Integrand size = 92, antiderivative size = 21 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10}{e^3 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \]
Time = 0.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10}{e^3 \left (4+e^{-\frac {4}{x}+x}+2 x\right )} \]
Integrate[(-20*x^2 + E^((-4 + x^2)/x)*(-40 - 10*x^2))/(E^(3 + (2*(-4 + x^2 ))/x)*x^2 + E^(3 + (-4 + x^2)/x)*(8*x^2 + 4*x^3) + E^3*(16*x^2 + 16*x^3 + 4*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.
\(\displaystyle \int \frac {e^{\frac {x^2-4}{x}} \left (-10 x^2-40\right )-20 x^2}{e^{\frac {2 \left (x^2-4\right )}{x}+3} x^2+e^{\frac {x^2-4}{x}+3} \left (4 x^3+8 x^2\right )+e^3 \left (4 x^4+16 x^3+16 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {10 e^{\frac {4}{x}-3} \left (-2 e^{4/x} x^2-e^x \left (x^2+4\right )\right )}{x^2 \left (2 e^{4/x} (x+2)+e^x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 \int -\frac {e^{\frac {4}{x}-3} \left (2 e^{4/x} x^2+e^x \left (x^2+4\right )\right )}{x^2 \left (2 e^{4/x} (x+2)+e^x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -10 \int \frac {e^{\frac {4}{x}-3} \left (2 e^{4/x} x^2+e^x \left (x^2+4\right )\right )}{x^2 \left (2 e^{4/x} (x+2)+e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -10 \int \left (\frac {e^{\frac {4}{x}-3} \left (x^2+4\right )}{x^2 \left (2 e^{4/x} x+4 e^{4/x}+e^x\right )}-\frac {2 e^{\frac {8}{x}-3} \left (x^3+x^2+4 x+8\right )}{x^2 \left (2 e^{4/x} x+4 e^{4/x}+e^x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 \left (-16 \int \frac {e^{\frac {8}{x}-3}}{x^2 \left (2 e^{4/x} x+4 e^{4/x}+e^x\right )^2}dx+\frac {4 \int \frac {e^{4/x}}{x^2 \left (2 e^{4/x} x+4 e^{4/x}+e^x\right )}dx}{e^3}-2 \int \frac {e^{\frac {8}{x}-3}}{\left (2 e^{4/x} x+4 e^{4/x}+e^x\right )^2}dx-8 \int \frac {e^{\frac {8}{x}-3}}{x \left (2 e^{4/x} x+4 e^{4/x}+e^x\right )^2}dx-2 \int \frac {e^{\frac {8}{x}-3} x}{\left (2 e^{4/x} x+4 e^{4/x}+e^x\right )^2}dx+\frac {\int \frac {e^{4/x}}{2 e^{4/x} x+4 e^{4/x}+e^x}dx}{e^3}\right )\) |
Int[(-20*x^2 + E^((-4 + x^2)/x)*(-40 - 10*x^2))/(E^(3 + (2*(-4 + x^2))/x)* x^2 + E^(3 + (-4 + x^2)/x)*(8*x^2 + 4*x^3) + E^3*(16*x^2 + 16*x^3 + 4*x^4) ),x]
3.30.81.3.1 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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10
method | result | size |
risch | \(\frac {10 \,{\mathrm e}^{-3}}{4+{\mathrm e}^{\frac {\left (-2+x \right ) \left (2+x \right )}{x}}+2 x}\) | \(23\) |
norman | \(\frac {10 \,{\mathrm e}^{-3}}{4+{\mathrm e}^{\frac {x^{2}-4}{x}}+2 x}\) | \(24\) |
parallelrisch | \(\frac {10 \,{\mathrm e}^{-3}}{4+{\mathrm e}^{\frac {x^{2}-4}{x}}+2 x}\) | \(24\) |
int(((-10*x^2-40)*exp((x^2-4)/x)-20*x^2)/(x^2*exp(3)*exp((x^2-4)/x)^2+(4*x ^3+8*x^2)*exp(3)*exp((x^2-4)/x)+(4*x^4+16*x^3+16*x^2)*exp(3)),x,method=_RE TURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10}{2 \, {\left (x + 2\right )} e^{3} + e^{\left (\frac {x^{2} + 3 \, x - 4}{x}\right )}} \]
integrate(((-10*x^2-40)*exp((x^2-4)/x)-20*x^2)/(x^2*exp(3)*exp((x^2-4)/x)^ 2+(4*x^3+8*x^2)*exp(3)*exp((x^2-4)/x)+(4*x^4+16*x^3+16*x^2)*exp(3)),x, alg orithm=\
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10}{2 x e^{3} + e^{3} e^{\frac {x^{2} - 4}{x}} + 4 e^{3}} \]
integrate(((-10*x**2-40)*exp((x**2-4)/x)-20*x**2)/(x**2*exp(3)*exp((x**2-4 )/x)**2+(4*x**3+8*x**2)*exp(3)*exp((x**2-4)/x)+(4*x**4+16*x**3+16*x**2)*ex p(3)),x)
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10 \, e^{\frac {4}{x}}}{2 \, {\left (x e^{3} + 2 \, e^{3}\right )} e^{\frac {4}{x}} + e^{\left (x + 3\right )}} \]
integrate(((-10*x^2-40)*exp((x^2-4)/x)-20*x^2)/(x^2*exp(3)*exp((x^2-4)/x)^ 2+(4*x^3+8*x^2)*exp(3)*exp((x^2-4)/x)+(4*x^4+16*x^3+16*x^2)*exp(3)),x, alg orithm=\
Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (19) = 38\).
Time = 0.27 (sec) , antiderivative size = 555, normalized size of antiderivative = 26.43 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10 \, {\left (2 \, x^{4} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + x^{3} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 6 \, x^{3} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} - x^{2} e^{\left (x + 6\right )} + 2 \, x^{2} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 12 \, x^{2} e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + 4 \, x e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 32 \, x e^{\left (\frac {3 \, x + 4}{x} + 3\right )} + 8 \, e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x}\right )} + 32 \, e^{\left (\frac {3 \, x + 4}{x} + 3\right )}\right )}}{4 \, x^{5} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 2 \, x^{4} e^{\left (x + 9\right )} + 2 \, x^{4} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 20 \, x^{4} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + x^{3} e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 6 \, x^{3} e^{\left (x + 9\right )} + 6 \, x^{3} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 48 \, x^{3} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + x^{2} e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 12 \, x^{2} e^{\left (x + 9\right )} + 12 \, x^{2} e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 112 \, x^{2} e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 4 \, x e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 32 \, x e^{\left (x + 9\right )} + 32 \, x e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 192 \, x e^{\left (\frac {3 \, x + 4}{x} + 6\right )} + 8 \, e^{\left (x + \frac {x^{2} + 3 \, x - 4}{x} + 6\right )} + 32 \, e^{\left (x + 9\right )} + 32 \, e^{\left (\frac {x^{2} + 3 \, x - 4}{x} + \frac {3 \, x + 4}{x} + 3\right )} + 128 \, e^{\left (\frac {3 \, x + 4}{x} + 6\right )}} \]
integrate(((-10*x^2-40)*exp((x^2-4)/x)-20*x^2)/(x^2*exp(3)*exp((x^2-4)/x)^ 2+(4*x^3+8*x^2)*exp(3)*exp((x^2-4)/x)+(4*x^4+16*x^3+16*x^2)*exp(3)),x, alg orithm=\
10*(2*x^4*e^((3*x + 4)/x + 3) + x^3*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x) + 6*x^3*e^((3*x + 4)/x + 3) - x^2*e^(x + 6) + 2*x^2*e^((x^2 + 3*x - 4)/x + ( 3*x + 4)/x) + 12*x^2*e^((3*x + 4)/x + 3) + 4*x*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x) + 32*x*e^((3*x + 4)/x + 3) + 8*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x ) + 32*e^((3*x + 4)/x + 3))/(4*x^5*e^((3*x + 4)/x + 6) + 2*x^4*e^(x + 9) + 2*x^4*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x + 3) + 20*x^4*e^((3*x + 4)/x + 6 ) + x^3*e^(x + (x^2 + 3*x - 4)/x + 6) + 6*x^3*e^(x + 9) + 6*x^3*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x + 3) + 48*x^3*e^((3*x + 4)/x + 6) + x^2*e^(x + (x ^2 + 3*x - 4)/x + 6) + 12*x^2*e^(x + 9) + 12*x^2*e^((x^2 + 3*x - 4)/x + (3 *x + 4)/x + 3) + 112*x^2*e^((3*x + 4)/x + 6) + 4*x*e^(x + (x^2 + 3*x - 4)/ x + 6) + 32*x*e^(x + 9) + 32*x*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x + 3) + 1 92*x*e^((3*x + 4)/x + 6) + 8*e^(x + (x^2 + 3*x - 4)/x + 6) + 32*e^(x + 9) + 32*e^((x^2 + 3*x - 4)/x + (3*x + 4)/x + 3) + 128*e^((3*x + 4)/x + 6))
Time = 10.84 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-20 x^2+e^{\frac {-4+x^2}{x}} \left (-40-10 x^2\right )}{e^{3+\frac {2 \left (-4+x^2\right )}{x}} x^2+e^{3+\frac {-4+x^2}{x}} \left (8 x^2+4 x^3\right )+e^3 \left (16 x^2+16 x^3+4 x^4\right )} \, dx=\frac {10}{4\,{\mathrm {e}}^3+2\,x\,{\mathrm {e}}^3+{\mathrm {e}}^3\,{\mathrm {e}}^{-\frac {4}{x}}\,{\mathrm {e}}^x} \]