3.58.54 \(\int \frac {16+8 x+e^{\frac {x^2}{2}} (4+98 x-4 x^2-x^3)}{9216-768 x-176 x^2+8 x^3+x^4} \, dx\)

Optimal. Leaf size=24 \[ \frac {4+e^{\frac {x^2}{2}}}{(8-x) (12+x)} \]

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Rubi [B]  time = 0.48, antiderivative size = 53, normalized size of antiderivative = 2.21, number of steps used = 9, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6742, 74, 2220, 2204} \begin {gather*} \frac {e^{\frac {x^2}{2}}}{20 (x+12)}+\frac {e^{\frac {x^2}{2}}}{20 (8-x)}+\frac {4}{(x+12) (8-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 + 8*x + E^(x^2/2)*(4 + 98*x - 4*x^2 - x^3))/(9216 - 768*x - 176*x^2 + 8*x^3 + x^4),x]

[Out]

E^(x^2/2)/(20*(8 - x)) + E^(x^2/2)/(20*(12 + x)) + 4/((8 - x)*(12 + x))

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2220

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2)*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[(f*(e + f*x)^(m +
1)*F^(a + b*(c + d*x)^2))/((m + 1)*f^2), x] + (-Dist[(2*b*d^2*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 2)*F^
(a + b*(c + d*x)^2), x], x] + Dist[(2*b*d*(d*e - c*f)*Log[F])/(f^2*(m + 1)), Int[(e + f*x)^(m + 1)*F^(a + b*(c
+ d*x)^2), x], x]) /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && LtQ[m, -1]

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 \left (\frac {8 (2+x)}{(-8+x)^2 (12+x)^2}-\frac {e^{\frac {x^2}{2}} \left (-4-98 x+4 x^2+x^3\right )}{(-8+x)^2 (12+x)^2}\right ) \, dx\\ &=8 \int \frac {2+x}{(-8+x)^2 (12+x)^2} \, dx-\int \frac {e^{\frac {x^2}{2}} \left (-4-98 x+4 x^2+x^3\right )}{(-8+x)^2 (12+x)^2} \, dx\\ &=\frac {4}{(8-x) (12+x)}-\int \left (-\frac {e^{\frac {x^2}{2}}}{20 (-8+x)^2}+\frac {2 e^{\frac {x^2}{2}}}{5 (-8+x)}+\frac {e^{\frac {x^2}{2}}}{20 (12+x)^2}+\frac {3 e^{\frac {x^2}{2}}}{5 (12+x)}\right ) \, dx\\ &=\frac {4}{(8-x) (12+x)}+\frac {1}{20} \int \frac {e^{\frac {x^2}{2}}}{(-8+x)^2} \, dx-\frac {1}{20} \int \frac {e^{\frac {x^2}{2}}}{(12+x)^2} \, dx-\frac {2}{5} \int \frac {e^{\frac {x^2}{2}}}{-8+x} \, dx-\frac {3}{5} \int \frac {e^{\frac {x^2}{2}}}{12+x} \, dx\\ &=\frac {e^{\frac {x^2}{2}}}{20 (8-x)}+\frac {e^{\frac {x^2}{2}}}{20 (12+x)}+\frac {4}{(8-x) (12+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 23, normalized size = 0.96 \begin {gather*} -\frac {4+e^{\frac {x^2}{2}}}{-96+4 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 + 8*x + E^(x^2/2)*(4 + 98*x - 4*x^2 - x^3))/(9216 - 768*x - 176*x^2 + 8*x^3 + x^4),x]

[Out]

-((4 + E^(x^2/2))/(-96 + 4*x + x^2))

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fricas [A]  time = 0.58, size = 20, normalized size = 0.83 \begin {gather*} -\frac {e^{\left (\frac {1}{2} \, x^{2}\right )} + 4}{x^{2} + 4 \, x - 96} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2+98*x+4)*exp(1/2*x^2)+8*x+16)/(x^4+8*x^3-176*x^2-768*x+9216),x, algorithm="fricas")

[Out]

-(e^(1/2*x^2) + 4)/(x^2 + 4*x - 96)

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giac [A]  time = 0.16, size = 20, normalized size = 0.83 \begin {gather*} -\frac {e^{\left (\frac {1}{2} \, x^{2}\right )} + 4}{x^{2} + 4 \, x - 96} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2+98*x+4)*exp(1/2*x^2)+8*x+16)/(x^4+8*x^3-176*x^2-768*x+9216),x, algorithm="giac")

[Out]

-(e^(1/2*x^2) + 4)/(x^2 + 4*x - 96)

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maple [A]  time = 0.15, size = 22, normalized size = 0.92

method	result	size
norman	\(\frac {-4-{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}+4 x -96}\)	\(22\)
risch	\(-\frac {4}{x^{2}+4 x -96}-\frac {{\mathrm e}^{\frac {x^{2}}{2}}}{x^{2}+4 x -96}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^3-4*x^2+98*x+4)*exp(1/2*x^2)+8*x+16)/(x^4+8*x^3-176*x^2-768*x+9216),x,method=_RETURNVERBOSE)

[Out]

(-4-exp(1/2*x^2))/(x^2+4*x-96)

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maxima [B]  time = 0.40, size = 49, normalized size = 2.04 \begin {gather*} -\frac {2 \, {\left (x + 2\right )}}{25 \, {\left (x^{2} + 4 \, x - 96\right )}} + \frac {2 \, {\left (x - 48\right )}}{25 \, {\left (x^{2} + 4 \, x - 96\right )}} - \frac {e^{\left (\frac {1}{2} \, x^{2}\right )}}{x^{2} + 4 \, x - 96} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^3-4*x^2+98*x+4)*exp(1/2*x^2)+8*x+16)/(x^4+8*x^3-176*x^2-768*x+9216),x, algorithm="maxima")

[Out]

-2/25*(x + 2)/(x^2 + 4*x - 96) + 2/25*(x - 48)/(x^2 + 4*x - 96) - e^(1/2*x^2)/(x^2 + 4*x - 96)

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mupad [B]  time = 0.15, size = 20, normalized size = 0.83 \begin {gather*} -\frac {{\mathrm {e}}^{\frac {x^2}{2}}+4}{x^2+4\,x-96} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + exp(x^2/2)*(98*x - 4*x^2 - x^3 + 4) + 16)/(8*x^3 - 176*x^2 - 768*x + x^4 + 9216),x)

[Out]

-(exp(x^2/2) + 4)/(4*x + x^2 - 96)

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sympy [A]  time = 0.13, size = 26, normalized size = 1.08 \begin {gather*} - \frac {e^{\frac {x^{2}}{2}}}{x^{2} + 4 x - 96} - \frac {4}{x^{2} + 4 x - 96} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**3-4*x**2+98*x+4)*exp(1/2*x**2)+8*x+16)/(x**4+8*x**3-176*x**2-768*x+9216),x)

[Out]

-exp(x**2/2)/(x**2 + 4*x - 96) - 4/(x**2 + 4*x - 96)

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