3.45.57 \(\int \frac {1}{625} e^{\frac {1}{625} (625-450 x+81 x^2)} (-450+162 x) \, dx\)

Optimal. Leaf size=13 \[ -5+e^{\left (-1+\frac {9 x}{25}\right )^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 2227, 2209} \begin {gather*} e^{\frac {1}{625} (25-9 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((625 - 450*x + 81*x^2)/625)*(-450 + 162*x))/625,x]

[Out]

E^((25 - 9*x)^2/625)

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}{625} \int e^{\frac {1}{625} \left (625-450 x+81 x^2\right )} (-450+162 x) \, dx\\ &=\frac {1}{625} \int e^{\frac {1}{625} (-25+9 x)^2} (-450+162 x) \, dx\\ &=e^{\frac {1}{625} (25-9 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} e^{\frac {1}{625} (25-9 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((625 - 450*x + 81*x^2)/625)*(-450 + 162*x))/625,x]

[Out]

E^((25 - 9*x)^2/625)

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fricas [A]  time = 0.48, size = 11, normalized size = 0.85 \begin {gather*} e^{\left (\frac {81}{625} \, x^{2} - \frac {18}{25} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/625*(162*x-450)*exp(81/625*x^2-18/25*x+1),x, algorithm="fricas")

[Out]

e^(81/625*x^2 - 18/25*x + 1)

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giac [A]  time = 0.13, size = 11, normalized size = 0.85 \begin {gather*} e^{\left (\frac {81}{625} \, x^{2} - \frac {18}{25} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/625*(162*x-450)*exp(81/625*x^2-18/25*x+1),x, algorithm="giac")

[Out]

e^(81/625*x^2 - 18/25*x + 1)

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maple [A]  time = 0.03, size = 11, normalized size = 0.85

method	result	size
risch	\({\mathrm e}^{\frac {\left (9 x -25\right )^{2}}{625}}\)	\(11\)
gosper	\({\mathrm e}^{\frac {81}{625} x^{2}-\frac {18}{25} x +1}\)	\(12\)
default	\({\mathrm e}^{\frac {81}{625} x^{2}-\frac {18}{25} x +1}\)	\(12\)
norman	\({\mathrm e}^{\frac {81}{625} x^{2}-\frac {18}{25} x +1}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/625*(162*x-450)*exp(81/625*x^2-18/25*x+1),x,method=_RETURNVERBOSE)

[Out]

exp(1/625*(9*x-25)^2)

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maxima [A]  time = 0.35, size = 11, normalized size = 0.85 \begin {gather*} e^{\left (\frac {81}{625} \, x^{2} - \frac {18}{25} \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/625*(162*x-450)*exp(81/625*x^2-18/25*x+1),x, algorithm="maxima")

[Out]

e^(81/625*x^2 - 18/25*x + 1)

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mupad [B]  time = 3.21, size = 13, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-\frac {18\,x}{25}}\,\mathrm {e}\,{\mathrm {e}}^{\frac {81\,x^2}{625}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((81*x^2)/625 - (18*x)/25 + 1)*(162*x - 450))/625,x)

[Out]

exp(-(18*x)/25)*exp(1)*exp((81*x^2)/625)

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sympy [A]  time = 0.10, size = 14, normalized size = 1.08 \begin {gather*} e^{\frac {81 x^{2}}{625} - \frac {18 x}{25} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/625*(162*x-450)*exp(81/625*x**2-18/25*x+1),x)

[Out]

exp(81*x**2/625 - 18*x/25 + 1)

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