Integrand size = 63, antiderivative size = 25 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=x \left (-2-x+5 e^{-x} x\right ) \log \left (\frac {4-x}{x}\right ) \]
Time = 5.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=x \left (-2-x+5 e^{-x} x\right ) \log \left (-1+\frac {4}{x}\right ) \]
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]
Time = 0.87 (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 |
\(\displaystyle 5 e^{-x} x^2 \log \left (\frac {4}{x}-1\right )-(x+1)^2 \log \left (\frac {4}{x}-1\right )+\log (4-x)-\log (x)\) |
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]
3.20.48.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(24)=48\).
Time = 0.41 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08
method | result | size |
parallelrisch | \(\frac {\left (-16 \ln \left (-\frac {x -4}{x}\right ) {\mathrm e}^{x} x^{2}+80 \ln \left (-\frac {x -4}{x}\right ) x^{2}-32 \ln \left (-\frac {x -4}{x}\right ) {\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{16}\) | \(52\) |
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 | \(-24 \ln \left (\frac {4}{x}\right )+\ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) \left (\frac {4}{x}+1\right ) x^{2}+2 \ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) x +5 x^{2} \ln \left (\frac {-x +4}{x}\right ) {\mathrm e}^{-x}-24 \ln \left (x -4\right )\) | \(79\) |
parts | \(-24 \ln \left (\frac {4}{x}\right )+\ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) \left (\frac {4}{x}+1\right ) x^{2}+2 \ln \left (-1+\frac {4}{x}\right ) \left (-1+\frac {4}{x}\right ) x +5 x^{2} \ln \left (\frac {-x +4}{x}\right ) {\mathrm e}^{-x}-24 \ln \left (x -4\right )\) | \(79\) |
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}-5 i {\mathrm e}^{-x} \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \pi \,x^{2}+i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right ) \operatorname {csgn}\left (i \left (x -4\right )\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) x \pi +\frac {5 i {\mathrm e}^{-x} \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \pi \,x^{2}}{2}-\frac {i x^{2} \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3}}{2}+5 i {\mathrm e}^{-x} \pi \,x^{2}-i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3} x \pi -\frac {5 i {\mathrm e}^{-x} \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^{2}}{2}-i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) x \pi -i \pi \,x^{2}+\frac {i x^{2} \pi \,\operatorname {csgn}\left (i \left (x -4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-2 i x \pi +i x^{2} \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2}+\frac {5 i {\mathrm e}^{-x} \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{3} \pi \,x^{2}}{2}-\frac {i x^{2} \pi \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-\frac {i x^{2} \pi \,\operatorname {csgn}\left (i \left (x -4\right )\right ) \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2}}{2}+2 i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} x \pi -i \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right ) x \pi +\frac {5 i {\mathrm e}^{-x} \operatorname {csgn}\left (\frac {i \left (x -4\right )}{x}\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi \,x^{2}}{2}+x^{2} \ln \left (x \right )+2 x \ln \left (x \right )-5 \,{\mathrm e}^{-x} \ln \left (x \right ) x^{2}\) | \(449\) |
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)
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx={\left (5 \, x^{2} - {\left (x^{2} + 2 \, x\right )} e^{x}\right )} e^{\left (-x\right )} \log \left (-\frac {x - 4}{x}\right ) \]
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)/(x-4)/exp(x),x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=5 x^{2} e^{- x} \log {\left (\frac {4 - x}{x} \right )} + \left (- x^{2} - 2 x\right ) \log {\left (\frac {4 - x}{x} \right )} \]
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)/(x-4)/exp(x),x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=-5 \, x^{2} e^{\left (-x\right )} \log \left (x\right ) + {\left (x^{2} + 2 \, x\right )} \log \left (x\right ) + {\left (5 \, x^{2} e^{\left (-x\right )} - x^{2} - 2 \, x\right )} \log \left (-x + 4\right ) \]
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)/(x-4)/exp(x),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=5 \, x^{2} e^{\left (-x\right )} \log \left (-\frac {x - 4}{x}\right ) - x^{2} \log \left (-\frac {x - 4}{x}\right ) - 2 \, x \log \left (-\frac {x - 4}{x}\right ) \]
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)/(x-4)/exp(x),x, algorithm=\
Timed out. \[ \int \frac {e^{-x} \left (e^x (-8-4 x)+20 x+\left (-40 x+30 x^2-5 x^3+e^x \left (8+6 x-2 x^2\right )\right ) \log \left (\frac {4-x}{x}\right )\right )}{-4+x} \, dx=\int -\frac {{\mathrm {e}}^{-x}\,\left (\ln \left (-\frac {x-4}{x}\right )\,\left (40\,x-{\mathrm {e}}^x\,\left (-2\,x^2+6\,x+8\right )-30\,x^2+5\,x^3\right )-20\,x+{\mathrm {e}}^x\,\left (4\,x+8\right )\right )}{x-4} \,d x \]
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)