3.70.32 \(\int 250 e^{-1+\frac {25 x^2}{e}} x \, dx\)

Optimal. Leaf size=12 \[ 5 e^{\frac {25 x^2}{e}} \]

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Rubi [A]  time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 2209} \begin {gather*} 5 e^{\frac {25 x^2}{e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[250*E^(-1 + (25*x^2)/E)*x,x]

[Out]

5*E^((25*x^2)/E)

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=250 \int e^{-1+\frac {25 x^2}{e}} x \, dx\\ &=5 e^{\frac {25 x^2}{e}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 1.00 \begin {gather*} 5 e^{\frac {25 x^2}{e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[250*E^(-1 + (25*x^2)/E)*x,x]

[Out]

5*E^((25*x^2)/E)

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fricas [A]  time = 0.66, size = 18, normalized size = 1.50 \begin {gather*} 5 \, e^{\left ({\left (25 \, x^{2} - e\right )} e^{\left (-1\right )} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(250*x*exp(25*x^2/exp(1))/exp(1),x, algorithm="fricas")

[Out]

5*e^((25*x^2 - e)*e^(-1) + 1)

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giac [A]  time = 0.14, size = 10, normalized size = 0.83 \begin {gather*} 5 \, e^{\left (25 \, x^{2} e^{\left (-1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(250*x*exp(25*x^2/exp(1))/exp(1),x, algorithm="giac")

[Out]

5*e^(25*x^2*e^(-1))

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maple [A]  time = 0.03, size = 11, normalized size = 0.92

method	result	size
risch	\(5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(11\)
gosper	\(5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(13\)
derivativedivides	\(5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(13\)
default	\(5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(13\)
norman	\(5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(13\)
meijerg	\(-5+5 \,{\mathrm e}^{25 x^{2} {\mathrm e}^{-1}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(250*x*exp(25*x^2/exp(1))/exp(1),x,method=_RETURNVERBOSE)

[Out]

5*exp(25*x^2*exp(-1))

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maxima [A]  time = 0.38, size = 10, normalized size = 0.83 \begin {gather*} 5 \, e^{\left (25 \, x^{2} e^{\left (-1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(250*x*exp(25*x^2/exp(1))/exp(1),x, algorithm="maxima")

[Out]

5*e^(25*x^2*e^(-1))

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mupad [B]  time = 0.03, size = 10, normalized size = 0.83 \begin {gather*} 5\,{\mathrm {e}}^{25\,x^2\,{\mathrm {e}}^{-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(250*x*exp(25*x^2*exp(-1))*exp(-1),x)

[Out]

5*exp(25*x^2*exp(-1))

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sympy [A]  time = 0.09, size = 10, normalized size = 0.83 \begin {gather*} 5 e^{\frac {25 x^{2}}{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(250*x*exp(25*x**2/exp(1))/exp(1),x)

[Out]

5*exp(25*x**2*exp(-1))

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