Integrand size = 68, antiderivative size = 30 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {5 e^{-x} \left (e^4+e^{1+2 x}\right )^2}{x \left (x+x^2\right )} \end {dmath*}
Time = 1.74 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {5 e^{2-x} \left (e^3+e^{2 x}\right )^2}{x^2 (1+x)} \end {dmath*}
Integrate[(E^8*(-10 - 20*x - 5*x^2) + E^(5 + 2*x)*(-20 - 20*x + 10*x^2) + E^(2 + 4*x)*(-10 + 15*x^2))/(E^x*(x^3 + 2*x^4 + x^5)),x]
Leaf count is larger than twice the leaf count of optimal. \(109\) vs. \(2(30)=60\).
Time = 2.50 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.63, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2026, 2007, 7292, 27, 25, 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.
\(\displaystyle \int \frac {e^{-x} \left (e^8 \left (-5 x^2-20 x-10\right )+e^{2 x+5} \left (10 x^2-20 x-20\right )+e^{4 x+2} \left (15 x^2-10\right )\right )}{x^5+2 x^4+x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (e^8 \left (-5 x^2-20 x-10\right )+e^{2 x+5} \left (10 x^2-20 x-20\right )+e^{4 x+2} \left (15 x^2-10\right )\right )}{x^3 \left (x^2+2 x+1\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-x} \left (e^8 \left (-5 x^2-20 x-10\right )+e^{2 x+5} \left (10 x^2-20 x-20\right )+e^{4 x+2} \left (15 x^2-10\right )\right )}{x^3 (x+1)^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e^{2-x} \left (e^{2 x}+e^3\right ) \left (3 e^{2 x} x^2-e^3 x^2-4 e^3 x-2 e^{2 x}-2 e^3\right )}{x^3 (x+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {e^{2-x} \left (e^3+e^{2 x}\right ) \left (-3 e^{2 x} x^2+e^3 x^2+4 e^3 x+2 e^{2 x}+2 e^3\right )}{x^3 (x+1)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {e^{2-x} \left (e^3+e^{2 x}\right ) \left (-3 e^{2 x} x^2+e^3 x^2+4 e^3 x+2 e^{2 x}+2 e^3\right )}{x^3 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (-\frac {2 e^{x+5} \left (x^2-2 x-2\right )}{x^3 (x+1)^2}+\frac {e^{8-x} \left (x^2+4 x+2\right )}{x^3 (x+1)^2}-\frac {e^{3 x+2} \left (3 x^2-2\right )}{x^3 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (-\frac {e^{8-x}}{x^2}-\frac {2 e^{x+5}}{x^2}-\frac {e^{3 x+2}}{x^2}-\frac {e^{8-x}}{x+1}-\frac {2 e^{x+5}}{x+1}-\frac {e^{3 x+2}}{x+1}+\frac {e^{8-x}}{x}+\frac {2 e^{x+5}}{x}+\frac {e^{3 x+2}}{x}\right )\) |
Int[(E^8*(-10 - 20*x - 5*x^2) + E^(5 + 2*x)*(-20 - 20*x + 10*x^2) + E^(2 + 4*x)*(-10 + 15*x^2))/(E^x*(x^3 + 2*x^4 + x^5)),x]
-5*(-(E^(8 - x)/x^2) - (2*E^(5 + x))/x^2 - E^(2 + 3*x)/x^2 + E^(8 - x)/x + (2*E^(5 + x))/x + E^(2 + 3*x)/x - E^(8 - x)/(1 + x) - (2*E^(5 + x))/(1 + x) - E^(2 + 3*x)/(1 + x))
3.1.13.3.1 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_.)*(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 = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37
method | result | size |
norman | \(\frac {\left (5 \,{\mathrm e}^{8}+5 \,{\mathrm e}^{2} {\mathrm e}^{4 x}+10 \,{\mathrm e}^{4} {\mathrm e} \,{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}}{x^{2} \left (1+x \right )}\) | \(41\) |
parallelrisch | \(\frac {\left (5 \,{\mathrm e}^{8}+10 \,{\mathrm e}^{4} {\mathrm e}^{1+2 x}+5 \,{\mathrm e}^{4 x +2}\right ) {\mathrm e}^{-x}}{\left (1+x \right ) x^{2}}\) | \(41\) |
risch | \(\frac {5 \,{\mathrm e}^{2+3 x}}{x^{2} \left (1+x \right )}+\frac {10 \,{\mathrm e}^{5+x}}{x^{2} \left (1+x \right )}+\frac {5 \,{\mathrm e}^{8-x}}{x^{2} \left (1+x \right )}\) | \(48\) |
default | \(-10 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{-x} \left (7 x^{2}+4 x -1\right )}{2 x^{2} \left (1+x \right )}-\frac {11 \,\operatorname {Ei}_{1}\left (x \right )}{2}+2 \,{\mathrm e} \,\operatorname {Ei}_{1}\left (1+x \right )\right )-10 \,{\mathrm e}^{2} \left (-\frac {{\mathrm e}^{3 x}}{2 x^{2}}+\frac {{\mathrm e}^{3 x}}{2 x}-\frac {3 \,\operatorname {Ei}_{1}\left (-3 x \right )}{2}+\frac {{\mathrm e}^{3 x}}{1+x}+6 \,{\mathrm e}^{-3} \operatorname {Ei}_{1}\left (-3 x -3\right )\right )-20 \,{\mathrm e}^{8} \left (-\frac {{\mathrm e}^{-x} \left (1+2 x \right )}{x \left (1+x \right )}+3 \,\operatorname {Ei}_{1}\left (x \right )-{\mathrm e} \,\operatorname {Ei}_{1}\left (1+x \right )\right )-5 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{-x}}{1+x}-\operatorname {Ei}_{1}\left (x \right )\right )-20 \,{\mathrm e} \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}+\frac {3 \,{\mathrm e}^{x}}{2 x}-\frac {3 \,\operatorname {Ei}_{1}\left (-x \right )}{2}+\frac {{\mathrm e}^{x}}{1+x}+4 \,{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-x \right )\right )+15 \,{\mathrm e}^{2} \left (-\operatorname {Ei}_{1}\left (-3 x \right )+\frac {{\mathrm e}^{3 x}}{1+x}+4 \,{\mathrm e}^{-3} \operatorname {Ei}_{1}\left (-3 x -3\right )\right )-20 \,{\mathrm e} \,{\mathrm e}^{4} \left (\operatorname {Ei}_{1}\left (-x \right )-\frac {{\mathrm e}^{x}}{1+x}-3 \,{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-x \right )-\frac {{\mathrm e}^{x}}{x}\right )+10 \,{\mathrm e} \,{\mathrm e}^{4} \left (-\operatorname {Ei}_{1}\left (-x \right )+\frac {{\mathrm e}^{x}}{1+x}+2 \,{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-x \right )\right )\) | \(316\) |
int(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5*x^2-2 0*x-10)*exp(4)^2)/(x^5+2*x^4+x^3)/exp(x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {5 \, {\left (e^{2} + 2 \, e^{\left (-2 \, x + 5\right )} + e^{\left (-4 \, x + 8\right )}\right )} e^{\left (3 \, x\right )}}{x^{3} + x^{2}} \end {dmath*}
integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5 *x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x^3)/exp(x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (22) = 44\).
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.47 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {\left (5 x^{6} e^{2} + 10 x^{5} e^{2} + 5 x^{4} e^{2}\right ) e^{3 x} + \left (10 x^{6} e^{5} + 20 x^{5} e^{5} + 10 x^{4} e^{5}\right ) e^{x} + \left (5 x^{6} e^{8} + 10 x^{5} e^{8} + 5 x^{4} e^{8}\right ) e^{- x}}{x^{9} + 3 x^{8} + 3 x^{7} + x^{6}} \end {dmath*}
integrate(((15*x**2-10)*exp(1+2*x)**2+(10*x**2-20*x-20)*exp(4)*exp(1+2*x)+ (-5*x**2-20*x-10)*exp(4)**2)/(x**5+2*x**4+x**3)/exp(x),x)
((5*x**6*exp(2) + 10*x**5*exp(2) + 5*x**4*exp(2))*exp(3*x) + (10*x**6*exp( 5) + 20*x**5*exp(5) + 10*x**4*exp(5))*exp(x) + (5*x**6*exp(8) + 10*x**5*ex p(8) + 5*x**4*exp(8))*exp(-x))/(x**9 + 3*x**8 + 3*x**7 + x**6)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {5 \, {\left (e^{\left (3 \, x + 2\right )} + 2 \, e^{\left (x + 5\right )} + e^{\left (-x + 8\right )}\right )}}{x^{3} + x^{2}} \end {dmath*}
integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5 *x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x^3)/exp(x),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {5 \, {\left (e^{\left (3 \, x + 2\right )} + 2 \, e^{\left (x + 5\right )} + e^{\left (-x + 8\right )}\right )}}{x^{3} + x^{2}} \end {dmath*}
integrate(((15*x^2-10)*exp(1+2*x)^2+(10*x^2-20*x-20)*exp(4)*exp(1+2*x)+(-5 *x^2-20*x-10)*exp(4)^2)/(x^5+2*x^4+x^3)/exp(x),x, algorithm=\
Time = 13.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \begin {dmath*} \int \frac {e^{-x} \left (e^8 \left (-10-20 x-5 x^2\right )+e^{5+2 x} \left (-20-20 x+10 x^2\right )+e^{2+4 x} \left (-10+15 x^2\right )\right )}{x^3+2 x^4+x^5} \, dx=\frac {10\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{x^3+x^2}+\frac {5\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^2}{x^3+x^2}+\frac {5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^8}{x^3+x^2} \end {dmath*}
int(-(exp(-x)*(exp(8)*(20*x + 5*x^2 + 10) - exp(4*x + 2)*(15*x^2 - 10) + e xp(4)*exp(2*x + 1)*(20*x - 10*x^2 + 20)))/(x^3 + 2*x^4 + x^5),x)