Integrand size = 63, antiderivative size = 25 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=x \left (-2-x+5 e^{-x} x\right ) \log \left (\frac {4-x}{x}\right ) \] Output:
(5*x/exp(x)-x-2)*ln((4-x)/x)*x
Time = 5.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=x \left (-2-x+5 e^{-x} x\right ) \log \left (-1+\frac {4}{x}\right ) \] Input:
Integrate[(E^x*(-8 - 4*x) + 20*x + (-40*x + 30*x^2 - 5*x^3 + E^x*(8 + 6*x - 2*x^2))*Log[(4 - x)/x])/(E^x*(-4 + x)),x]
Output:
x*(-2 - x + (5*x)/E^x)*Log[-1 + 4/x]
Time = 0.89 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {e^{-x} \left (\left (-5 x^3+30 x^2+e^x \left (-2 x^2+6 x+8\right )-40 x\right ) \log \left (\frac {4-x}{x}\right )+e^x (-4 x-8)+20 x\right )}{x-4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 e^{-x} x \left (x^2 \log \left (\frac {4}{x}-1\right )-6 x \log \left (\frac {4}{x}-1\right )+8 \log \left (\frac {4}{x}-1\right )-4\right )}{x-4}-\frac {2 \left (x^2 \log \left (\frac {4}{x}-1\right )+2 x-3 x \log \left (\frac {4}{x}-1\right )-4 \log \left (\frac {4}{x}-1\right )+4\right )}{x-4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 e^{-x} x^2 \log \left (\frac {4}{x}-1\right )-(x+1)^2 \log \left (\frac {4}{x}-1\right )+\log (4-x)-\log (x)\) |
Input:
Int[(E^x*(-8 - 4*x) + 20*x + (-40*x + 30*x^2 - 5*x^3 + E^x*(8 + 6*x - 2*x^ 2))*Log[(4 - x)/x])/(E^x*(-4 + x)),x]
Output:
(5*x^2*Log[-1 + 4/x])/E^x - (1 + x)^2*Log[-1 + 4/x] + Log[4 - x] - Log[x]
Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(24)=48\).
Time = 93.49 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08
method | result | size |
parallelrisch | \(\frac {\left (-16 \,{\mathrm e}^{x} \ln \left (-\frac {x -4}{x}\right ) x^{2}+80 \ln \left (-\frac {x -4}{x}\right ) x^{2}-32 \,{\mathrm e}^{x} \ln \left (-\frac {x -4}{x}\right ) x \right ) {\mathrm e}^{-x}}{16}\) | \(52\) |
norman | \(\left (5 x^{2} \ln \left (\frac {-x +4}{x}\right )-2 \,{\mathrm e}^{x} x \ln \left (\frac {-x +4}{x}\right )-{\mathrm e}^{x} x^{2} \ln \left (\frac {-x +4}{x}\right )\right ) {\mathrm e}^{-x}\) | \(54\) |
default | \(-24 \ln \left (\frac {4}{x}\right )+2 \ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) x +\ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) \left (1+\frac {4}{x}\right ) x^{2}+5 x^{2} \ln \left (\frac {-x +4}{x}\right ) {\mathrm e}^{-x}-24 \ln \left (x -4\right )\) | \(79\) |
parts | \(-24 \ln \left (\frac {4}{x}\right )+2 \ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) x +\ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) \left (1+\frac {4}{x}\right ) x^{2}+5 x^{2} \ln \left (\frac {-x +4}{x}\right ) {\mathrm e}^{-x}-24 \ln \left (x -4\right )\) | \(79\) |
risch | \(-x^{2} \ln \left (x -4\right )-2 x \ln \left (x -4\right )+5 \,{\mathrm e}^{-x} \ln \left (x -4\right ) x^{2}-i \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} x \pi +i x^{2} \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \pi -i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) x \pi +5 i \pi \,{\mathrm e}^{-x} x^{2}-\frac {i x^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \pi }{2}+\frac {5 i \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) {\mathrm e}^{-x} x^{2}}{2}+2 i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} x \pi -2 i x \pi +\frac {5 i \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) {\mathrm e}^{-x} x^{2}}{2}-i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3} x \pi -\frac {5 i \pi \,\operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x -4\right )\right ) {\mathrm e}^{-x} x^{2}}{2}-\frac {i x^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \pi }{2}+i \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right ) \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi x -\frac {i x^{2} \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3} \pi }{2}+\frac {5 i \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3} {\mathrm e}^{-x} x^{2}}{2}-i x^{2} \pi -5 i \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} {\mathrm e}^{-x} x^{2}+\frac {i x^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right ) \pi }{2}+x^{2} \ln \left (x \right )+2 x \ln \left (x \right )-5 \ln \left (x \right ) {\mathrm e}^{-x} x^{2}\) | \(449\) |
orering | \(\text {Expression too large to display}\) | \(3181\) |
Input:
int((((-2*x^2+6*x+8)*exp(x)-5*x^3+30*x^2-40*x)*ln((-x+4)/x)+(-4*x-8)*exp(x )+20*x)/(x-4)/exp(x),x,method=_RETURNVERBOSE)
Output:
1/16*(-16*exp(x)*ln(-(x-4)/x)*x^2+80*ln(-(x-4)/x)*x^2-32*exp(x)*ln(-(x-4)/ x)*x)/exp(x)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx={\left (5 \, x^{2} - {\left (x^{2} + 2 \, x\right )} e^{x}\right )} e^{\left (-x\right )} \log \left (-\frac {x - 4}{x}\right ) \] Input:
integrate((((-2*x^2+6*x+8)*exp(x)-5*x^3+30*x^2-40*x)*log((-x+4)/x)+(-4*x-8 )*exp(x)+20*x)/(-4+x)/exp(x),x, algorithm="fricas")
Output:
(5*x^2 - (x^2 + 2*x)*e^x)*e^(-x)*log(-(x - 4)/x)
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=5 x^{2} e^{- x} \log {\left (\frac {4 - x}{x} \right )} + \left (- x^{2} - 2 x\right ) \log {\left (\frac {4 - x}{x} \right )} \] Input:
integrate((((-2*x**2+6*x+8)*exp(x)-5*x**3+30*x**2-40*x)*ln((-x+4)/x)+(-4*x -8)*exp(x)+20*x)/(-4+x)/exp(x),x)
Output:
5*x**2*exp(-x)*log((4 - x)/x) + (-x**2 - 2*x)*log((4 - x)/x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=-5 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) + {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + {\left (5 \, x^{2} e^{\left (-x\right )} - x^{2} - 2 \, x\right )} \log \left (-x + 4\right ) \] Input:
integrate((((-2*x^2+6*x+8)*exp(x)-5*x^3+30*x^2-40*x)*log((-x+4)/x)+(-4*x-8 )*exp(x)+20*x)/(-4+x)/exp(x),x, algorithm="maxima")
Output:
-5*x^2*e^(-x)*log(x) + (x^2 + 2*x)*log(x) + (5*x^2*e^(-x) - x^2 - 2*x)*log (-x + 4)
Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=5 \, x^{2} e^{\left (-x\right )} \log \left (-\frac {x - 4}{x}\right ) - x^{2} \log \left (-\frac {x - 4}{x}\right ) - 2 \, x \log \left (-\frac {x - 4}{x}\right ) \] Input:
integrate((((-2*x^2+6*x+8)*exp(x)-5*x^3+30*x^2-40*x)*log((-x+4)/x)+(-4*x-8 )*exp(x)+20*x)/(-4+x)/exp(x),x, algorithm="giac")
Output:
5*x^2*e^(-x)*log(-(x - 4)/x) - x^2*log(-(x - 4)/x) - 2*x*log(-(x - 4)/x)
Timed out. \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=\int -\frac {{\mathrm {e}}^{-x}\,\left (\ln \left (-\frac {x-4}{x}\right )\,\left (40\,x-{\mathrm {e}}^x\,\left (-2\,x^2+6\,x+8\right )-30\,x^2+5\,x^3\right )-20\,x+{\mathrm {e}}^x\,\left (4\,x+8\right )\right )}{x-4} \,d x \] Input:
int(-(exp(-x)*(log(-(x - 4)/x)*(40*x - exp(x)*(6*x - 2*x^2 + 8) - 30*x^2 + 5*x^3) - 20*x + exp(x)*(4*x + 8)))/(x - 4),x)
Output:
int(-(exp(-x)*(log(-(x - 4)/x)*(40*x - exp(x)*(6*x - 2*x^2 + 8) - 30*x^2 + 5*x^3) - 20*x + exp(x)*(4*x + 8)))/(x - 4), x)
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=\frac {\mathrm {log}\left (\frac {-x +4}{x}\right ) x \left (-e^{x} x -2 e^{x}+5 x \right )}{e^{x}} \] Input:
int((((-2*x^2+6*x+8)*exp(x)-5*x^3+30*x^2-40*x)*log((-x+4)/x)+(-4*x-8)*exp( x)+20*x)/(-4+x)/exp(x),x)
Output:
(log(( - x + 4)/x)*x*( - e**x*x - 2*e**x + 5*x))/e**x