Integrand size = 82, antiderivative size = 26 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x (1+x)}} \]
Time = 5.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x (1+x)}} \]
Integrate[(E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))*(-4 - 8*x + 2*x^3 + x^4 + E^x*(4*x + 4*x^2 + x^4 + x^5)))/(3*x^2 + 6*x^3 + 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.
\(\displaystyle \int \frac {e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5} \left (x^4+2 x^3+e^x \left (x^5+x^4+4 x^2+4 x\right )-8 x-4\right )}{3 x^4+6 x^3+3 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5} \left (x^4+2 x^3+e^x \left (x^5+x^4+4 x^2+4 x\right )-8 x-4\right )}{x^2 \left (3 x^2+6 x+3\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5} \left (x^4+2 x^3+e^x \left (x^5+x^4+4 x^2+4 x\right )-8 x-4\right )}{x^2 \left (\sqrt {3} x+\sqrt {3}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (x^3+4\right ) \exp \left (\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+x+e^x-5\right )}{3 (x+1) x}+\frac {e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5} x^2}{3 (x+1)^2}+\frac {2 e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5} x}{3 (x+1)^2}-\frac {8 e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5}}{3 (x+1)^2 x}-\frac {4 e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x^2+3 x}+e^x-5}}{3 (x+1)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {e^{e^x-5} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} \left (e^x x^5+\left (e^x+1\right ) x^4+2 x^3+4 e^x x^2+4 \left (e^x-2\right ) x-4\right )}{3 x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} \left (-e^x x^5-\left (1+e^x\right ) x^4-2 x^3-4 e^x x^2+4 \left (2-e^x\right ) x+4\right )}{x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} \left (-e^x x^5-\left (1+e^x\right ) x^4-2 x^3-4 e^x x^2+4 \left (2-e^x\right ) x+4\right )}{x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} \int \left (-\frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} x^2}{(x+1)^2}-\frac {2 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} x}{(x+1)^2}-\frac {\exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right ) \left (x^3+4\right )}{(x+1) x}+\frac {8 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{(x+1)^2 x}+\frac {4 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{(x+1)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -\frac {1}{3} \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} \left (-e^x x^5-\left (1+e^x\right ) x^4-2 x^3-4 e^x x^2-4 \left (-2+e^x\right ) x+4\right )}{x^2 (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} \int \left (-\frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} x^2}{(x+1)^2}-\frac {2 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5} x}{(x+1)^2}-\frac {\exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right ) \left (x^3+4\right )}{(x+1) x}+\frac {8 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{(x+1)^2 x}+\frac {4 e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{(x+1)^2 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\int \exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right )dx+4 \int \frac {\exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right )}{x}dx+\int \exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right ) xdx-3 \int \frac {\exp \left (x+e^x-5+\frac {e^{-5+e^x} \left (x^3+4\right )}{3 (x+1) x}\right )}{x+1}dx+\int e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}dx-8 \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{-x-1}dx+3 \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{(x+1)^2}dx-8 \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{x+1}dx-4 \int \frac {e^{\frac {e^{-5+e^x} \left (x^3+4\right )}{3 x (x+1)}+e^x-5}}{x^2}dx\right )\) |
Int[(E^(-5 + E^x + (E^(-5 + E^x)*(4 + x^3))/(3*x + 3*x^2))*(-4 - 8*x + 2*x ^3 + x^4 + E^x*(4*x + 4*x^2 + x^4 + x^5)))/(3*x^2 + 6*x^3 + 3*x^4),x]
3.26.97.3.1 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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 20.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}-5} \left (x^{3}+4\right )}{3 \left (1+x \right ) x}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x}-5} \left (x^{3}+4\right )}{3 \left (1+x \right ) x}}\) | \(22\) |
int(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp((x^3+4) *exp(exp(x)-5)/(3*x^2+3*x))/(3*x^4+6*x^3+3*x^2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (21) = 42\).
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\left (-\frac {15 \, x^{2} - 3 \, {\left (x^{2} + x\right )} e^{x} - {\left (x^{3} + 4\right )} e^{\left (e^{x} - 5\right )} + 15 \, x}{3 \, {\left (x^{2} + x\right )}} - e^{x} + 5\right )} \]
integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp(( x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3*x^4+6*x^3+3*x^2),x, algorithm=\
Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\frac {\left (x^{3} + 4\right ) e^{e^{x} - 5}}{3 x^{2} + 3 x}} \]
integrate(((x**5+x**4+4*x**2+4*x)*exp(x)+x**4+2*x**3-8*x-4)*exp(exp(x)-5)* exp((x**3+4)*exp(exp(x)-5)/(3*x**2+3*x))/(3*x**4+6*x**3+3*x**2),x)
Time = 0.42 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=e^{\left (\frac {1}{3} \, x e^{\left (e^{x} - 5\right )} + \frac {4 \, e^{\left (e^{x} - 5\right )}}{3 \, x} - \frac {e^{\left (e^{x}\right )}}{x e^{5} + e^{5}} - \frac {1}{3} \, e^{\left (e^{x} - 5\right )}\right )} \]
integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp(( x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3*x^4+6*x^3+3*x^2),x, algorithm=\
\[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx=\int { \frac {{\left (x^{4} + 2 \, x^{3} + {\left (x^{5} + x^{4} + 4 \, x^{2} + 4 \, x\right )} e^{x} - 8 \, x - 4\right )} e^{\left (\frac {{\left (x^{3} + 4\right )} e^{\left (e^{x} - 5\right )}}{3 \, {\left (x^{2} + x\right )}} + e^{x} - 5\right )}}{3 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}} \,d x } \]
integrate(((x^5+x^4+4*x^2+4*x)*exp(x)+x^4+2*x^3-8*x-4)*exp(exp(x)-5)*exp(( x^3+4)*exp(exp(x)-5)/(3*x^2+3*x))/(3*x^4+6*x^3+3*x^2),x, algorithm=\
integrate(1/3*(x^4 + 2*x^3 + (x^5 + x^4 + 4*x^2 + 4*x)*e^x - 8*x - 4)*e^(1 /3*(x^3 + 4)*e^(e^x - 5)/(x^2 + x) + e^x - 5)/(x^4 + 2*x^3 + x^2), x)
Time = 11.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-5+e^x+\frac {e^{-5+e^x} \left (4+x^3\right )}{3 x+3 x^2}} \left (-4-8 x+2 x^3+x^4+e^x \left (4 x+4 x^2+x^4+x^5\right )\right )}{3 x^2+6 x^3+3 x^4} \, dx={\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}}{3\,x^2+3\,x}}\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-5}}{3\,x^2+3\,x}} \]