3.70.27 \(\int \frac {12-12 x^2}{e (16-1160 x+21057 x^2-1160 x^3+16 x^4)} \, dx\)

Optimal. Leaf size=19 \[ \frac {3 x}{e (4-x+4 (-36+x) x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 1680, 1814, 8} \begin {gather*} -\frac {48 x}{e \left (20961-(145-8 x)^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 - 12*x^2)/(E*(16 - 1160*x + 21057*x^2 - 1160*x^3 + 16*x^4)),x]

[Out]

(-48*x)/(E*(20961 - (145 - 8*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} &=\frac {\int \frac {12-12 x^2}{16-1160 x+21057 x^2-1160 x^3+16 x^4} \, dx}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {48 \left (-20961-2320 x-64 x^2\right )}{\left (20961-64 x^2\right )^2} \, dx,x,-\frac {145}{8}+x\right )}{e}\\ &=\frac {48 \operatorname {Subst}\left (\int \frac {-20961-2320 x-64 x^2}{\left (20961-64 x^2\right )^2} \, dx,x,-\frac {145}{8}+x\right )}{e}\\ &=-\frac {48 x}{e \left (20961-(145-8 x)^2\right )}-\frac {8 \operatorname {Subst}\left (\int 0 \, dx,x,-\frac {145}{8}+x\right )}{6987 e}\\ &=-\frac {48 x}{e \left (20961-(145-8 x)^2\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 18, normalized size = 0.95 \begin {gather*} \frac {3 x}{e \left (4-145 x+4 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 - 12*x^2)/(E*(16 - 1160*x + 21057*x^2 - 1160*x^3 + 16*x^4)),x]

[Out]

(3*x)/(E*(4 - 145*x + 4*x^2))

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 17, normalized size = 0.89 \begin {gather*} \frac {3 \, x e^{\left (-1\right )}}{4 \, x^{2} - 145 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+12)/(16*x^4-1160*x^3+21057*x^2-1160*x+16)/exp(1),x, algorithm="fricas")

[Out]

3*x*e^(-1)/(4*x^2 - 145*x + 4)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 16, normalized size = 0.84 \begin {gather*} \frac {3 \, e^{\left (-1\right )}}{4 \, x + \frac {4}{x} - 145} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+12)/(16*x^4-1160*x^3+21057*x^2-1160*x+16)/exp(1),x, algorithm="giac")

[Out]

3*e^(-1)/(4*x + 4/x - 145)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 16, normalized size = 0.84

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{-1} x}{4 \left (x^{2}-\frac {145}{4} x +1\right )}\)	\(16\)
default	\(\frac {3 \,{\mathrm e}^{-1} x}{4 \left (x^{2}-\frac {145}{4} x +1\right )}\)	\(18\)
gosper	\(\frac {3 x \,{\mathrm e}^{-1}}{4 x^{2}-145 x +4}\)	\(20\)
norman	\(\frac {3 x \,{\mathrm e}^{-1}}{4 x^{2}-145 x +4}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*x^2+12)/(16*x^4-1160*x^3+21057*x^2-1160*x+16)/exp(1),x,method=_RETURNVERBOSE)

[Out]

3/4*exp(-1)*x/(x^2-145/4*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 17, normalized size = 0.89 \begin {gather*} \frac {3 \, x e^{\left (-1\right )}}{4 \, x^{2} - 145 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+12)/(16*x^4-1160*x^3+21057*x^2-1160*x+16)/exp(1),x, algorithm="maxima")

[Out]

3*x*e^(-1)/(4*x^2 - 145*x + 4)

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 17, normalized size = 0.89 \begin {gather*} \frac {3\,x\,{\mathrm {e}}^{-1}}{4\,x^2-145\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-1)*(12*x^2 - 12))/(21057*x^2 - 1160*x - 1160*x^3 + 16*x^4 + 16),x)

[Out]

(3*x*exp(-1))/(4*x^2 - 145*x + 4)

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 22, normalized size = 1.16 \begin {gather*} \frac {3 x}{4 e x^{2} - 145 e x + 4 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x**2+12)/(16*x**4-1160*x**3+21057*x**2-1160*x+16)/exp(1),x)

[Out]

3*x/(4*E*x**2 - 145*E*x + 4*E)

________________________________________________________________________________________