\(\int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} (16-10 e^{10}-5 x+10 \log (25 x))}{20-20 x+5 x^2} \, dx\) [788]

Optimal result

Integrand size = 54, antiderivative size = 26 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\frac {x \left (e^{10}-\frac {6}{5 x}-\log (25 x)\right )}{-2+x}} \] Output:

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

Mathematica [A] (verified)

Time = 2.70 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=5^{-1+\frac {2+x}{2-x}} e^{\frac {6-5 e^{10} x}{10-5 x}} x^{-\frac {x}{-2+x}} \] Input:

Integrate[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 
 5*x + 10*Log[25*x]))/(20 - 20*x + 5*x^2),x]
 

Output:

(5^(-1 + (2 + x)/(2 - x))*E^((6 - 5*E^10*x)/(10 - 5*x)))/x^(x/(-2 + 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.

\(\displaystyle \int \frac {e^{\frac {5 e^{10} x-5 x \log (25 x)-6}{5 x-10}} \left (-5 x+10 \log (25 x)-10 e^{10}+16\right )}{5 x^2-20 x+20} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 20 \int \frac {25^{\frac {5 x}{10-5 x}-1} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {5 x}{10-5 x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{4 (2-x)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 5 \int \frac {25^{\frac {x}{2-x}-1} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{(2-x)^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle 5 \int \frac {25^{-\frac {2 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} x^{\frac {x}{2-x}} \left (-5 x+10 \log (25 x)+2 \left (8-5 e^{10}\right )\right )}{(2-x)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 5 \int \left (\frac {2\ 5^{1-\frac {4 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} \log (25 x) x^{\frac {x}{2-x}}}{(x-2)^2}+\frac {25^{-\frac {2 (x-1)}{x-2}} e^{\frac {6-5 e^{10} x}{5 (2-x)}} \left (-5 x-10 e^{10}+16\right ) x^{\frac {x}{2-x}}}{(x-2)^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(E^((-6 + 5*E^10*x - 5*x*Log[25*x])/(-10 + 5*x))*(16 - 10*E^10 - 5*x + 
 10*Log[25*x]))/(20 - 20*x + 5*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
risch	\({\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}\)	\(23\)
parallelrisch	\({\mathrm e}^{-\frac {-5 x \,{\mathrm e}^{10}+5 x \ln \left (25 x \right )+6}{5 \left (-2+x \right )}}\)	\(25\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}-2 \,{\mathrm e}^{\frac {-5 x \ln \left (25 x \right )+5 x \,{\mathrm e}^{10}-6}{5 x -10}}}{-2+x}\)	\(62\)

Input:

int((10*ln(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)^2-6)/(5 
*x-10))/(5*x^2-20*x+20),x,method=_RETURNVERBOSE)
 

Output:

exp(1/5*(-5*x*ln(25*x)+5*x*exp(10)-6)/(-2+x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )} \] Input:

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) 
^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm="fricas")
 

Output:

e^(1/5*(5*x*e^10 - 5*x*log(25*x) - 6)/(x - 2))
 

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=e^{\frac {- 5 x \log {\left (25 x \right )} + 5 x e^{10} - 6}{5 x - 10}} \] Input:

integrate((10*ln(25*x)-10*exp(5)**2-5*x+16)*exp((-5*x*ln(25*x)+5*x*exp(5)* 
*2-6)/(5*x-10))/(5*x**2-20*x+20),x)
 

Output:

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

Maxima [F]

\[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\int { -\frac {{\left (5 \, x + 10 \, e^{10} - 10 \, \log \left (25 \, x\right ) - 16\right )} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )}}{5 \, {\left (x^{2} - 4 \, x + 4\right )}} \,d x } \] Input:

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) 
^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm="maxima")
 

Output:

-1/5*integrate((5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x* 
log(25*x) - 6)/(x - 2))/(x^2 - 4*x + 4), x)
 

Giac [F]

\[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\int { -\frac {{\left (5 \, x + 10 \, e^{10} - 10 \, \log \left (25 \, x\right ) - 16\right )} e^{\left (\frac {5 \, x e^{10} - 5 \, x \log \left (25 \, x\right ) - 6}{5 \, {\left (x - 2\right )}}\right )}}{5 \, {\left (x^{2} - 4 \, x + 4\right )}} \,d x } \] Input:

integrate((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5) 
^2-6)/(5*x-10))/(5*x^2-20*x+20),x, algorithm="giac")
 

Output:

integrate(-1/5*(5*x + 10*e^10 - 10*log(25*x) - 16)*e^(1/5*(5*x*e^10 - 5*x* 
log(25*x) - 6)/(x - 2))/(x^2 - 4*x + 4), x)
 

Mupad [B] (verification not implemented)

Time = 2.87 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{10}}{5\,x-10}}\,{\mathrm {e}}^{-\frac {6}{5\,x-10}}}{5^{\frac {10\,x}{5\,x-10}}\,x^{\frac {5\,x}{5\,x-10}}} \] Input:

int(-(exp(-(5*x*log(25*x) - 5*x*exp(10) + 6)/(5*x - 10))*(5*x - 10*log(25* 
x) + 10*exp(10) - 16))/(5*x^2 - 20*x + 20),x)
 

Output:

(exp((5*x*exp(10))/(5*x - 10))*exp(-6/(5*x - 10)))/(5^((10*x)/(5*x - 10))* 
x^((5*x)/(5*x - 10)))
 

Reduce [F]

\[ \int \frac {e^{\frac {-6+5 e^{10} x-5 x \log (25 x)}{-10+5 x}} \left (16-10 e^{10}-5 x+10 \log (25 x)\right )}{20-20 x+5 x^2} \, dx=\text {too large to display} \] Input:

int((10*log(25*x)-10*exp(5)^2-5*x+16)*exp((-5*x*log(25*x)+5*x*exp(5)^2-6)/ 
(5*x-10))/(5*x^2-20*x+20),x)
 

Output:

(e**(e**10)*(50*e**((2*e**10)/(x - 2))*e**10 - 25*e**((2*e**10)/(x - 2))*x 
 + 70*e**((2*e**10)/(x - 2)) - 100*e**((10*log(25*x) + 6)/(5*x - 10))*int( 
e**((2*e**10)/(x - 2))/(e**((10*log(25*x) + 6)/(5*x - 10))*x**3 - 4*e**((1 
0*log(25*x) + 6)/(5*x - 10))*x**2 + 4*e**((10*log(25*x) + 6)/(5*x - 10))*x 
),x)*e**20*x + 200*e**((10*log(25*x) + 6)/(5*x - 10))*int(e**((2*e**10)/(x 
 - 2))/(e**((10*log(25*x) + 6)/(5*x - 10))*x**3 - 4*e**((10*log(25*x) + 6) 
/(5*x - 10))*x**2 + 4*e**((10*log(25*x) + 6)/(5*x - 10))*x),x)*e**20 + 20* 
e**((10*log(25*x) + 6)/(5*x - 10))*int(e**((2*e**10)/(x - 2))/(e**((10*log 
(25*x) + 6)/(5*x - 10))*x**3 - 4*e**((10*log(25*x) + 6)/(5*x - 10))*x**2 + 
 4*e**((10*log(25*x) + 6)/(5*x - 10))*x),x)*e**10*x - 40*e**((10*log(25*x) 
 + 6)/(5*x - 10))*int(e**((2*e**10)/(x - 2))/(e**((10*log(25*x) + 6)/(5*x 
- 10))*x**3 - 4*e**((10*log(25*x) + 6)/(5*x - 10))*x**2 + 4*e**((10*log(25 
*x) + 6)/(5*x - 10))*x),x)*e**10 + 224*e**((10*log(25*x) + 6)/(5*x - 10))* 
int(e**((2*e**10)/(x - 2))/(e**((10*log(25*x) + 6)/(5*x - 10))*x**3 - 4*e* 
*((10*log(25*x) + 6)/(5*x - 10))*x**2 + 4*e**((10*log(25*x) + 6)/(5*x - 10 
))*x),x)*x - 448*e**((10*log(25*x) + 6)/(5*x - 10))*int(e**((2*e**10)/(x - 
 2))/(e**((10*log(25*x) + 6)/(5*x - 10))*x**3 - 4*e**((10*log(25*x) + 6)/( 
5*x - 10))*x**2 + 4*e**((10*log(25*x) + 6)/(5*x - 10))*x),x) - 50*e**((10* 
log(25*x) + 6)/(5*x - 10))*int(e**((2*e**10)/(x - 2))/(e**((10*log(25*x) + 
 6)/(5*x - 10))*x**2 - 4*e**((10*log(25*x) + 6)/(5*x - 10))*x + 4*e**((...