3.14.71 \(\int \frac {1}{125} e^{-x+x^2} (8-10 x+17 x^2-2 x^3) \, dx\)

Optimal. Leaf size=19 \[ \frac {1}{125} e^{-x+x^2} (8-x) x \]

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Rubi [A]  time = 0.27, antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 24, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 6742, 2234, 2204, 2240, 2241} \begin {gather*} \frac {8}{125} e^{x^2-x} x-\frac {1}{125} e^{x^2-x} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-x + x^2)*(8 - 10*x + 17*x^2 - 2*x^3))/125,x]

[Out]

(8*E^(-x + x^2)*x)/125 - (E^(-x + x^2)*x^2)/125

Rule 12

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

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 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2240

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

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
 b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[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} &=\frac {1}{125} \int e^{-x+x^2} \left (8-10 x+17 x^2-2 x^3\right ) \, dx\\ &=\frac {1}{125} \int \left (8 e^{-x+x^2}-10 e^{-x+x^2} x+17 e^{-x+x^2} x^2-2 e^{-x+x^2} x^3\right ) \, dx\\ &=-\left (\frac {2}{125} \int e^{-x+x^2} x^3 \, dx\right )+\frac {8}{125} \int e^{-x+x^2} \, dx-\frac {2}{25} \int e^{-x+x^2} x \, dx+\frac {17}{125} \int e^{-x+x^2} x^2 \, dx\\ &=-\frac {1}{25} e^{-x+x^2}+\frac {17}{250} e^{-x+x^2} x-\frac {1}{125} e^{-x+x^2} x^2-\frac {1}{125} \int e^{-x+x^2} x^2 \, dx+\frac {2}{125} \int e^{-x+x^2} x \, dx-\frac {1}{25} \int e^{-x+x^2} \, dx-\frac {17}{250} \int e^{-x+x^2} \, dx+\frac {17}{250} \int e^{-x+x^2} x \, dx+\frac {8 \int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{125 \sqrt [4]{e}}\\ &=\frac {1}{500} e^{-x+x^2}+\frac {8}{125} e^{-x+x^2} x-\frac {1}{125} e^{-x+x^2} x^2+\frac {4 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-1+2 x)\right )}{125 \sqrt [4]{e}}+\frac {1}{250} \int e^{-x+x^2} \, dx-\frac {1}{250} \int e^{-x+x^2} x \, dx+\frac {1}{125} \int e^{-x+x^2} \, dx+\frac {17}{500} \int e^{-x+x^2} \, dx-\frac {\int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{25 \sqrt [4]{e}}-\frac {17 \int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{250 \sqrt [4]{e}}\\ &=\frac {8}{125} e^{-x+x^2} x-\frac {1}{125} e^{-x+x^2} x^2-\frac {11 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-1+2 x)\right )}{500 \sqrt [4]{e}}-\frac {1}{500} \int e^{-x+x^2} \, dx+\frac {\int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{250 \sqrt [4]{e}}+\frac {\int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{125 \sqrt [4]{e}}+\frac {17 \int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{500 \sqrt [4]{e}}\\ &=\frac {8}{125} e^{-x+x^2} x-\frac {1}{125} e^{-x+x^2} x^2+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} (-1+2 x)\right )}{1000 \sqrt [4]{e}}-\frac {\int e^{\frac {1}{4} (-1+2 x)^2} \, dx}{500 \sqrt [4]{e}}\\ &=\frac {8}{125} e^{-x+x^2} x-\frac {1}{125} e^{-x+x^2} x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 15, normalized size = 0.79 \begin {gather*} -\frac {1}{125} e^{(-1+x) x} (-8+x) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + x^2)*(8 - 10*x + 17*x^2 - 2*x^3))/125,x]

[Out]

-1/125*(E^((-1 + x)*x)*(-8 + x)*x)

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fricas [A]  time = 0.65, size = 17, normalized size = 0.89 \begin {gather*} -\frac {1}{125} \, {\left (x^{2} - 8 \, x\right )} e^{\left (x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-2*x^3+17*x^2-10*x+8)*exp(x^2)/exp(x),x, algorithm="fricas")

[Out]

-1/125*(x^2 - 8*x)*e^(x^2 - x)

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giac [A]  time = 0.42, size = 22, normalized size = 1.16 \begin {gather*} -\frac {1}{500} \, {\left ({\left (2 \, x - 1\right )}^{2} - 28 \, x - 1\right )} e^{\left (x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-2*x^3+17*x^2-10*x+8)*exp(x^2)/exp(x),x, algorithm="giac")

[Out]

-1/500*((2*x - 1)^2 - 28*x - 1)*e^(x^2 - x)

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maple [A]  time = 0.04, size = 13, normalized size = 0.68




method result size



risch \(-\frac {x \left (-8+x \right ) {\mathrm e}^{x \left (x -1\right )}}{125}\) \(13\)
gosper \(-\frac {{\mathrm e}^{x^{2}} x \left (-8+x \right ) {\mathrm e}^{-x}}{125}\) \(15\)
norman \(\left (-\frac {x^{2} {\mathrm e}^{x^{2}}}{125}+\frac {8 \,{\mathrm e}^{x^{2}} x}{125}\right ) {\mathrm e}^{-x}\) \(23\)
default \(\frac {8 x \,{\mathrm e}^{x^{2}-x}}{125}-\frac {x^{2} {\mathrm e}^{x^{2}-x}}{125}\) \(26\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/125*(-2*x^3+17*x^2-10*x+8)*exp(x^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

-1/125*x*(-8+x)*exp(x*(x-1))

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maxima [C]  time = 0.65, size = 250, normalized size = 13.16 \begin {gather*} -\frac {4}{125} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - \frac {1}{2} i\right ) e^{\left (-\frac {1}{4}\right )} + \frac {1}{1000} \, {\left (\frac {12 \, {\left (2 \, x - 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )}{\left (-{\left (2 \, x - 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x - 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x - 1\right )}^{2}}} - 6 \, e^{\left (\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )} + 8 \, \Gamma \left (2, -\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )\right )} e^{\left (-\frac {1}{4}\right )} - \frac {17}{1000} \, {\left (\frac {4 \, {\left (2 \, x - 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )}{\left (-{\left (2 \, x - 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x - 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x - 1\right )}^{2}}} - 4 \, e^{\left (\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} - \frac {1}{50} \, {\left (\frac {\sqrt {\pi } {\left (2 \, x - 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x - 1\right )}^{2}}} + 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x - 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-2*x^3+17*x^2-10*x+8)*exp(x^2)/exp(x),x, algorithm="maxima")

[Out]

-4/125*I*sqrt(pi)*erf(I*x - 1/2*I)*e^(-1/4) + 1/1000*(12*(2*x - 1)^3*gamma(3/2, -1/4*(2*x - 1)^2)/(-(2*x - 1)^
2)^(3/2) - sqrt(pi)*(2*x - 1)*(erf(1/2*sqrt(-(2*x - 1)^2)) - 1)/sqrt(-(2*x - 1)^2) - 6*e^(1/4*(2*x - 1)^2) + 8
*gamma(2, -1/4*(2*x - 1)^2))*e^(-1/4) - 17/1000*(4*(2*x - 1)^3*gamma(3/2, -1/4*(2*x - 1)^2)/(-(2*x - 1)^2)^(3/
2) - sqrt(pi)*(2*x - 1)*(erf(1/2*sqrt(-(2*x - 1)^2)) - 1)/sqrt(-(2*x - 1)^2) - 4*e^(1/4*(2*x - 1)^2))*e^(-1/4)
 - 1/50*(sqrt(pi)*(2*x - 1)*(erf(1/2*sqrt(-(2*x - 1)^2)) - 1)/sqrt(-(2*x - 1)^2) + 2*e^(1/4*(2*x - 1)^2))*e^(-
1/4)

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mupad [B]  time = 0.07, size = 14, normalized size = 0.74 \begin {gather*} -\frac {x\,{\mathrm {e}}^{x^2-x}\,\left (x-8\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*exp(x^2)*(10*x - 17*x^2 + 2*x^3 - 8))/125,x)

[Out]

-(x*exp(x^2 - x)*(x - 8))/125

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sympy [A]  time = 0.18, size = 19, normalized size = 1.00 \begin {gather*} \frac {\left (- x^{2} e^{- x} + 8 x e^{- x}\right ) e^{x^{2}}}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/125*(-2*x**3+17*x**2-10*x+8)*exp(x**2)/exp(x),x)

[Out]

(-x**2*exp(-x) + 8*x*exp(-x))*exp(x**2)/125

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