Integrand size = 184, antiderivative size = 28 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5 e^{-x}}{x \left (4+\log \left (4-\frac {e^x (-8+x)}{x}\right )\right )} \]
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5 e^{-x}}{x \left (4+\log \left (4+e^x \left (-1+\frac {8}{x}\right )\right )\right )} \]
Integrate[(80*x + 80*x^2 + E^x*(120 + 180*x - 25*x^2) + (20*x + 20*x^2 + E ^x*(40 + 35*x - 5*x^2))*Log[(E^x*(8 - x) + 4*x)/x])/(-64*E^x*x^3 + E^(2*x) *(-128*x^2 + 16*x^3) + (-32*E^x*x^3 + E^(2*x)*(-64*x^2 + 8*x^3))*Log[(E^x* (8 - x) + 4*x)/x] + (-4*E^x*x^3 + E^(2*x)*(-8*x^2 + x^3))*Log[(E^x*(8 - x) + 4*x)/x]^2),x]
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.
| \(\displaystyle \int \frac {80 x^2+e^x \left (-25 x^2+180 x+120\right )+\left (20 x^2+e^x \left (-5 x^2+35 x+40\right )+20 x\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+80 x}{-64 e^x x^3+e^{2 x} \left (16 x^3-128 x^2\right )+\left (e^{2 x} \left (x^3-8 x^2\right )-4 e^x x^3\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )+\left (e^{2 x} \left (8 x^3-64 x^2\right )-32 e^x x^3\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{-x} \left (5 e^x \left (-5 x^2+36 x+24\right )+80 x (x+1)-5 \left (e^x (x-8)-4 x\right ) (x+1) \log \left (e^x \left (\frac {8}{x}-1\right )+4\right )\right )}{\left (e^x (x-8)-4 x\right ) x^2 \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {20 e^{-x} \left (x^2-8 x+8\right )}{(x-8) x \left (e^x x-4 x-8 e^x\right ) \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}-\frac {5 e^{-x} \left (5 x^2+x^2 \log \left (e^x \left (\frac {8}{x}-1\right )+4\right )-36 x-7 x \log \left (e^x \left (\frac {8}{x}-1\right )+4\right )-8 \log \left (e^x \left (\frac {8}{x}-1\right )+4\right )-24\right )}{(x-8) x^2 \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \int \frac {e^{-x}}{x^2 \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx-5 \int \frac {e^{-x}}{x^2 \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )}dx-\frac {5}{8} \int \frac {e^{-x}}{(x-8) \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx-\frac {35}{8} \int \frac {e^{-x}}{x \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx-20 \int \frac {e^{-x}}{\left (e^x x-4 x-8 e^x\right ) \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx-20 \int \frac {e^{-x}}{(x-8) \left (e^x x-4 x-8 e^x\right ) \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx+20 \int \frac {e^{-x}}{x \left (e^x x-4 x-8 e^x\right ) \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )^2}dx-5 \int \frac {e^{-x}}{x \left (\log \left (e^x \left (\frac {8}{x}-1\right )+4\right )+4\right )}dx\) |
Int[(80*x + 80*x^2 + E^x*(120 + 180*x - 25*x^2) + (20*x + 20*x^2 + E^x*(40 + 35*x - 5*x^2))*Log[(E^x*(8 - x) + 4*x)/x])/(-64*E^x*x^3 + E^(2*x)*(-128 *x^2 + 16*x^3) + (-32*E^x*x^3 + E^(2*x)*(-64*x^2 + 8*x^3))*Log[(E^x*(8 - x ) + 4*x)/x] + (-4*E^x*x^3 + E^(2*x)*(-8*x^2 + x^3))*Log[(E^x*(8 - x) + 4*x )/x]^2),x]
3.6.44.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.94 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
| method | result | size |
| parallelrisch | \(\frac {5 \,{\mathrm e}^{-x}}{x \left (\ln \left (\frac {\left (8-x \right ) {\mathrm e}^{x}+4 x}{x}\right )+4\right )}\) | \(31\) |
| risch | \(-\frac {10 i {\mathrm e}^{-x}}{x \left (\pi \,\operatorname {csgn}\left (i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\pi {\operatorname {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )}^{3}-2 \pi {\operatorname {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \pi +2 i \ln \left (x \right )-2 i \ln \left (x \left ({\mathrm e}^{x}-4\right )-8 \,{\mathrm e}^{x}\right )-8 i\right )}\) | \(193\) |
int((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*ln(((8-x)*exp(x)+4*x)/x)+(-25*x ^2+180*x+120)*exp(x)+80*x^2+80*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x^3)*ln( ((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*ln(((8-x)* exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x,method=_RETURNVE RBOSE)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5}{x e^{x} \log \left (-\frac {{\left (x - 8\right )} e^{x} - 4 \, x}{x}\right ) + 4 \, x e^{x}} \]
integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x) +(-25*x^2+180*x+120)*exp(x)+80*x^2+80*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x ^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*lo g(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algori thm=\
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5}{x e^{x} \log {\left (\frac {4 x + \left (8 - x\right ) e^{x}}{x} \right )} + 4 x e^{x}} \]
integrate((((-5*x**2+35*x+40)*exp(x)+20*x**2+20*x)*ln(((8-x)*exp(x)+4*x)/x )+(-25*x**2+180*x+120)*exp(x)+80*x**2+80*x)/(((x**3-8*x**2)*exp(x)**2-4*ex p(x)*x**3)*ln(((8-x)*exp(x)+4*x)/x)**2+((8*x**3-64*x**2)*exp(x)**2-32*exp( x)*x**3)*ln(((8-x)*exp(x)+4*x)/x)+(16*x**3-128*x**2)*exp(x)**2-64*exp(x)*x **3),x)
Time = 0.56 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5}{x e^{x} \log \left (-{\left (x - 8\right )} e^{x} + 4 \, x\right ) - {\left (x \log \left (x\right ) - 4 \, x\right )} e^{x}} \]
integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x) +(-25*x^2+180*x+120)*exp(x)+80*x^2+80*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x ^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*lo g(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algori thm=\
Time = 0.59 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5}{x e^{x} \log \left (-\frac {x e^{x} - 4 \, x - 8 \, e^{x}}{x}\right ) + 4 \, x e^{x}} \]
integrate((((-5*x^2+35*x+40)*exp(x)+20*x^2+20*x)*log(((8-x)*exp(x)+4*x)/x) +(-25*x^2+180*x+120)*exp(x)+80*x^2+80*x)/(((x^3-8*x^2)*exp(x)^2-4*exp(x)*x ^3)*log(((8-x)*exp(x)+4*x)/x)^2+((8*x^3-64*x^2)*exp(x)^2-32*exp(x)*x^3)*lo g(((8-x)*exp(x)+4*x)/x)+(16*x^3-128*x^2)*exp(x)^2-64*exp(x)*x^3),x, algori thm=\
Time = 8.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {80 x+80 x^2+e^x \left (120+180 x-25 x^2\right )+\left (20 x+20 x^2+e^x \left (40+35 x-5 x^2\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )}{-64 e^x x^3+e^{2 x} \left (-128 x^2+16 x^3\right )+\left (-32 e^x x^3+e^{2 x} \left (-64 x^2+8 x^3\right )\right ) \log \left (\frac {e^x (8-x)+4 x}{x}\right )+\left (-4 e^x x^3+e^{2 x} \left (-8 x^2+x^3\right )\right ) \log ^2\left (\frac {e^x (8-x)+4 x}{x}\right )} \, dx=\frac {5\,{\mathrm {e}}^{-x}}{x\,\left (\ln \left (\frac {4\,x-{\mathrm {e}}^x\,\left (x-8\right )}{x}\right )+4\right )} \]
int(-(80*x + log((4*x - exp(x)*(x - 8))/x)*(20*x + exp(x)*(35*x - 5*x^2 + 40) + 20*x^2) + exp(x)*(180*x - 25*x^2 + 120) + 80*x^2)/(64*x^3*exp(x) + l og((4*x - exp(x)*(x - 8))/x)*(32*x^3*exp(x) + exp(2*x)*(64*x^2 - 8*x^3)) + exp(2*x)*(128*x^2 - 16*x^3) + log((4*x - exp(x)*(x - 8))/x)^2*(4*x^3*exp( x) + exp(2*x)*(8*x^2 - x^3))),x)