Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 e^{-\frac {5+\log (x)}{5-x}} \left (-2-e^x+2 x\right )}{x} \]
Time = 5.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-2 e^{\frac {5}{-5+x}} \left (2+e^x-2 x\right ) x^{\frac {6-x}{-5+x}} \]
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]
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.
\(\displaystyle \int \frac {e^{-\frac {-\log (x)-5}{x-5}} \left (-12 x^2+\left (-4 x^2+2 e^x x+4 x\right ) \log (x)+e^x \left (-2 x^3+22 x^2-62 x+60\right )-44 x+120\right )}{x^4-10 x^3+25 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-\frac {-\log (x)-5}{x-5}} \left (-12 x^2+\left (-4 x^2+2 e^x x+4 x\right ) \log (x)+e^x \left (-2 x^3+22 x^2-62 x+60\right )-44 x+120\right )}{x^2 \left (x^2-10 x+25\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2} \left (-6 x^2-22 x+e^x \left (-x^3+11 x^2-31 x+30\right )+\left (-2 x^2+e^x x+2 x\right ) \log (x)+60\right )}{2 (5-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2} \left (-6 x^2-22 x+e^x \left (-x^3+11 x^2-31 x+30\right )+\left (-2 x^2+e^x x+2 x\right ) \log (x)+60\right )}{(5-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (-\frac {e^{x-\frac {5}{5-x}} \left (x^3-11 x^2-\log (x) x+31 x-30\right ) x^{\frac {1}{x-5}-2}}{(x-5)^2}+\frac {60 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}+\frac {2 e^{-\frac {5}{5-x}} \log (x) x^{\frac {1}{x-5}-1}}{(x-5)^2}-\frac {22 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}-\frac {2 e^{-\frac {5}{5-x}} \log (x) x^{\frac {1}{x-5}}}{(x-5)^2}-\frac {6 e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (60 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}dx+30 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-2}}{(x-5)^2}dx-22 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-31 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-2 \int \frac {\int \frac {e^{\frac {5}{x-5}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx}{x}dx-\int \frac {\int \frac {e^{x+\frac {5}{x-5}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx}{x}dx+2 \log (x) \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx+\log (x) \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}-1}}{(x-5)^2}dx-6 \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx+11 \int \frac {e^{x-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx-\int \frac {e^{x-\frac {5}{5-x}} x^{\frac {x-4}{x-5}}}{(5-x)^2}dx+2 \int \frac {\int \frac {e^{\frac {5}{x-5}} x^{\frac {1}{x-5}}}{(x-5)^2}dx}{x}dx-2 \log (x) \int \frac {e^{-\frac {5}{5-x}} x^{\frac {1}{x-5}}}{(x-5)^2}dx\right )\) |
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 ]
3.19.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
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 | \(\frac {\left (30000+6000 x^{2}-3000 \,{\mathrm e}^{x} x -36000 x +15000 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {5+\ln \left (x \right )}{-5+x}}}{1500 x \left (-5+x \right )}\) | \(44\) |
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 )
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right ) + 5}{x - 5}\right )}}{x} \]
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)/exp((-5-log(x))/(-5+x)),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {\left (4 x - 2 e^{x} - 4\right ) e^{- \frac {- \log {\left (x \right )} - 5}{x - 5}}}{x} \]
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)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x - e^{x} - 2\right )} e^{\left (\frac {\log \left (x\right )}{x - 5} + \frac {5}{x - 5}\right )}}{x} \]
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)/exp((-5-log(x))/(-5+x)),x, algorithm=\
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 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=\frac {2 \, {\left (2 \, x e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )} - e^{\left (\frac {x^{2} - 4 \, x + \log \left (x\right )}{x - 5}\right )} - 2 \, e^{\left (\frac {x + \log \left (x\right )}{x - 5}\right )}\right )} e^{\left (-1\right )}}{x} \]
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)/exp((-5-log(x))/(-5+x)),x, algorithm=\
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
Time = 13.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {-5-\log (x)}{-5+x}} \left (120-44 x-12 x^2+e^x \left (60-62 x+22 x^2-2 x^3\right )+\left (4 x+2 e^x x-4 x^2\right ) \log (x)\right )}{25 x^2-10 x^3+x^4} \, dx=-x^{\frac {1}{x-5}-1}\,{\mathrm {e}}^{\frac {5}{x-5}}\,\left (2\,{\mathrm {e}}^x-4\,x+4\right ) \]