\(\int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx\) [541]

Optimal result

Integrand size = 29, antiderivative size = 10 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4+4 e^x x} \]

[Out]

1/(4*exp(x)*x+4)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6820, 12, 6818} \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 \left (e^x x+1\right )} \]

[In]

Int[(E^x*(-1 - x))/(4 + 8*E^x*x + 4*E^(2*x)*x^2),x]

[Out]

1/(4*(1 + E^x*x))

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x (-1-x)}{4 \left (1+e^x x\right )^2} \, dx \\ & = \frac {1}{4} \int \frac {e^x (-1-x)}{\left (1+e^x x\right )^2} \, dx \\ & = \frac {1}{4 \left (1+e^x x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 \left (1+e^x x\right )} \]

[In]

Integrate[(E^x*(-1 - x))/(4 + 8*E^x*x + 4*E^(2*x)*x^2),x]

[Out]

1/(4*(1 + E^x*x))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
norman	\(\frac {1}{4 \,{\mathrm e}^{x} x +4}\)	\(11\)
risch	\(\frac {1}{4 \,{\mathrm e}^{x} x +4}\)	\(11\)
parallelrisch	\(\frac {1}{4 \,{\mathrm e}^{x} x +4}\)	\(11\)

[In]

int((-1-x)*exp(x)/(4*exp(x)^2*x^2+8*exp(x)*x+4),x,method=_RETURNVERBOSE)

[Out]

1/4/(exp(x)*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 \, {\left (x e^{x} + 1\right )}} \]

[In]

integrate((-1-x)*exp(x)/(4*exp(x)^2*x^2+8*exp(x)*x+4),x, algorithm="fricas")

[Out]

1/4/(x*e^x + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 x e^{x} + 4} \]

[In]

integrate((-1-x)*exp(x)/(4*exp(x)**2*x**2+8*exp(x)*x+4),x)

[Out]

1/(4*x*exp(x) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 \, {\left (x e^{x} + 1\right )}} \]

[In]

integrate((-1-x)*exp(x)/(4*exp(x)^2*x^2+8*exp(x)*x+4),x, algorithm="maxima")

[Out]

1/4/(x*e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4 \, {\left (x e^{x} + 1\right )}} \]

[In]

integrate((-1-x)*exp(x)/(4*exp(x)^2*x^2+8*exp(x)*x+4),x, algorithm="giac")

[Out]

1/4/(x*e^x + 1)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90 \[ \int \frac {e^x (-1-x)}{4+8 e^x x+4 e^{2 x} x^2} \, dx=\frac {1}{4\,x\,{\mathrm {e}}^x+4} \]

[In]

int(-(exp(x)*(x + 1))/(4*x^2*exp(2*x) + 8*x*exp(x) + 4),x)

[Out]

1/(4*x*exp(x) + 4)