\(\int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} (-156 x+104 x^2-26 x^3)+e^{\frac {-3 x+x^2}{-2+x}} (6 x-4 x^2+x^3) \log (x)}{-104 x+104 x^2-26 x^3+(4 x-4 x^2+x^3) \log (x)} \, dx\) [6024]

Optimal result

Integrand size = 102, antiderivative size = 23 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{x+\frac {x}{2-x}}+\log \left ((26-\log (x))^2\right ) \]

[Out]

exp(x+x/(2-x))+ln((26-ln(x))^2)

Rubi [F]

\[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=\int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx \]

[In]

Int[(8 - 8*x + 2*x^2 + E^((-3*x + x^2)/(-2 + x))*(-156*x + 104*x^2 - 26*x^3) + E^((-3*x + x^2)/(-2 + x))*(6*x
- 4*x^2 + x^3)*Log[x])/(-104*x + 104*x^2 - 26*x^3 + (4*x - 4*x^2 + x^3)*Log[x]),x]

[Out]

E^(((3 - x)*x)/(2 - x)) - 8*Defer[Int][1/((-2 + x)^2*(-26 + Log[x])), x] + 8*Defer[Int][1/((-2 + x)^2*x*(-26 +
Log[x])), x] + 2*Defer[Int][x/((-2 + x)^2*(-26 + Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-8+8 x-2 x^2-e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )-e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{(2-x)^2 x (26-\log (x))} \, dx \\ & = \int \left (\frac {e^{\frac {(-3+x) x}{-2+x}} \left (6-4 x+x^2\right )}{(-2+x)^2}-\frac {8}{(-2+x)^2 (-26+\log (x))}+\frac {8}{(-2+x)^2 x (-26+\log (x))}+\frac {2 x}{(-2+x)^2 (-26+\log (x))}\right ) \, dx \\ & = 2 \int \frac {x}{(-2+x)^2 (-26+\log (x))} \, dx-8 \int \frac {1}{(-2+x)^2 (-26+\log (x))} \, dx+8 \int \frac {1}{(-2+x)^2 x (-26+\log (x))} \, dx+\int \frac {e^{\frac {(-3+x) x}{-2+x}} \left (6-4 x+x^2\right )}{(-2+x)^2} \, dx \\ & = e^{\frac {(3-x) x}{2-x}}+2 \int \frac {x}{(-2+x)^2 (-26+\log (x))} \, dx-8 \int \frac {1}{(-2+x)^2 (-26+\log (x))} \, dx+8 \int \frac {1}{(-2+x)^2 x (-26+\log (x))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{-1-\frac {2}{-2+x}+x}+2 \log (26-\log (x)) \]

[In]

Integrate[(8 - 8*x + 2*x^2 + E^((-3*x + x^2)/(-2 + x))*(-156*x + 104*x^2 - 26*x^3) + E^((-3*x + x^2)/(-2 + x))
*(6*x - 4*x^2 + x^3)*Log[x])/(-104*x + 104*x^2 - 26*x^3 + (4*x - 4*x^2 + x^3)*Log[x]),x]

[Out]

E^(-1 - 2/(-2 + x) + x) + 2*Log[26 - Log[x]]

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\({\mathrm e}^{\frac {x \left (-3+x \right )}{-2+x}}+2 \ln \left (-26+\ln \left (x \right )\right )\)	\(20\)
parallelrisch	\({\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}+2 \ln \left (-26+\ln \left (x \right )\right )\)	\(23\)
default	\(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\)	\(48\)
norman	\(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\)	\(48\)
parts	\(2 \ln \left (-26+\ln \left (x \right )\right )+\frac {x \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}-2 \,{\mathrm e}^{\frac {x^{2}-3 x}{-2+x}}}{-2+x}\)	\(48\)

[In]

int(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*ln(x)+(-26*x^3+104*x^2-156*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*x+8)/((
x^3-4*x^2+4*x)*ln(x)-26*x^3+104*x^2-104*x),x,method=_RETURNVERBOSE)

[Out]

exp(x*(-3+x)/(-2+x))+2*ln(-26+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (\frac {x^{2} - 3 \, x}{x - 2}\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \]

[In]

integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-156*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*
x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+104*x^2-104*x),x, algorithm="fricas")

[Out]

e^((x^2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\frac {x^{2} - 3 x}{x - 2}} + 2 \log {\left (\log {\left (x \right )} - 26 \right )} \]

[In]

integrate(((x**3-4*x**2+6*x)*exp((x**2-3*x)/(-2+x))*ln(x)+(-26*x**3+104*x**2-156*x)*exp((x**2-3*x)/(-2+x))+2*x
**2-8*x+8)/((x**3-4*x**2+4*x)*ln(x)-26*x**3+104*x**2-104*x),x)

[Out]

exp((x**2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (x - \frac {2}{x - 2} - 1\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \]

[In]

integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-156*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*
x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+104*x^2-104*x),x, algorithm="maxima")

[Out]

e^(x - 2/(x - 2) - 1) + 2*log(log(x) - 26)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=e^{\left (\frac {x^{2} - 3 \, x}{x - 2}\right )} + 2 \, \log \left (\log \left (x\right ) - 26\right ) \]

[In]

integrate(((x^3-4*x^2+6*x)*exp((x^2-3*x)/(-2+x))*log(x)+(-26*x^3+104*x^2-156*x)*exp((x^2-3*x)/(-2+x))+2*x^2-8*
x+8)/((x^3-4*x^2+4*x)*log(x)-26*x^3+104*x^2-104*x),x, algorithm="giac")

[Out]

e^((x^2 - 3*x)/(x - 2)) + 2*log(log(x) - 26)

Mupad [B] (verification not implemented)

Time = 13.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {8-8 x+2 x^2+e^{\frac {-3 x+x^2}{-2+x}} \left (-156 x+104 x^2-26 x^3\right )+e^{\frac {-3 x+x^2}{-2+x}} \left (6 x-4 x^2+x^3\right ) \log (x)}{-104 x+104 x^2-26 x^3+\left (4 x-4 x^2+x^3\right ) \log (x)} \, dx=2\,\ln \left (\ln \left (x\right )-26\right )+{\mathrm {e}}^{-\frac {3\,x}{x-2}}\,{\mathrm {e}}^{\frac {x^2}{x-2}} \]

[In]

int(-(2*x^2 - exp(-(3*x - x^2)/(x - 2))*(156*x - 104*x^2 + 26*x^3) - 8*x + exp(-(3*x - x^2)/(x - 2))*log(x)*(6
*x - 4*x^2 + x^3) + 8)/(104*x - log(x)*(4*x - 4*x^2 + x^3) - 104*x^2 + 26*x^3),x)

[Out]

2*log(log(x) - 26) + exp(-(3*x)/(x - 2))*exp(x^2/(x - 2))