3.25.58 \(\int \frac {e^{-\frac {4 x^2}{25}} (-25 x^2-32 x^3+8 x^4+e^{\frac {4 x^2}{25}} (200+450 x^2-675 x^4))}{225 x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {(-4+x) \left (2-e^{-\frac {4 x^2}{25}} x\right )}{9 x}-x \left (-2+x^2\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 43, normalized size of antiderivative = 1.30, number of steps used = 10, number of rules used = 6, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 6688, 2226, 2205, 2209, 2212} \begin {gather*} -x^3-\frac {1}{9} e^{-\frac {4 x^2}{25}} x+\frac {4}{9} e^{-\frac {4 x^2}{25}}+2 x-\frac {8}{9 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25*x^2 - 32*x^3 + 8*x^4 + E^((4*x^2)/25)*(200 + 450*x^2 - 675*x^4))/(225*E^((4*x^2)/25)*x^2),x]

[Out]

4/(9*E^((4*x^2)/25)) - 8/(9*x) + 2*x - x/(9*E^((4*x^2)/25)) - x^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2205

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

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]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{225} \int \frac {e^{-\frac {4 x^2}{25}} \left (-25 x^2-32 x^3+8 x^4+e^{\frac {4 x^2}{25}} \left (200+450 x^2-675 x^4\right )\right )}{x^2} \, dx\\ &=\frac {1}{225} \int \left (25 \left (18+\frac {8}{x^2}-27 x^2\right )+e^{-\frac {4 x^2}{25}} \left (-25-32 x+8 x^2\right )\right ) \, dx\\ &=\frac {1}{225} \int e^{-\frac {4 x^2}{25}} \left (-25-32 x+8 x^2\right ) \, dx+\frac {1}{9} \int \left (18+\frac {8}{x^2}-27 x^2\right ) \, dx\\ &=-\frac {8}{9 x}+2 x-x^3+\frac {1}{225} \int \left (-25 e^{-\frac {4 x^2}{25}}-32 e^{-\frac {4 x^2}{25}} x+8 e^{-\frac {4 x^2}{25}} x^2\right ) \, dx\\ &=-\frac {8}{9 x}+2 x-x^3+\frac {8}{225} \int e^{-\frac {4 x^2}{25}} x^2 \, dx-\frac {1}{9} \int e^{-\frac {4 x^2}{25}} \, dx-\frac {32}{225} \int e^{-\frac {4 x^2}{25}} x \, dx\\ &=\frac {4}{9} e^{-\frac {4 x^2}{25}}-\frac {8}{9 x}+2 x-\frac {1}{9} e^{-\frac {4 x^2}{25}} x-x^3-\frac {5}{36} \sqrt {\pi } \text {erf}\left (\frac {2 x}{5}\right )+\frac {1}{9} \int e^{-\frac {4 x^2}{25}} \, dx\\ &=\frac {4}{9} e^{-\frac {4 x^2}{25}}-\frac {8}{9 x}+2 x-\frac {1}{9} e^{-\frac {4 x^2}{25}} x-x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 43, normalized size = 1.30 \begin {gather*} \frac {4}{9} e^{-\frac {4 x^2}{25}}-\frac {8}{9 x}+2 x-\frac {1}{9} e^{-\frac {4 x^2}{25}} x-x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x^2 - 32*x^3 + 8*x^4 + E^((4*x^2)/25)*(200 + 450*x^2 - 675*x^4))/(225*E^((4*x^2)/25)*x^2),x]

[Out]

4/(9*E^((4*x^2)/25)) - 8/(9*x) + 2*x - x/(9*E^((4*x^2)/25)) - x^3

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fricas [A]  time = 0.55, size = 37, normalized size = 1.12 \begin {gather*} -\frac {{\left (x^{2} + {\left (9 \, x^{4} - 18 \, x^{2} + 8\right )} e^{\left (\frac {4}{25} \, x^{2}\right )} - 4 \, x\right )} e^{\left (-\frac {4}{25} \, x^{2}\right )}}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/225*((-675*x^4+450*x^2+200)*exp(4/25*x^2)+8*x^4-32*x^3-25*x^2)/x^2/exp(4/25*x^2),x, algorithm="fri
cas")

[Out]

-1/9*(x^2 + (9*x^4 - 18*x^2 + 8)*e^(4/25*x^2) - 4*x)*e^(-4/25*x^2)/x

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giac [A]  time = 0.42, size = 36, normalized size = 1.09 \begin {gather*} -\frac {9 \, x^{4} + x^{2} e^{\left (-\frac {4}{25} \, x^{2}\right )} - 18 \, x^{2} - 4 \, x e^{\left (-\frac {4}{25} \, x^{2}\right )} + 8}{9 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/225*((-675*x^4+450*x^2+200)*exp(4/25*x^2)+8*x^4-32*x^3-25*x^2)/x^2/exp(4/25*x^2),x, algorithm="gia
c")

[Out]

-1/9*(9*x^4 + x^2*e^(-4/25*x^2) - 18*x^2 - 4*x*e^(-4/25*x^2) + 8)/x

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maple [A]  time = 0.03, size = 28, normalized size = 0.85

method	result	size
risch	\(2 x -\frac {8}{9 x}-x^{3}+\frac {\left (-25 x +100\right ) {\mathrm e}^{-\frac {4 x^{2}}{25}}}{225}\)	\(28\)
default	\(2 x -\frac {8}{9 x}-x^{3}+\frac {4 \,{\mathrm e}^{-\frac {4 x^{2}}{25}}}{9}-\frac {x \,{\mathrm e}^{-\frac {4 x^{2}}{25}}}{9}\)	\(34\)
norman	\(\frac {\left (\frac {4 x}{9}-\frac {x^{2}}{9}+2 x^{2} {\mathrm e}^{\frac {4 x^{2}}{25}}-{\mathrm e}^{\frac {4 x^{2}}{25}} x^{4}-\frac {8 \,{\mathrm e}^{\frac {4 x^{2}}{25}}}{9}\right ) {\mathrm e}^{-\frac {4 x^{2}}{25}}}{x}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/225*((-675*x^4+450*x^2+200)*exp(4/25*x^2)+8*x^4-32*x^3-25*x^2)/x^2/exp(4/25*x^2),x,method=_RETURNVERBOSE
)

[Out]

2*x-8/9/x-x^3+1/225*(-25*x+100)*exp(-4/25*x^2)

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maxima [A]  time = 0.48, size = 31, normalized size = 0.94 \begin {gather*} -x^{3} - \frac {1}{9} \, x e^{\left (-\frac {4}{25} \, x^{2}\right )} + 2 \, x - \frac {8}{9 \, x} + \frac {4}{9} \, e^{\left (-\frac {4}{25} \, x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/225*((-675*x^4+450*x^2+200)*exp(4/25*x^2)+8*x^4-32*x^3-25*x^2)/x^2/exp(4/25*x^2),x, algorithm="max
ima")

[Out]

-x^3 - 1/9*x*e^(-4/25*x^2) + 2*x - 8/9/x + 4/9*e^(-4/25*x^2)

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mupad [B]  time = 0.11, size = 32, normalized size = 0.97 \begin {gather*} \frac {4\,{\mathrm {e}}^{-\frac {4\,x^2}{25}}}{9}-x\,\left (\frac {{\mathrm {e}}^{-\frac {4\,x^2}{25}}}{9}-2\right )-\frac {8}{9\,x}-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(4*x^2)/25)*((exp((4*x^2)/25)*(450*x^2 - 675*x^4 + 200))/225 - x^2/9 - (32*x^3)/225 + (8*x^4)/225))/
x^2,x)

[Out]

(4*exp(-(4*x^2)/25))/9 - x*(exp(-(4*x^2)/25)/9 - 2) - 8/(9*x) - x^3

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sympy [A]  time = 0.15, size = 24, normalized size = 0.73 \begin {gather*} - x^{3} + 2 x + \frac {\left (4 - x\right ) e^{- \frac {4 x^{2}}{25}}}{9} - \frac {8}{9 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/225*((-675*x**4+450*x**2+200)*exp(4/25*x**2)+8*x**4-32*x**3-25*x**2)/x**2/exp(4/25*x**2),x)

[Out]

-x**3 + 2*x + (4 - x)*exp(-4*x**2/25)/9 - 8/(9*x)

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