\(\int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (240-60 x-15 x^2)+(120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (-60 x+15 x^2)) \log (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x)}{(8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} (-4 x+x^2)) \log ^2(-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x)} \, dx\) [1643]

Optimal result

Integrand size = 171, antiderivative size = 25 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 x}{\log \left (-2+e^{-\frac {2 (2+x)}{x}} (-4+x)-x\right )} \] Output:

15*x/ln(exp(ln(-4+x)-2*(2+x)/x)-2-x)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 x}{\log \left (-2+e^{-2-\frac {4}{x}} (-4+x)-x\right )} \] Input:

Integrate[(-60*x^2 + 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(240 - 60*x 
 - 15*x^2) + (120*x + 30*x^2 - 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*( 
-60*x + 15*x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x])/((8*x + 2 
*x^2 - x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-4*x + x^2))*Log[-2 + E^((- 
4 - 2*x + x*Log[-4 + x])/x) - x]^2),x]
 

Output:

(15*x)/Log[-2 + E^(-2 - 4/x)*(-4 + 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.

\(\displaystyle \int \frac {15 x^3-60 x^2+\left (-15 x^2-60 x+240\right ) e^{\frac {-2 x+x \log (x-4)-4}{x}}+\left (-15 x^3+30 x^2+\left (15 x^2-60 x\right ) e^{\frac {-2 x+x \log (x-4)-4}{x}}+120 x\right ) \log \left (-x+e^{\frac {-2 x+x \log (x-4)-4}{x}}-2\right )}{\left (-x^3+2 x^2+\left (x^2-4 x\right ) e^{\frac {-2 x+x \log (x-4)-4}{x}}+8 x\right ) \log ^2\left (-x+e^{\frac {-2 x+x \log (x-4)-4}{x}}-2\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {30 e^{\frac {4}{x}+2} \left (5 x^2-4 x-16\right )}{(x-4) x \left (e^{\frac {4}{x}+2} x-x+2 e^{\frac {4}{x}+2}+4\right ) \log ^2\left (e^{-\frac {4}{x}-2} (x-4)-x-2\right )}+\frac {15 \left (-x^2+x^2 \log \left (e^{-\frac {4}{x}-2} (x-4)-x-2\right )-4 x-4 x \log \left (e^{-\frac {4}{x}-2} (x-4)-x-2\right )+16\right )}{(x-4) x \log ^2\left (e^{-\frac {4}{x}-2} (x-4)-x-2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -15 \int \frac {1}{\log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx-60 \int \frac {1}{(x-4) \log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx-60 \int \frac {1}{x \log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx+150 \int \frac {e^{2+\frac {4}{x}}}{\left (e^{2+\frac {4}{x}} x-x+2 e^{2+\frac {4}{x}}+4\right ) \log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx+360 \int \frac {e^{2+\frac {4}{x}}}{(x-4) \left (e^{2+\frac {4}{x}} x-x+2 e^{2+\frac {4}{x}}+4\right ) \log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx+120 \int \frac {e^{2+\frac {4}{x}}}{x \left (e^{2+\frac {4}{x}} x-x+2 e^{2+\frac {4}{x}}+4\right ) \log ^2\left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx+15 \int \frac {1}{\log \left (e^{-2-\frac {4}{x}} (x-4)-x-2\right )}dx\)

Input:

Int[(-60*x^2 + 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(240 - 60*x - 15* 
x^2) + (120*x + 30*x^2 - 15*x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-60*x 
+ 15*x^2))*Log[-2 + E^((-4 - 2*x + x*Log[-4 + x])/x) - x])/((8*x + 2*x^2 - 
 x^3 + E^((-4 - 2*x + x*Log[-4 + x])/x)*(-4*x + x^2))*Log[-2 + E^((-4 - 2* 
x + x*Log[-4 + x])/x) - x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 9.70 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {15 x}{\ln \left (\left (x -4\right ) {\mathrm e}^{-\frac {2 \left (2+x \right )}{x}}-x -2\right )}\)	\(25\)
parallelrisch	\(\frac {15 x}{\ln \left ({\mathrm e}^{\frac {x \ln \left (x -4\right )-2 x -4}{x}}-x -2\right )}\)	\(28\)

Input:

int((((15*x^2-60*x)*exp((x*ln(x-4)-2*x-4)/x)-15*x^3+30*x^2+120*x)*ln(exp(( 
x*ln(x-4)-2*x-4)/x)-x-2)+(-15*x^2-60*x+240)*exp((x*ln(x-4)-2*x-4)/x)+15*x^ 
3-60*x^2)/((x^2-4*x)*exp((x*ln(x-4)-2*x-4)/x)-x^3+2*x^2+8*x)/ln(exp((x*ln( 
x-4)-2*x-4)/x)-x-2)^2,x,method=_RETURNVERBOSE)
 

Output:

15*x/ln((x-4)*exp(-2*(2+x)/x)-x-2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 \, x}{\log \left (-x + e^{\left (\frac {x \log \left (x - 4\right ) - 2 \, x - 4}{x}\right )} - 2\right )} \] Input:

integrate((((15*x^2-60*x)*exp((x*log(-4+x)-2*x-4)/x)-15*x^3+30*x^2+120*x)* 
log(exp((x*log(-4+x)-2*x-4)/x)-x-2)+(-15*x^2-60*x+240)*exp((x*log(-4+x)-2* 
x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(-4+x)-2*x-4)/x)-x^3+2*x^2+8*x 
)/log(exp((x*log(-4+x)-2*x-4)/x)-x-2)^2,x, algorithm="fricas")
 

Output:

15*x/log(-x + e^((x*log(x - 4) - 2*x - 4)/x) - 2)
 

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 x}{\log {\left (- x + e^{\frac {x \log {\left (x - 4 \right )} - 2 x - 4}{x}} - 2 \right )}} \] Input:

integrate((((15*x**2-60*x)*exp((x*ln(-4+x)-2*x-4)/x)-15*x**3+30*x**2+120*x 
)*ln(exp((x*ln(-4+x)-2*x-4)/x)-x-2)+(-15*x**2-60*x+240)*exp((x*ln(-4+x)-2* 
x-4)/x)+15*x**3-60*x**2)/((x**2-4*x)*exp((x*ln(-4+x)-2*x-4)/x)-x**3+2*x**2 
+8*x)/ln(exp((x*ln(-4+x)-2*x-4)/x)-x-2)**2,x)
 

Output:

15*x/log(-x + exp((x*log(x - 4) - 2*x - 4)/x) - 2)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 \, x^{2}}{x \log \left (-{\left (x e^{2} + 2 \, e^{2}\right )} e^{\frac {4}{x}} + x - 4\right ) - 2 \, x - 4} \] Input:

integrate((((15*x^2-60*x)*exp((x*log(-4+x)-2*x-4)/x)-15*x^3+30*x^2+120*x)* 
log(exp((x*log(-4+x)-2*x-4)/x)-x-2)+(-15*x^2-60*x+240)*exp((x*log(-4+x)-2* 
x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(-4+x)-2*x-4)/x)-x^3+2*x^2+8*x 
)/log(exp((x*log(-4+x)-2*x-4)/x)-x-2)^2,x, algorithm="maxima")
 

Output:

15*x^2/(x*log(-(x*e^2 + 2*e^2)*e^(4/x) + x - 4) - 2*x - 4)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 \, x^{2}}{x \log \left (-x e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} + x - 2 \, e^{\left (\frac {2 \, {\left (x + 2\right )}}{x}\right )} - 4\right ) - 2 \, x - 4} \] Input:

integrate((((15*x^2-60*x)*exp((x*log(-4+x)-2*x-4)/x)-15*x^3+30*x^2+120*x)* 
log(exp((x*log(-4+x)-2*x-4)/x)-x-2)+(-15*x^2-60*x+240)*exp((x*log(-4+x)-2* 
x-4)/x)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(-4+x)-2*x-4)/x)-x^3+2*x^2+8*x 
)/log(exp((x*log(-4+x)-2*x-4)/x)-x-2)^2,x, algorithm="giac")
 

Output:

15*x^2/(x*log(-x*e^(2*(x + 2)/x) + x - 2*e^(2*(x + 2)/x) - 4) - 2*x - 4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=-\int -\frac {\ln \left ({\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}-x-2\right )\,\left (120\,x-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (60\,x-15\,x^2\right )+30\,x^2-15\,x^3\right )-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (15\,x^2+60\,x-240\right )-60\,x^2+15\,x^3}{{\ln \left ({\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}-x-2\right )}^2\,\left (8\,x-{\mathrm {e}}^{-\frac {2\,x-x\,\ln \left (x-4\right )+4}{x}}\,\left (4\,x-x^2\right )+2\,x^2-x^3\right )} \,d x \] Input:

int((log(exp(-(2*x - x*log(x - 4) + 4)/x) - x - 2)*(120*x - exp(-(2*x - x* 
log(x - 4) + 4)/x)*(60*x - 15*x^2) + 30*x^2 - 15*x^3) - exp(-(2*x - x*log( 
x - 4) + 4)/x)*(60*x + 15*x^2 - 240) - 60*x^2 + 15*x^3)/(log(exp(-(2*x - x 
*log(x - 4) + 4)/x) - x - 2)^2*(8*x - exp(-(2*x - x*log(x - 4) + 4)/x)*(4* 
x - x^2) + 2*x^2 - x^3)),x)
 

Output:

-int(-(log(exp(-(2*x - x*log(x - 4) + 4)/x) - x - 2)*(120*x - exp(-(2*x - 
x*log(x - 4) + 4)/x)*(60*x - 15*x^2) + 30*x^2 - 15*x^3) - exp(-(2*x - x*lo 
g(x - 4) + 4)/x)*(60*x + 15*x^2 - 240) - 60*x^2 + 15*x^3)/(log(exp(-(2*x - 
 x*log(x - 4) + 4)/x) - x - 2)^2*(8*x - exp(-(2*x - x*log(x - 4) + 4)/x)*( 
4*x - x^2) + 2*x^2 - x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {-60 x^2+15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (240-60 x-15 x^2\right )+\left (120 x+30 x^2-15 x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-60 x+15 x^2\right )\right ) \log \left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )}{\left (8 x+2 x^2-x^3+e^{\frac {-4-2 x+x \log (-4+x)}{x}} \left (-4 x+x^2\right )\right ) \log ^2\left (-2+e^{\frac {-4-2 x+x \log (-4+x)}{x}}-x\right )} \, dx=\frac {15 x}{\mathrm {log}\left (\frac {-e^{\frac {4}{x}} e^{2} x -2 e^{\frac {4}{x}} e^{2}+x -4}{e^{\frac {4}{x}} e^{2}}\right )} \] Input:

int((((15*x^2-60*x)*exp((x*log(-4+x)-2*x-4)/x)-15*x^3+30*x^2+120*x)*log(ex 
p((x*log(-4+x)-2*x-4)/x)-x-2)+(-15*x^2-60*x+240)*exp((x*log(-4+x)-2*x-4)/x 
)+15*x^3-60*x^2)/((x^2-4*x)*exp((x*log(-4+x)-2*x-4)/x)-x^3+2*x^2+8*x)/log( 
exp((x*log(-4+x)-2*x-4)/x)-x-2)^2,x)
 

Output:

(15*x)/log(( - e**(4/x)*e**2*x - 2*e**(4/x)*e**2 + x - 4)/(e**(4/x)*e**2))