Integrand size = 77, antiderivative size = 31 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=x \left (-25-e^2-x+\frac {6 e^{-x}}{x^2 (x+x (1+x))}\right ) \] Output:
x*(6/exp(x)/x^2/(x*(1+x)+x)-x-exp(2)-25)
Time = 6.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=-\left (\left (25+e^2\right ) x\right )-x^2+\frac {6 e^{-x}}{x^2 (2+x)} \] Input:
Integrate[(-24 - 30*x - 6*x^2 + E^x*(-100*x^3 - 108*x^4 - 33*x^5 - 2*x^6 + E^2*(-4*x^3 - 4*x^4 - x^5)))/(E^x*(4*x^3 + 4*x^4 + x^5)),x]
Output:
-((25 + E^2)*x) - x^2 + 6/(E^x*x^2*(2 + x))
Time = 1.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {2026, 2007, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-x} \left (-6 x^2+e^x \left (-2 x^6-33 x^5-108 x^4-100 x^3+e^2 \left (-x^5-4 x^4-4 x^3\right )\right )-30 x-24\right )}{x^5+4 x^4+4 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (-6 x^2+e^x \left (-2 x^6-33 x^5-108 x^4-100 x^3+e^2 \left (-x^5-4 x^4-4 x^3\right )\right )-30 x-24\right )}{x^3 \left (x^2+4 x+4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-x} \left (-6 x^2+e^x \left (-2 x^6-33 x^5-108 x^4-100 x^3+e^2 \left (-x^5-4 x^4-4 x^3\right )\right )-30 x-24\right )}{x^3 (x+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {6 e^{-x} \left (x^2+5 x+4\right )}{(x+2)^2 x^3}-2 x-e^2-25\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x^2+\frac {3 e^{-x}}{x^2}-\left (25+e^2\right ) x+\frac {3 e^{-x}}{2 (x+2)}-\frac {3 e^{-x}}{2 x}\) |
Input:
Int[(-24 - 30*x - 6*x^2 + E^x*(-100*x^3 - 108*x^4 - 33*x^5 - 2*x^6 + E^2*( -4*x^3 - 4*x^4 - x^5)))/(E^x*(4*x^3 + 4*x^4 + x^5)),x]
Output:
3/(E^x*x^2) - 3/(2*E^x*x) - (25 + E^2)*x - x^2 + 3/(2*E^x*(2 + x))
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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])
Time = 0.58 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-{\mathrm e}^{2} x -x^{2}-25 x +\frac {6 \,{\mathrm e}^{-x}}{x^{2} \left (2+x \right )}\) | \(29\) |
norman | \(\frac {\left (6+\left (-27-{\mathrm e}^{2}\right ) x^{4} {\mathrm e}^{x}+\left (100+4 \,{\mathrm e}^{2}\right ) x^{2} {\mathrm e}^{x}-x^{5} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{2} \left (2+x \right )}\) | \(47\) |
parallelrisch | \(\frac {\left (6-x^{4} {\mathrm e}^{2} {\mathrm e}^{x}-x^{5} {\mathrm e}^{x}-27 \,{\mathrm e}^{x} x^{4}+4 x^{2} {\mathrm e}^{2} {\mathrm e}^{x}+100 \,{\mathrm e}^{x} x^{2}\right ) {\mathrm e}^{-x}}{x^{2} \left (2+x \right )}\) | \(55\) |
parts | \(-x^{2}-25 x -\frac {3 \,{\mathrm e}^{-x}}{2+x}-\frac {3 \,{\mathrm e}^{-x} \left (4 x^{2}+5 x -2\right )}{x^{2} \left (2+x \right )}+\frac {15 \,{\mathrm e}^{-x} \left (1+x \right )}{\left (2+x \right ) x}-{\mathrm e}^{2} x\) | \(67\) |
default | \(-x^{2}-25 x -\frac {3 \,{\mathrm e}^{-x} \left (4 x^{2}+5 x -2\right )}{x^{2} \left (2+x \right )}+\frac {4 \,{\mathrm e}^{2}}{2+x}-4 \,{\mathrm e}^{2} \left (\ln \left (2+x \right )+\frac {2}{2+x}\right )-{\mathrm e}^{2} \left (x -4 \ln \left (2+x \right )-\frac {4}{2+x}\right )+\frac {15 \,{\mathrm e}^{-x} \left (1+x \right )}{\left (2+x \right ) x}-\frac {3 \,{\mathrm e}^{-x}}{2+x}\) | \(106\) |
Input:
int((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2 -30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x,method=_RETURNVERBOSE)
Output:
-exp(2)*x-x^2-25*x+6/x^2/(2+x)*exp(-x)
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=-\frac {{\left ({\left (x^{5} + 27 \, x^{4} + 50 \, x^{3} + {\left (x^{4} + 2 \, x^{3}\right )} e^{2}\right )} e^{x} - 6\right )} e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \] Input:
integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x) -6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x, algorithm="fricas")
Output:
-((x^5 + 27*x^4 + 50*x^3 + (x^4 + 2*x^3)*e^2)*e^x - 6)*e^(-x)/(x^3 + 2*x^2 )
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=- x^{2} + x \left (-25 - e^{2}\right ) + \frac {6 e^{- x}}{x^{3} + 2 x^{2}} \] Input:
integrate((((-x**5-4*x**4-4*x**3)*exp(2)-2*x**6-33*x**5-108*x**4-100*x**3) *exp(x)-6*x**2-30*x-24)/(x**5+4*x**4+4*x**3)/exp(x),x)
Output:
-x**2 + x*(-25 - exp(2)) + 6*exp(-x)/(x**3 + 2*x**2)
Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=-\frac {x^{5} + x^{4} {\left (e^{2} + 27\right )} + 2 \, x^{3} {\left (e^{2} + 25\right )} - 6 \, e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \] Input:
integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x) -6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x, algorithm="maxima")
Output:
-(x^5 + x^4*(e^2 + 27) + 2*x^3*(e^2 + 25) - 6*e^(-x))/(x^3 + 2*x^2)
Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=-\frac {x^{5} + x^{4} e^{2} + 27 \, x^{4} + 2 \, x^{3} e^{2} + 50 \, x^{3} - 6 \, e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \] Input:
integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x) -6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x, algorithm="giac")
Output:
-(x^5 + x^4*e^2 + 27*x^4 + 2*x^3*e^2 + 50*x^3 - 6*e^(-x))/(x^3 + 2*x^2)
Time = 2.78 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {6}{2\,x^2\,{\mathrm {e}}^x+x^3\,{\mathrm {e}}^x}-\frac {x^5+\left ({\mathrm {e}}^2+27\right )\,x^4+\left (2\,{\mathrm {e}}^2+50\right )\,x^3}{x^3+2\,x^2} \] Input:
int(-(exp(-x)*(30*x + 6*x^2 + exp(x)*(exp(2)*(4*x^3 + 4*x^4 + x^5) + 100*x ^3 + 108*x^4 + 33*x^5 + 2*x^6) + 24))/(4*x^3 + 4*x^4 + x^5),x)
Output:
6/(2*x^2*exp(x) + x^3*exp(x)) - (x^3*(2*exp(2) + 50) + x^5 + x^4*(exp(2) + 27))/(2*x^2 + x^3)
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00 \[ \int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{4 x^3+4 x^4+x^5} \, dx=\frac {-e^{x} e^{2} x^{4}-2 e^{x} e^{2} x^{3}-e^{x} x^{5}-27 e^{x} x^{4}-50 e^{x} x^{3}+6}{e^{x} x^{2} \left (x +2\right )} \] Input:
int((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2 -30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x)
Output:
( - e**x*e**2*x**4 - 2*e**x*e**2*x**3 - e**x*x**5 - 27*e**x*x**4 - 50*e**x *x**3 + 6)/(e**x*x**2*(x + 2))