3.41.60 \(\int \frac {e^3 (4+x^2)}{16+16 x-4 x^2-4 x^3+x^4} \, dx\)

Optimal. Leaf size=21 \[ 5+e-\frac {e^3}{-1+x-\frac {4+x}{x}} \]

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {12, 1680, 1814, 8} \begin {gather*} \frac {e^3 x}{5-(1-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(4 + x^2))/(16 + 16*x - 4*x^2 - 4*x^3 + x^4),x]

[Out]

(E^3*x)/(5 - (1 - x)^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^3 \int \frac {4+x^2}{16+16 x-4 x^2-4 x^3+x^4} \, dx\\ &=e^3 \operatorname {Subst}\left (\int \frac {5+2 x+x^2}{\left (5-x^2\right )^2} \, dx,x,-1+x\right )\\ &=\frac {e^3 x}{5-(1-x)^2}-\frac {1}{10} e^3 \operatorname {Subst}(\int 0 \, dx,x,-1+x)\\ &=\frac {e^3 x}{5-(1-x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} -\frac {e^3 x}{-4-2 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(4 + x^2))/(16 + 16*x - 4*x^2 - 4*x^3 + x^4),x]

[Out]

-((E^3*x)/(-4 - 2*x + x^2))

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fricas [A]  time = 1.00, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x e^{3}}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)*exp(3)/(x^4-4*x^3-4*x^2+16*x+16),x, algorithm="fricas")

[Out]

-x*e^3/(x^2 - 2*x - 4)

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giac [A]  time = 0.64, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x e^{3}}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)*exp(3)/(x^4-4*x^3-4*x^2+16*x+16),x, algorithm="giac")

[Out]

-x*e^3/(x^2 - 2*x - 4)

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maple [A]  time = 0.02, size = 16, normalized size = 0.76

method	result	size
gosper	\(-\frac {x \,{\mathrm e}^{3}}{x^{2}-2 x -4}\)	\(16\)
default	\(-\frac {x \,{\mathrm e}^{3}}{x^{2}-2 x -4}\)	\(16\)
norman	\(-\frac {x \,{\mathrm e}^{3}}{x^{2}-2 x -4}\)	\(16\)
risch	\(-\frac {x \,{\mathrm e}^{3}}{x^{2}-2 x -4}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+4)*exp(3)/(x^4-4*x^3-4*x^2+16*x+16),x,method=_RETURNVERBOSE)

[Out]

-x*exp(3)/(x^2-2*x-4)

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maxima [A]  time = 0.36, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x e^{3}}{x^{2} - 2 \, x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)*exp(3)/(x^4-4*x^3-4*x^2+16*x+16),x, algorithm="maxima")

[Out]

-x*e^3/(x^2 - 2*x - 4)

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mupad [B]  time = 2.98, size = 16, normalized size = 0.76 \begin {gather*} \frac {x\,{\mathrm {e}}^3}{-x^2+2\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3)*(x^2 + 4))/(16*x - 4*x^2 - 4*x^3 + x^4 + 16),x)

[Out]

(x*exp(3))/(2*x - x^2 + 4)

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sympy [A]  time = 0.13, size = 14, normalized size = 0.67 \begin {gather*} - \frac {x e^{3}}{x^{2} - 2 x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+4)*exp(3)/(x**4-4*x**3-4*x**2+16*x+16),x)

[Out]

-x*exp(3)/(x**2 - 2*x - 4)

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