\(\int \frac {e^8 (5+8 x-2 x^2)+(5+8 x-2 x^2) \log (5)}{-15 x^3+3 x^4+(-30 x^2+6 x^3) \log (e^{5+x} (5 x-x^2))+(-15 x+3 x^2) \log ^2(e^{5+x} (5 x-x^2))} \, dx\) [1527]

Optimal result

Integrand size = 98, antiderivative size = 27 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^8+\log (5)}{3 \left (x+\log \left (e^{5+x} (5-x) x\right )\right )} \] Output:

1/3/(ln((5-x)*x*exp(5+x))+x)*(exp(4)^2+ln(5))
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^8+\log (5)}{3 \left (x+\log \left (-e^{5+x} (-5+x) x\right )\right )} \] Input:

Integrate[(E^8*(5 + 8*x - 2*x^2) + (5 + 8*x - 2*x^2)*Log[5])/(-15*x^3 + 3* 
x^4 + (-30*x^2 + 6*x^3)*Log[E^(5 + x)*(5*x - x^2)] + (-15*x + 3*x^2)*Log[E 
^(5 + x)*(5*x - x^2)]^2),x]
 

Output:

(E^8 + Log[5])/(3*(x + Log[-(E^(5 + x)*(-5 + 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.

\(\displaystyle \int \frac {e^8 \left (-2 x^2+8 x+5\right )+\left (-2 x^2+8 x+5\right ) \log (5)}{3 x^4-15 x^3+\left (3 x^2-15 x\right ) \log ^2\left (e^{x+5} \left (5 x-x^2\right )\right )+\left (6 x^3-30 x^2\right ) \log \left (e^{x+5} \left (5 x-x^2\right )\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (-2 x^2+8 x+5\right ) \left (e^8+\log (5)\right )}{3 x^4-15 x^3+\left (3 x^2-15 x\right ) \log ^2\left (e^{x+5} \left (5 x-x^2\right )\right )+\left (6 x^3-30 x^2\right ) \log \left (e^{x+5} \left (5 x-x^2\right )\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \left (e^8+\log (5)\right ) \int -\frac {-2 x^2+8 x+5}{3 \left (-x^4+5 x^3+\left (5 x-x^2\right ) \log ^2\left (e^{x+5} \left (5 x-x^2\right )\right )+2 \left (5 x^2-x^3\right ) \log \left (e^{x+5} \left (5 x-x^2\right )\right )\right )}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{3} \left (e^8+\log (5)\right ) \int \frac {-2 x^2+8 x+5}{-x^4+5 x^3+\left (5 x-x^2\right ) \log ^2\left (e^{x+5} \left (5 x-x^2\right )\right )+2 \left (5 x^2-x^3\right ) \log \left (e^{x+5} \left (5 x-x^2\right )\right )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle -\frac {1}{3} \left (e^8+\log (5)\right ) \int \frac {-2 x^2+8 x+5}{(5-x) x \left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{3} \left (e^8+\log (5)\right ) \int \left (\frac {1}{x \left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}+\frac {2}{\left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}+\frac {1}{\left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2 (x-5)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{3} \left (e^8+\log (5)\right ) \left (2 \int \frac {1}{\left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}dx+\int \frac {1}{(x-5) \left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}dx+\int \frac {1}{x \left (x+\log \left (-e^{x+5} (x-5) x\right )\right )^2}dx\right )\)

Input:

Int[(E^8*(5 + 8*x - 2*x^2) + (5 + 8*x - 2*x^2)*Log[5])/(-15*x^3 + 3*x^4 + 
(-30*x^2 + 6*x^3)*Log[E^(5 + x)*(5*x - x^2)] + (-15*x + 3*x^2)*Log[E^(5 + 
x)*(5*x - x^2)]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\ln \left (5\right )+{\mathrm e}^{8}}{3 x +3 \ln \left (\left (-x^{2}+5 x \right ) {\mathrm e}^{5+x}\right )}\)	\(29\)
risch	\(\frac {2 i \ln \left (5\right )}{3 \left (\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )-\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{3}+2 \pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2} \operatorname {csgn}\left (i x \right )-2 \pi +2 i \ln \left (x \right )+2 i \ln \left (-5+x \right )+2 i \ln \left ({\mathrm e}^{5+x}\right )+2 i x \right )}+\frac {2 i {\mathrm e}^{8}}{3 \left (\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )-\pi \,\operatorname {csgn}\left (i \left (-5+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}+\pi \,\operatorname {csgn}\left (i {\mathrm e}^{5+x} \left (-5+x \right )\right ) \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{3}+2 \pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2}-\pi \operatorname {csgn}\left (i x \left (-5+x \right ) {\mathrm e}^{5+x}\right )^{2} \operatorname {csgn}\left (i x \right )-2 \pi +2 i \ln \left (x \right )+2 i \ln \left (-5+x \right )+2 i \ln \left ({\mathrm e}^{5+x}\right )+2 i x \right )}\)	\(480\)

Input:

int(((-2*x^2+8*x+5)*ln(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*ln((-x^2+ 
5*x)*exp(5+x))^2+(6*x^3-30*x^2)*ln((-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x,me 
thod=_RETURNVERBOSE)
 

Output:

1/3*(exp(4)^2+ln(5))/(x+ln((-x^2+5*x)*exp(5+x)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (x + \log \left (-{\left (x^{2} - 5 \, x\right )} e^{\left (x + 5\right )}\right )\right )}} \] Input:

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*lo 
g((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15* 
x^3),x, algorithm="fricas")
 

Output:

1/3*(e^8 + log(5))/(x + log(-(x^2 - 5*x)*e^(x + 5)))
 

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {\log {\left (5 \right )} + e^{8}}{3 x + 3 \log {\left (\left (- x^{2} + 5 x\right ) e^{x + 5} \right )}} \] Input:

integrate(((-2*x**2+8*x+5)*ln(5)+(-2*x**2+8*x+5)*exp(4)**2)/((3*x**2-15*x) 
*ln((-x**2+5*x)*exp(5+x))**2+(6*x**3-30*x**2)*ln((-x**2+5*x)*exp(5+x))+3*x 
**4-15*x**3),x)
 

Output:

(log(5) + exp(8))/(3*x + 3*log((-x**2 + 5*x)*exp(x + 5)))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (2 \, x + \log \left (x\right ) + \log \left (-x + 5\right ) + 5\right )}} \] Input:

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*lo 
g((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15* 
x^3),x, algorithm="maxima")
 

Output:

1/3*(e^8 + log(5))/(2*x + log(x) + log(-x + 5) + 5)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {e^{8} + \log \left (5\right )}{3 \, {\left (2 \, x + \log \left (-x^{2} + 5 \, x\right ) + 5\right )}} \] Input:

integrate(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*lo 
g((-x^2+5*x)*exp(5+x))^2+(6*x^3-30*x^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15* 
x^3),x, algorithm="giac")
 

Output:

1/3*(e^8 + log(5))/(2*x + log(-x^2 + 5*x) + 5)
 

Mupad [B] (verification not implemented)

Time = 3.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {\frac {{\mathrm {e}}^8}{3}+\frac {\ln \left (5\right )}{3}}{x+\ln \left ({\mathrm {e}}^5\,{\mathrm {e}}^x\,\left (5\,x-x^2\right )\right )} \] Input:

int(-(exp(8)*(8*x - 2*x^2 + 5) + log(5)*(8*x - 2*x^2 + 5))/(log(exp(x + 5) 
*(5*x - x^2))^2*(15*x - 3*x^2) + log(exp(x + 5)*(5*x - x^2))*(30*x^2 - 6*x 
^3) + 15*x^3 - 3*x^4),x)
 

Output:

(exp(8)/3 + log(5)/3)/(x + log(exp(5)*exp(x)*(5*x - x^2)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {e^8 \left (5+8 x-2 x^2\right )+\left (5+8 x-2 x^2\right ) \log (5)}{-15 x^3+3 x^4+\left (-30 x^2+6 x^3\right ) \log \left (e^{5+x} \left (5 x-x^2\right )\right )+\left (-15 x+3 x^2\right ) \log ^2\left (e^{5+x} \left (5 x-x^2\right )\right )} \, dx=\frac {\mathrm {log}\left (5\right )+e^{8}}{3 \,\mathrm {log}\left (-e^{x} e^{5} x^{2}+5 e^{x} e^{5} x \right )+3 x} \] Input:

int(((-2*x^2+8*x+5)*log(5)+(-2*x^2+8*x+5)*exp(4)^2)/((3*x^2-15*x)*log((-x^ 
2+5*x)*exp(5+x))^2+(6*x^3-30*x^2)*log((-x^2+5*x)*exp(5+x))+3*x^4-15*x^3),x 
)
 

Output:

(log(5) + e**8)/(3*(log( - e**x*e**5*x**2 + 5*e**x*e**5*x) + x))