3.102.92 \(\int \frac {24-4 e^{\frac {x^2}{2}} x^3}{3 x^2} \, dx\)

Optimal. Leaf size=25 \[ -\frac {4 \left (6+x+\left (-e+e^{\frac {x^2}{2}}\right ) x\right )}{3 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 0.76, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 14, 2209} \begin {gather*} -\frac {4}{3} e^{\frac {x^2}{2}}-\frac {8}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24 - 4*E^(x^2/2)*x^3)/(3*x^2),x]

[Out]

(-4*E^(x^2/2))/3 - 8/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2209

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {24-4 e^{\frac {x^2}{2}} x^3}{x^2} \, dx\\ &=\frac {1}{3} \int \left (\frac {24}{x^2}-4 e^{\frac {x^2}{2}} x\right ) \, dx\\ &=-\frac {8}{x}-\frac {4}{3} \int e^{\frac {x^2}{2}} x \, dx\\ &=-\frac {4}{3} e^{\frac {x^2}{2}}-\frac {8}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.76 \begin {gather*} -\frac {4}{3} e^{\frac {x^2}{2}}-\frac {8}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 - 4*E^(x^2/2)*x^3)/(3*x^2),x]

[Out]

(-4*E^(x^2/2))/3 - 8/x

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fricas [A]  time = 0.57, size = 15, normalized size = 0.60 \begin {gather*} -\frac {4 \, {\left (x e^{\left (\frac {1}{2} \, x^{2}\right )} + 6\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-4*x^3*exp(1/2*x^2)+24)/x^2,x, algorithm="fricas")

[Out]

-4/3*(x*e^(1/2*x^2) + 6)/x

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giac [A]  time = 0.21, size = 15, normalized size = 0.60 \begin {gather*} -\frac {4 \, {\left (x e^{\left (\frac {1}{2} \, x^{2}\right )} + 6\right )}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-4*x^3*exp(1/2*x^2)+24)/x^2,x, algorithm="giac")

[Out]

-4/3*(x*e^(1/2*x^2) + 6)/x

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maple [A]  time = 0.05, size = 15, normalized size = 0.60

method	result	size
default	\(-\frac {8}{x}-\frac {4 \,{\mathrm e}^{\frac {x^{2}}{2}}}{3}\)	\(15\)
risch	\(-\frac {8}{x}-\frac {4 \,{\mathrm e}^{\frac {x^{2}}{2}}}{3}\)	\(15\)
norman	\(\frac {-8-\frac {4 x \,{\mathrm e}^{\frac {x^{2}}{2}}}{3}}{x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-4*x^3*exp(1/2*x^2)+24)/x^2,x,method=_RETURNVERBOSE)

[Out]

-8/x-4/3*exp(1/2*x^2)

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maxima [A]  time = 0.37, size = 14, normalized size = 0.56 \begin {gather*} -\frac {8}{x} - \frac {4}{3} \, e^{\left (\frac {1}{2} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-4*x^3*exp(1/2*x^2)+24)/x^2,x, algorithm="maxima")

[Out]

-8/x - 4/3*e^(1/2*x^2)

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mupad [B]  time = 6.42, size = 14, normalized size = 0.56 \begin {gather*} -\frac {4\,{\mathrm {e}}^{\frac {x^2}{2}}}{3}-\frac {8}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*x^3*exp(x^2/2))/3 - 8)/x^2,x)

[Out]

- (4*exp(x^2/2))/3 - 8/x

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sympy [A]  time = 0.09, size = 14, normalized size = 0.56 \begin {gather*} - \frac {4 e^{\frac {x^{2}}{2}}}{3} - \frac {8}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-4*x**3*exp(1/2*x**2)+24)/x**2,x)

[Out]

-4*exp(x**2/2)/3 - 8/x

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