3.50.31 \(\int \frac {1}{5} (-10 x+e^{1-2 x+x^2} (-8+8 x)) \, dx\)

Optimal. Leaf size=27 \[ \frac {4}{5} \left (5+e^{\frac {\left (x-x^2\right )^2}{x^2}}\right )-x^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2227, 2209} \begin {gather*} \frac {4}{5} e^{(x-1)^2}-x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x + E^(1 - 2*x + x^2)*(-8 + 8*x))/5,x]

[Out]

(4*E^(-1 + x)^2)/5 - x^2

Rule 12

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

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 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \left (-10 x+e^{1-2 x+x^2} (-8+8 x)\right ) \, dx\\ &=-x^2+\frac {1}{5} \int e^{1-2 x+x^2} (-8+8 x) \, dx\\ &=-x^2+\frac {1}{5} \int e^{(-1+x)^2} (-8+8 x) \, dx\\ &=\frac {4}{5} e^{(-1+x)^2}-x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.63 \begin {gather*} \frac {4}{5} e^{(-1+x)^2}-x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x + E^(1 - 2*x + x^2)*(-8 + 8*x))/5,x]

[Out]

(4*E^(-1 + x)^2)/5 - x^2

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fricas [A]  time = 0.89, size = 17, normalized size = 0.63 \begin {gather*} -x^{2} + \frac {4}{5} \, e^{\left (x^{2} - 2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(8*x-8)*exp(x^2-2*x+1)-2*x,x, algorithm="fricas")

[Out]

-x^2 + 4/5*e^(x^2 - 2*x + 1)

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giac [A]  time = 0.12, size = 17, normalized size = 0.63 \begin {gather*} -x^{2} + \frac {4}{5} \, e^{\left (x^{2} - 2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(8*x-8)*exp(x^2-2*x+1)-2*x,x, algorithm="giac")

[Out]

-x^2 + 4/5*e^(x^2 - 2*x + 1)

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maple [A]  time = 0.03, size = 15, normalized size = 0.56

method	result	size
risch	\(-x^{2}+\frac {4 \,{\mathrm e}^{\left (x -1\right )^{2}}}{5}\)	\(15\)
default	\(-x^{2}+\frac {4 \,{\mathrm e}^{x^{2}-2 x +1}}{5}\)	\(18\)
norman	\(-x^{2}+\frac {4 \,{\mathrm e}^{x^{2}-2 x +1}}{5}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(8*x-8)*exp(x^2-2*x+1)-2*x,x,method=_RETURNVERBOSE)

[Out]

-x^2+4/5*exp((x-1)^2)

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maxima [A]  time = 0.36, size = 17, normalized size = 0.63 \begin {gather*} -x^{2} + \frac {4}{5} \, e^{\left (x^{2} - 2 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(8*x-8)*exp(x^2-2*x+1)-2*x,x, algorithm="maxima")

[Out]

-x^2 + 4/5*e^(x^2 - 2*x + 1)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.67 \begin {gather*} \frac {4\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{x^2}\,\mathrm {e}}{5}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2 - 2*x + 1)*(8*x - 8))/5 - 2*x,x)

[Out]

(4*exp(-2*x)*exp(x^2)*exp(1))/5 - x^2

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sympy [A]  time = 0.11, size = 15, normalized size = 0.56 \begin {gather*} - x^{2} + \frac {4 e^{x^{2} - 2 x + 1}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(8*x-8)*exp(x**2-2*x+1)-2*x,x)

[Out]

-x**2 + 4*exp(x**2 - 2*x + 1)/5

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