\(\int \frac {e^{-2 e} (-10-12 x-6 x^2-x^3)}{8+12 x+6 x^2+x^3} \, dx\) [3754]

Optimal result

Integrand size = 36, antiderivative size = 24 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=e^{-2 e} \left (4-x+\frac {1}{3} \left (-4+\frac {3}{(2+x)^2}\right )\right ) \]

[Out]

(8/3+1/(2+x)^2-x)/exp(exp(1))^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 2099} \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {e^{-2 e}}{(x+2)^2}-e^{-2 e} x \]

[In]

Int[(-10 - 12*x - 6*x^2 - x^3)/(E^(2*E)*(8 + 12*x + 6*x^2 + x^3)),x]

[Out]

-(x/E^(2*E)) + 1/(E^(2*E)*(2 + x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = e^{-2 e} \int \frac {-10-12 x-6 x^2-x^3}{8+12 x+6 x^2+x^3} \, dx \\ & = e^{-2 e} \int \left (-1-\frac {2}{(2+x)^3}\right ) \, dx \\ & = -e^{-2 e} x+\frac {e^{-2 e}}{(2+x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-e^{-2 e} \left (x-\frac {1}{(2+x)^2}\right ) \]

[In]

Integrate[(-10 - 12*x - 6*x^2 - x^3)/(E^(2*E)*(8 + 12*x + 6*x^2 + x^3)),x]

[Out]

-((x - (2 + x)^(-2))/E^(2*E))

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
default	\({\mathrm e}^{-2 \,{\mathrm e}} \left (-x +\frac {1}{\left (2+x \right )^{2}}\right )\)	\(16\)
gosper	\(-\frac {\left (x^{3}-12 x -17\right ) {\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\)	\(26\)
risch	\(-{\mathrm e}^{-2 \,{\mathrm e}} x +\frac {{\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\)	\(26\)
parallelrisch	\(-\frac {\left (x^{3}-12 x -17\right ) {\mathrm e}^{-2 \,{\mathrm e}}}{x^{2}+4 x +4}\)	\(26\)
norman	\(\frac {\left (12 \,{\mathrm e}^{-{\mathrm e}} x -{\mathrm e}^{-{\mathrm e}} x^{3}+17 \,{\mathrm e}^{-{\mathrm e}}\right ) {\mathrm e}^{-{\mathrm e}}}{\left (2+x \right )^{2}}\)	\(38\)

[In]

int((-x^3-6*x^2-12*x-10)/(x^3+6*x^2+12*x+8)/exp(exp(1))^2,x,method=_RETURNVERBOSE)

[Out]

1/exp(exp(1))^2*(-x+1/(2+x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-\frac {{\left (x^{3} + 4 \, x^{2} + 4 \, x - 1\right )} e^{\left (-2 \, e\right )}}{x^{2} + 4 \, x + 4} \]

[In]

integrate((-x^3-6*x^2-12*x-10)/(x^3+6*x^2+12*x+8)/exp(exp(1))^2,x, algorithm="fricas")

[Out]

-(x^3 + 4*x^2 + 4*x - 1)*e^(-2*e)/(x^2 + 4*x + 4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=- \frac {x}{e^{2 e}} + \frac {1}{x^{2} e^{2 e} + 4 x e^{2 e} + 4 e^{2 e}} \]

[In]

integrate((-x**3-6*x**2-12*x-10)/(x**3+6*x**2+12*x+8)/exp(exp(1))**2,x)

[Out]

-x*exp(-2*E) + 1/(x**2*exp(2*E) + 4*x*exp(2*E) + 4*exp(2*E))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-{\left (x - \frac {1}{x^{2} + 4 \, x + 4}\right )} e^{\left (-2 \, e\right )} \]

[In]

integrate((-x^3-6*x^2-12*x-10)/(x^3+6*x^2+12*x+8)/exp(exp(1))^2,x, algorithm="maxima")

[Out]

-(x - 1/(x^2 + 4*x + 4))*e^(-2*e)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=-{\left (x - \frac {1}{{\left (x + 2\right )}^{2}}\right )} e^{\left (-2 \, e\right )} \]

[In]

integrate((-x^3-6*x^2-12*x-10)/(x^3+6*x^2+12*x+8)/exp(exp(1))^2,x, algorithm="giac")

[Out]

-(x - 1/(x + 2)^2)*e^(-2*e)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-2 e} \left (-10-12 x-6 x^2-x^3\right )}{8+12 x+6 x^2+x^3} \, dx=\frac {{\mathrm {e}}^{-2\,\mathrm {e}}}{{\left (x+2\right )}^2}-x\,{\mathrm {e}}^{-2\,\mathrm {e}} \]

[In]

int(-(exp(-2*exp(1))*(12*x + 6*x^2 + x^3 + 10))/(12*x + 6*x^2 + x^3 + 8),x)

[Out]

exp(-2*exp(1))/(x + 2)^2 - x*exp(-2*exp(1))