3.100.18 \(\int -2 e^{\frac {1}{2} (-237-2 x^2+2 \log (2))} x \, dx\)

Optimal. Leaf size=13 \[ 2 e^{-\frac {237}{2}-x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.38, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 2225, 2209} \begin {gather*} e^{\frac {1}{2} (2 \log (2)-237)-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-2*E^((-237 - 2*x^2 + 2*Log[2])/2)*x,x]

[Out]

E^(-x^2 + (-237 + 2*Log[2])/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 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (2 \int e^{\frac {1}{2} \left (-237-2 x^2+2 \log (2)\right )} x \, dx\right )\\ &=-\left (2 \int e^{-x^2+\frac {1}{2} (-237+2 \log (2))} x \, dx\right )\\ &=e^{-x^2+\frac {1}{2} (-237+2 \log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} 2 e^{-\frac {237}{2}-x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-2*E^((-237 - 2*x^2 + 2*Log[2])/2)*x,x]

[Out]

2*E^(-237/2 - x^2)

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fricas [A]  time = 1.05, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (-x^{2} + \log \relax (2) - \frac {237}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*exp(log(2)-x^2-237/2),x, algorithm="fricas")

[Out]

e^(-x^2 + log(2) - 237/2)

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giac [A]  time = 0.17, size = 10, normalized size = 0.77 \begin {gather*} e^{\left (-x^{2} + \log \relax (2) - \frac {237}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*exp(log(2)-x^2-237/2),x, algorithm="giac")

[Out]

e^(-x^2 + log(2) - 237/2)

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maple [A]  time = 0.02, size = 11, normalized size = 0.85

method	result	size
gosper	\({\mathrm e}^{\ln \relax (2)-x^{2}-\frac {237}{2}}\)	\(11\)
derivativedivides	\({\mathrm e}^{\ln \relax (2)-x^{2}-\frac {237}{2}}\)	\(11\)
default	\({\mathrm e}^{\ln \relax (2)-x^{2}-\frac {237}{2}}\)	\(11\)
norman	\({\mathrm e}^{\ln \relax (2)-x^{2}-\frac {237}{2}}\)	\(11\)
risch	\(2 \,{\mathrm e}^{-\frac {237}{2}-x^{2}}\)	\(11\)
meijerg	\(-{\mathrm e}^{\ln \relax (2)-\frac {237}{2}} \left (1-{\mathrm e}^{-x^{2}}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*x*exp(ln(2)-x^2-237/2),x,method=_RETURNVERBOSE)

[Out]

exp(ln(2)-x^2-237/2)

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maxima [A]  time = 0.34, size = 10, normalized size = 0.77 \begin {gather*} 2 \, e^{\left (-x^{2} - \frac {237}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*exp(log(2)-x^2-237/2),x, algorithm="maxima")

[Out]

2*e^(-x^2 - 237/2)

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mupad [B]  time = 7.02, size = 10, normalized size = 0.77 \begin {gather*} 2\,{\mathrm {e}}^{-\frac {237}{2}}\,{\mathrm {e}}^{-x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*x*exp(log(2) - x^2 - 237/2),x)

[Out]

2*exp(-237/2)*exp(-x^2)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.77 \begin {gather*} 2 e^{- x^{2} - \frac {237}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*x*exp(ln(2)-x**2-237/2),x)

[Out]

2*exp(-x**2 - 237/2)

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