Integrand size = 83, antiderivative size = 28 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{-5+\frac {2}{x^2}+\frac {4}{x (3+x)}}}{35 (5+x)} \end {dmath*}
Time = 1.45 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{-5+\frac {2}{x^2}+\frac {4}{3 x}-\frac {4}{3 (3+x)}}}{35 (5+x)} \end {dmath*}
Integrate[(E^((6 + 6*x - 15*x^2 - 5*x^3)/(3*x^2 + x^3))*(-180 - 216*x - 96 *x^2 - 21*x^3 - 6*x^4 - x^5))/(7875*x^3 + 8400*x^4 + 3290*x^5 + 560*x^6 + 35*x^7),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.
\(\displaystyle \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{35 x^7+560 x^6+3290 x^5+8400 x^4+7875 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{x^3 \left (35 x^4+560 x^3+3290 x^2+8400 x+7875\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{140 x^3 (x+3)}+\frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{140 x^3 (x+5)}+\frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{140 x^3 (x+3)^2}+\frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^3+3 x^2}} \left (-x^5-6 x^4-21 x^3-96 x^2-216 x-180\right )}{140 x^3 (x+5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{175} \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{x^3}dx-\frac {8 \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{x^2}dx}{2625}+\frac {8 \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{x}dx}{13125}+\frac {2}{105} \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{(x+3)^2}dx-\frac {1}{105} \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{x+3}dx-\frac {1}{35} \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{(x+5)^2}dx+\frac {39 \int \frac {e^{\frac {-5 x^3-15 x^2+6 x+6}{x^2 (x+3)}}}{x+5}dx}{4375}\) |
Int[(E^((6 + 6*x - 15*x^2 - 5*x^3)/(3*x^2 + x^3))*(-180 - 216*x - 96*x^2 - 21*x^3 - 6*x^4 - x^5))/(7875*x^3 + 8400*x^4 + 3290*x^5 + 560*x^6 + 35*x^7 ),x]
3.2.14.3.1 Defintions of rubi rules used
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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
gosper | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+15 x^{2}-6 x -6}{x^{2} \left (3+x \right )}}}{175+35 x}\) | \(34\) |
risch | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+15 x^{2}-6 x -6}{x^{2} \left (3+x \right )}}}{175+35 x}\) | \(34\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {5 x^{3}+15 x^{2}-6 x -6}{x^{2} \left (3+x \right )}}}{175+35 x}\) | \(34\) |
norman | \(\frac {\frac {3 x^{2} {\mathrm e}^{\frac {-5 x^{3}-15 x^{2}+6 x +6}{x^{3}+3 x^{2}}}}{35}+\frac {x^{3} {\mathrm e}^{\frac {-5 x^{3}-15 x^{2}+6 x +6}{x^{3}+3 x^{2}}}}{35}}{x^{2} \left (x^{2}+8 x +15\right )}\) | \(82\) |
int((-x^5-6*x^4-21*x^3-96*x^2-216*x-180)*exp((-5*x^3-15*x^2+6*x+6)/(x^3+3* x^2))/(35*x^7+560*x^6+3290*x^5+8400*x^4+7875*x^3),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{\left (-\frac {5 \, x^{3} + 15 \, x^{2} - 6 \, x - 6}{x^{3} + 3 \, x^{2}}\right )}}{35 \, {\left (x + 5\right )}} \end {dmath*}
integrate((-x^5-6*x^4-21*x^3-96*x^2-216*x-180)*exp((-5*x^3-15*x^2+6*x+6)/( x^3+3*x^2))/(35*x^7+560*x^6+3290*x^5+8400*x^4+7875*x^3),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{\frac {- 5 x^{3} - 15 x^{2} + 6 x + 6}{x^{3} + 3 x^{2}}}}{35 x + 175} \end {dmath*}
integrate((-x**5-6*x**4-21*x**3-96*x**2-216*x-180)*exp((-5*x**3-15*x**2+6* x+6)/(x**3+3*x**2))/(35*x**7+560*x**6+3290*x**5+8400*x**4+7875*x**3),x)
Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{\left (-\frac {4}{3 \, {\left (x + 3\right )}} + \frac {4}{3 \, x} + \frac {2}{x^{2}}\right )}}{35 \, {\left (x e^{5} + 5 \, e^{5}\right )}} \end {dmath*}
integrate((-x^5-6*x^4-21*x^3-96*x^2-216*x-180)*exp((-5*x^3-15*x^2+6*x+6)/( x^3+3*x^2))/(35*x^7+560*x^6+3290*x^5+8400*x^4+7875*x^3),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {e^{\left (-\frac {5 \, x^{3} + 15 \, x^{2} - 6 \, x - 6}{x^{3} + 3 \, x^{2}}\right )}}{35 \, {\left (x + 5\right )}} \end {dmath*}
integrate((-x^5-6*x^4-21*x^3-96*x^2-216*x-180)*exp((-5*x^3-15*x^2+6*x+6)/( x^3+3*x^2))/(35*x^7+560*x^6+3290*x^5+8400*x^4+7875*x^3),x, algorithm=\
Time = 13.54 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \begin {dmath*} \int \frac {e^{\frac {6+6 x-15 x^2-5 x^3}{3 x^2+x^3}} \left (-180-216 x-96 x^2-21 x^3-6 x^4-x^5\right )}{7875 x^3+8400 x^4+3290 x^5+560 x^6+35 x^7} \, dx=\frac {{\mathrm {e}}^{\frac {6}{x^2+3\,x}}\,{\mathrm {e}}^{-\frac {5\,x}{x+3}}\,{\mathrm {e}}^{\frac {6}{x^3+3\,x^2}}\,{\mathrm {e}}^{-\frac {15}{x+3}}}{35\,\left (x+5\right )} \end {dmath*}