3.95.38 \(\int \frac {1+e^x (-3-x)}{16+8 x+x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {-1-e^x}{4+x} \]

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Rubi [A]  time = 0.08, antiderivative size = 18, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {27, 6742, 2197} \begin {gather*} -\frac {e^x}{x+4}-\frac {1}{x+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^x*(-3 - x))/(16 + 8*x + x^2),x]

[Out]

-(4 + x)^(-1) - E^x/(4 + 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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+e^x (-3-x)}{(4+x)^2} \, dx\\ &=\int \left (\frac {1}{(4+x)^2}-\frac {e^x (3+x)}{(4+x)^2}\right ) \, dx\\ &=-\frac {1}{4+x}-\int \frac {e^x (3+x)}{(4+x)^2} \, dx\\ &=-\frac {1}{4+x}-\frac {e^x}{4+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 12, normalized size = 0.92 \begin {gather*} -\frac {1+e^x}{4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^x*(-3 - x))/(16 + 8*x + x^2),x]

[Out]

-((1 + E^x)/(4 + x))

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fricas [A]  time = 0.59, size = 11, normalized size = 0.85 \begin {gather*} -\frac {e^{x} + 1}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3-x)*exp(x)+1)/(x^2+8*x+16),x, algorithm="fricas")

[Out]

-(e^x + 1)/(x + 4)

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giac [A]  time = 0.19, size = 11, normalized size = 0.85 \begin {gather*} -\frac {e^{x} + 1}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3-x)*exp(x)+1)/(x^2+8*x+16),x, algorithm="giac")

[Out]

-(e^x + 1)/(x + 4)

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maple [A]  time = 0.62, size = 13, normalized size = 1.00

method	result	size
norman	\(\frac {-{\mathrm e}^{x}-1}{4+x}\)	\(13\)
default	\(-\frac {1}{4+x}-\frac {{\mathrm e}^{x}}{4+x}\)	\(18\)
risch	\(-\frac {1}{4+x}-\frac {{\mathrm e}^{x}}{4+x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3-x)*exp(x)+1)/(x^2+8*x+16),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {x e^{x}}{x^{2} + 8 \, x + 16} + \frac {3 \, e^{\left (-4\right )} E_{2}\left (-x - 4\right )}{x + 4} - \frac {1}{x + 4} - \int \frac {{\left (x - 4\right )} e^{x}}{x^{3} + 12 \, x^{2} + 48 \, x + 64}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3-x)*exp(x)+1)/(x^2+8*x+16),x, algorithm="maxima")

[Out]

-x*e^x/(x^2 + 8*x + 16) + 3*e^(-4)*exp_integral_e(2, -x - 4)/(x + 4) - 1/(x + 4) - integrate((x - 4)*e^x/(x^3
+ 12*x^2 + 48*x + 64), x)

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mupad [B]  time = 7.04, size = 11, normalized size = 0.85 \begin {gather*} -\frac {{\mathrm {e}}^x+1}{x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x + 3) - 1)/(8*x + x^2 + 16),x)

[Out]

-(exp(x) + 1)/(x + 4)

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sympy [A]  time = 0.11, size = 12, normalized size = 0.92 \begin {gather*} - \frac {e^{x}}{x + 4} - \frac {1}{x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3-x)*exp(x)+1)/(x**2+8*x+16),x)

[Out]

-exp(x)/(x + 4) - 1/(x + 4)

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