\(\int \frac {e^{-\frac {-5-\log (x)}{-5+x}} (120-44 x-12 x^2+e^x (60-62 x+22 x^2-2 x^3)+(4 x+2 e^x x-4 x^2) \log (x))}{25 x^2-10 x^3+x^4} \, dx\) [7846]

Optimal result

Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 e^{-\frac {5+\log (x)}{5-x}} \left (-2-e^x+2 x\right )}{x} \]

[Out]

2/x*(2*x-2-exp(x))/exp((5+ln(x))/(5-x))

Rubi [F]

\[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx \]

[In]

Int[(120 - 44*x - 12*x^2 + E^x*(60 - 62*x + 22*x^2 - 2*x^3) + (4*x + 2*E^x*x - 4*x^2)*Log[x])/(E^((-5 - Log[x]
)/(-5 + x))*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

-16*Defer[Int][1/(E^((-5 - Log[x])/(-5 + x))*(-5 + x)^2), x] + 2*Defer[Int][E^((5 - 5*x + x^2 + Log[x])/(-5 +
x))/(-5 + x)^2, x] - (4*Defer[Int][1/(E^((-5 - Log[x])/(-5 + x))*(-5 + x)), x])/25 - (12*Defer[Int][E^((5 - 5*
x + x^2 + Log[x])/(-5 + x))/(-5 + x), x])/25 + (24*Defer[Int][1/(E^((-5 - Log[x])/(-5 + x))*x^2), x])/5 + (12*
Defer[Int][E^((5 - 5*x + x^2 + Log[x])/(-5 + x))/x^2, x])/5 + (4*Defer[Int][1/(E^((-5 - Log[x])/(-5 + x))*x),
x])/25 - (38*Defer[Int][E^((5 - 5*x + x^2 + Log[x])/(-5 + x))/x, x])/25 - (16*Defer[Int][Log[x]/(E^((-5 - Log[
x])/(-5 + x))*(-5 + x)^2), x])/5 + (2*Defer[Int][(E^((5 - 5*x + x^2 + Log[x])/(-5 + x))*Log[x])/(-5 + x)^2, x]
)/5 - (4*Defer[Int][Log[x]/(E^((-5 - Log[x])/(-5 + x))*(-5 + x)), x])/25 - (2*Defer[Int][(E^((5 - 5*x + x^2 +
Log[x])/(-5 + x))*Log[x])/(-5 + x), x])/25 + (4*Defer[Int][Log[x]/(E^((-5 - Log[x])/(-5 + x))*x), x])/25 + (2*
Defer[Int][(E^((5 - 5*x + x^2 + Log[x])/(-5 + x))*Log[x])/x, x])/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{x^2 \left (25-10 x+x^2\right )} \, dx \\ & = \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{(-5+x)^2 x^2} \, dx \\ & = \int \left (-\frac {12 e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2}+\frac {120 e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2 x^2}-\frac {44 e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2 x}-\frac {4 e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2}+\frac {4 e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2 x}-\frac {2 e^{x-\frac {-5-\log (x)}{-5+x}} \left (-30+31 x-11 x^2+x^3-x \log (x)\right )}{(-5+x)^2 x^2}\right ) \, dx \\ & = -\left (2 \int \frac {e^{x-\frac {-5-\log (x)}{-5+x}} \left (-30+31 x-11 x^2+x^3-x \log (x)\right )}{(-5+x)^2 x^2} \, dx\right )-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2 x} \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-44 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2 x} \, dx+120 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2 x^2} \, dx \\ & = -\left (2 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \left (-30+31 x-11 x^2+x^3-x \log (x)\right )}{(5-x)^2 x^2} \, dx\right )-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+4 \int \left (\frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{5 (-5+x)^2}-\frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{25 (-5+x)}+\frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{25 x}\right ) \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-44 \int \left (\frac {e^{-\frac {-5-\log (x)}{-5+x}}}{5 (-5+x)^2}-\frac {e^{-\frac {-5-\log (x)}{-5+x}}}{25 (-5+x)}+\frac {e^{-\frac {-5-\log (x)}{-5+x}}}{25 x}\right ) \, dx+120 \int \left (\frac {e^{-\frac {-5-\log (x)}{-5+x}}}{25 (-5+x)^2}-\frac {2 e^{-\frac {-5-\log (x)}{-5+x}}}{125 (-5+x)}+\frac {e^{-\frac {-5-\log (x)}{-5+x}}}{25 x^2}+\frac {2 e^{-\frac {-5-\log (x)}{-5+x}}}{125 x}\right ) \, dx \\ & = -\left (\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{-5+x} \, dx\right )+\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{x} \, dx+\frac {4}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx-\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx+\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-2 \int \left (-\frac {11 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2}-\frac {30 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2 x^2}+\frac {31 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2 x}+\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} x}{(-5+x)^2}-\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{(-5+x)^2 x}\right ) \, dx-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x^2} \, dx-\frac {44}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx \\ & = -\left (\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{-5+x} \, dx\right )+\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{x} \, dx+\frac {4}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx-\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx+\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-2 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} x}{(-5+x)^2} \, dx+2 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{(-5+x)^2 x} \, dx-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x^2} \, dx-\frac {44}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+22 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx+60 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2 x^2} \, dx-62 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2 x} \, dx \\ & = -\left (\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{-5+x} \, dx\right )+\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{x} \, dx+\frac {4}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx-\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx+\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-2 \int \left (\frac {5 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2}+\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{-5+x}\right ) \, dx+2 \int \left (\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{5 (-5+x)^2}-\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{25 (-5+x)}+\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{25 x}\right ) \, dx-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x^2} \, dx-\frac {44}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+22 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx+60 \int \left (\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{25 (-5+x)^2}-\frac {2 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{125 (-5+x)}+\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{25 x^2}+\frac {2 e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{125 x}\right ) \, dx-62 \int \left (\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{5 (-5+x)^2}-\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{25 (-5+x)}+\frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{25 x}\right ) \, dx \\ & = -\left (\frac {2}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{-5+x} \, dx\right )+\frac {2}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{x} \, dx-\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{-5+x} \, dx+\frac {4}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{x} \, dx+\frac {2}{5} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {4}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx-\frac {24}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{-5+x} \, dx+\frac {24}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{x} \, dx+\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx-\frac {44}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{-5+x} \, dx+\frac {48}{25} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x} \, dx-2 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{-5+x} \, dx+\frac {12}{5} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {12}{5} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{x^2} \, dx+\frac {62}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{-5+x} \, dx-\frac {62}{25} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{x} \, dx-4 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \log (x)}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx+\frac {24}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{x^2} \, dx-\frac {44}{5} \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-10 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx-12 \int \frac {e^{-\frac {-5-\log (x)}{-5+x}}}{(-5+x)^2} \, dx-\frac {62}{5} \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx+22 \int \frac {e^{\frac {5-5 x+x^2+\log (x)}{-5+x}}}{(-5+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-2 e^{\frac {5}{-5+x}} \left (2+e^x-2 x\right ) x^{\frac {6-x}{-5+x}} \]

[In]

Integrate[(120 - 44*x - 12*x^2 + E^x*(60 - 62*x + 22*x^2 - 2*x^3) + (4*x + 2*E^x*x - 4*x^2)*Log[x])/(E^((-5 -
Log[x])/(-5 + x))*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

-2*E^(5/(-5 + x))*(2 + E^x - 2*x)*x^((6 - x)/(-5 + x))

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {2 \left (2 x -2-{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {5+\ln \left (x \right )}{-5+x}}}{x}\)	\(26\)
parallelrisch	\(\frac {\left (30000+6000 x^{2}-3000 \,{\mathrm e}^{x} x -36000 x +15000 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {5+\ln \left (x \right )}{-5+x}}}{1500 x \left (-5+x \right )}\)	\(44\)

[In]

int(((2*exp(x)*x-4*x^2+4*x)*ln(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-44*x+120)/(x^4-10*x^3+25*x^2)/exp((-5-
ln(x))/(-5+x)),x,method=_RETURNVERBOSE)

[Out]

2/x*(2*x-2-exp(x))*exp((5+ln(x))/(-5+x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right ) + 5}{x - 5}\right )}}{x} \]

[In]

integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-44*x+120)/(x^4-10*x^3+25*x^2)/e
xp((-5-log(x))/(-5+x)),x, algorithm="fricas")

[Out]

2*(2*x - e^x - 2)*e^((log(x) + 5)/(x - 5))/x

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {\left (4 x - 2 e^{x} - 4\right ) e^{- \frac {- \log {\left (x \right )} - 5}{x - 5}}}{x} \]

[In]

integrate(((2*exp(x)*x-4*x**2+4*x)*ln(x)+(-2*x**3+22*x**2-62*x+60)*exp(x)-12*x**2-44*x+120)/(x**4-10*x**3+25*x
**2)/exp((-5-ln(x))/(-5+x)),x)

[Out]

(4*x - 2*exp(x) - 4)*exp(-(-log(x) - 5)/(x - 5))/x

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right )}{x - 5} + \frac {5}{x - 5}\right )}}{x} \]

[In]

integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-44*x+120)/(x^4-10*x^3+25*x^2)/e
xp((-5-log(x))/(-5+x)),x, algorithm="maxima")

[Out]

2*(2*x - e^x - 2)*e^(log(x)/(x - 5) + 5/(x - 5))/x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )} - e^{\left (\frac {x^{2} - 4 \, x + \log \left (x\right )}{x - 5}\right )} - 2 \, e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )}\right )} e^{\left (-1\right )}}{x} \]

[In]

integrate(((2*exp(x)*x-4*x^2+4*x)*log(x)+(-2*x^3+22*x^2-62*x+60)*exp(x)-12*x^2-44*x+120)/(x^4-10*x^3+25*x^2)/e
xp((-5-log(x))/(-5+x)),x, algorithm="giac")

[Out]

2*(2*x*e^((x + log(x))/(x - 5)) - e^((x^2 - 4*x + log(x))/(x - 5)) - 2*e^((x + log(x))/(x - 5)))*e^(-1)/x

Mupad [B] (verification not implemented)

Time = 13.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-x^{\frac {1}{x-5}-1}\,{\mathrm {e}}^{\frac {5}{x-5}}\,\left (2\,{\mathrm {e}}^x-4\,x+4\right ) \]

[In]

int(-(exp((log(x) + 5)/(x - 5))*(44*x - log(x)*(4*x + 2*x*exp(x) - 4*x^2) + 12*x^2 + exp(x)*(62*x - 22*x^2 + 2
*x^3 - 60) - 120))/(25*x^2 - 10*x^3 + x^4),x)

[Out]

-x^(1/(x - 5) - 1)*exp(5/(x - 5))*(2*exp(x) - 4*x + 4)