 
icas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {- 3 x^{2} \log {\left (x + 5 \right )}^{2} + 6 x^{2} \log {\left (x + 5 \right )} - 3 x^{2} + 3 x}{e^{2 - x} - 2} \] Input:

integrate((((-3*x**3-21*x**2-30*x)*exp(2-x)+12*x**2+60*x)*ln(5+x)**2+((6*x 
**3+36*x**2+60*x)*exp(2-x)-12*x**2-120*x)*ln(5+x)+(-3*x**3-12*x**2-12*x+15 
)*exp(2-x)+54*x-30)/((5+x)*exp(2-x)**2+(-4*x-20)*exp(2-x)+20+4*x),x)
 

Output:

(-3*x**2*log(x + 5)**2 + 6*x**2*log(x + 5) - 3*x**2 + 3*x)/(exp(2 - x) - 2 
)
 

Maxima [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} e^{x} \log \left (x + 5\right )^{2} - 2 \, x^{2} e^{x} \log \left (x + 5\right ) + {\left (x^{2} - x\right )} e^{x}\right )}}{e^{2} - 2 \, e^{x}} \] Input:

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 
36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 
-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="ma 
xima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=-\frac {3 \, {\left (x^{2} \log \left (x + 5\right )^{2} - 2 \, x^{2} \log \left (x + 5\right ) + x^{2} - x\right )}}{e^{\left (-x + 2\right )} - 2} \] Input:

integrate((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+ 
36*x^2+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2 
-x)+54*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x, algorithm="gi 
ac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3\,x\,{\mathrm {e}}^{x-2}\,\left (x\,{\ln \left (x+5\right )}^2-2\,x\,\ln \left (x+5\right )+x-1\right )}{2\,{\mathrm {e}}^{x-2}-1} \] Input:

int(-(exp(2 - x)*(12*x + 12*x^2 + 3*x^3 - 15) - log(x + 5)^2*(60*x - exp(2 
 - x)*(30*x + 21*x^2 + 3*x^3) + 12*x^2) - 54*x + log(x + 5)*(120*x - exp(2 
 - x)*(60*x + 36*x^2 + 6*x^3) + 12*x^2) + 30)/(4*x - exp(2 - x)*(4*x + 20) 
 + exp(4 - 2*x)*(x + 5) + 20),x)
 

Output:

(3*x*exp(x - 2)*(x - 2*x*log(x + 5) + x*log(x + 5)^2 - 1))/(2*exp(x - 2) - 
 1)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-30+54 x+e^{2-x} \left (15-12 x-12 x^2-3 x^3\right )+\left (-120 x-12 x^2+e^{2-x} \left (60 x+36 x^2+6 x^3\right )\right ) \log (5+x)+\left (60 x+12 x^2+e^{2-x} \left (-30 x-21 x^2-3 x^3\right )\right ) \log ^2(5+x)}{20+e^{2-x} (-20-4 x)+4 x+e^{4-2 x} (5+x)} \, dx=\frac {3 e^{x} x \left (\mathrm {log}\left (x +5\right )^{2} x -2 \,\mathrm {log}\left (x +5\right ) x +x -1\right )}{2 e^{x}-e^{2}} \] Input:

int((((-3*x^3-21*x^2-30*x)*exp(2-x)+12*x^2+60*x)*log(5+x)^2+((6*x^3+36*x^2 
+60*x)*exp(2-x)-12*x^2-120*x)*log(5+x)+(-3*x^3-12*x^2-12*x+15)*exp(2-x)+54 
*x-30)/((5+x)*exp(2-x)^2+(-4*x-20)*exp(2-x)+20+4*x),x)
 

Output:

(3*e**x*x*(log(x + 5)**2*x - 2*log(x + 5)*x + x - 1))/(2*e**x - e**2)