3.60.59 \(\int \frac {e^{1-10 x-x^2} (2-20 x-14 x^2-2 x^3)}{4+4 x+x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {e^{1-x (10+x)} x}{2+x} \]

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Rubi [B]  time = 0.07, antiderivative size = 35, normalized size of antiderivative = 2.06, number of steps used = 2, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 2288} \begin {gather*} \frac {e^{-x^2-10 x+1} \left (x^3+7 x^2+10 x\right )}{(x+2)^2 (x+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - 10*x - x^2)*(2 - 20*x - 14*x^2 - 2*x^3))/(4 + 4*x + x^2),x]

[Out]

(E^(1 - 10*x - x^2)*(10*x + 7*x^2 + x^3))/((2 + x)^2*(5 + x))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{1-10 x-x^2} \left (2-20 x-14 x^2-2 x^3\right )}{(2+x)^2} \, dx\\ &=\frac {e^{1-10 x-x^2} \left (10 x+7 x^2+x^3\right )}{(2+x)^2 (5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.29 \begin {gather*} \frac {2 e^{1-10 x-x^2} x}{4+2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - 10*x - x^2)*(2 - 20*x - 14*x^2 - 2*x^3))/(4 + 4*x + x^2),x]

[Out]

(2*E^(1 - 10*x - x^2)*x)/(4 + 2*x)

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fricas [A]  time = 0.79, size = 18, normalized size = 1.06 \begin {gather*} \frac {x e^{\left (-x^{2} - 10 \, x + 1\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3-14*x^2-20*x+2)*exp(-x^2-10*x+1)/(x^2+4*x+4),x, algorithm="fricas")

[Out]

x*e^(-x^2 - 10*x + 1)/(x + 2)

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giac [A]  time = 0.26, size = 18, normalized size = 1.06 \begin {gather*} \frac {x e^{\left (-x^{2} - 10 \, x + 1\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3-14*x^2-20*x+2)*exp(-x^2-10*x+1)/(x^2+4*x+4),x, algorithm="giac")

[Out]

x*e^(-x^2 - 10*x + 1)/(x + 2)

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maple [A]  time = 0.44, size = 19, normalized size = 1.12

method	result	size
gosper	\(\frac {x \,{\mathrm e}^{-x^{2}-10 x +1}}{2+x}\)	\(19\)
norman	\(\frac {x \,{\mathrm e}^{-x^{2}-10 x +1}}{2+x}\)	\(19\)
risch	\(\frac {x \,{\mathrm e}^{-x^{2}-10 x +1}}{2+x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^3-14*x^2-20*x+2)*exp(-x^2-10*x+1)/(x^2+4*x+4),x,method=_RETURNVERBOSE)

[Out]

x*exp(-x^2-10*x+1)/(2+x)

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maxima [A]  time = 0.41, size = 18, normalized size = 1.06 \begin {gather*} \frac {x e^{\left (-x^{2} - 10 \, x + 1\right )}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^3-14*x^2-20*x+2)*exp(-x^2-10*x+1)/(x^2+4*x+4),x, algorithm="maxima")

[Out]

x*e^(-x^2 - 10*x + 1)/(x + 2)

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mupad [B]  time = 0.16, size = 19, normalized size = 1.12 \begin {gather*} \frac {x\,{\mathrm {e}}^{-10\,x}\,\mathrm {e}\,{\mathrm {e}}^{-x^2}}{x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1 - x^2 - 10*x)*(20*x + 14*x^2 + 2*x^3 - 2))/(4*x + x^2 + 4),x)

[Out]

(x*exp(-10*x)*exp(1)*exp(-x^2))/(x + 2)

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sympy [A]  time = 0.12, size = 14, normalized size = 0.82 \begin {gather*} \frac {x e^{- x^{2} - 10 x + 1}}{x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**3-14*x**2-20*x+2)*exp(-x**2-10*x+1)/(x**2+4*x+4),x)

[Out]

x*exp(-x**2 - 10*x + 1)/(x + 2)

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