3.72.82 \(\int e^{x^2} (8 x-2 x \log (3)) \, dx\)

Optimal. Leaf size=19 \[ 2+\left (e^{x^2}-\log (2)\right ) (4-\log (3)) \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.63, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6, 12, 2209} \begin {gather*} e^{x^2} (4-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x^2*(8*x - 2*x*Log[3]),x]

[Out]

E^x^2*(4 - Log[3])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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} &=\int e^{x^2} x (8-2 \log (3)) \, dx\\ &=(8-2 \log (3)) \int e^{x^2} x \, dx\\ &=e^{x^2} (4-\log (3))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.58 \begin {gather*} -e^{x^2} (-4+\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*(8*x - 2*x*Log[3]),x]

[Out]

-(E^x^2*(-4 + Log[3]))

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fricas [A]  time = 0.68, size = 10, normalized size = 0.53 \begin {gather*} -{\left (\log \relax (3) - 4\right )} e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(3)+8*x)*exp(x^2),x, algorithm="fricas")

[Out]

-(log(3) - 4)*e^(x^2)

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giac [A]  time = 0.21, size = 10, normalized size = 0.53 \begin {gather*} -{\left (\log \relax (3) - 4\right )} e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(3)+8*x)*exp(x^2),x, algorithm="giac")

[Out]

-(log(3) - 4)*e^(x^2)

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maple [A]  time = 0.02, size = 11, normalized size = 0.58

method	result	size
gosper	\(-\left (-4+\ln \relax (3)\right ) {\mathrm e}^{x^{2}}\)	\(11\)
norman	\(\left (-\ln \relax (3)+4\right ) {\mathrm e}^{x^{2}}\)	\(12\)
default	\(4 \,{\mathrm e}^{x^{2}}-\ln \relax (3) {\mathrm e}^{x^{2}}\)	\(16\)
risch	\(4 \,{\mathrm e}^{x^{2}}-\ln \relax (3) {\mathrm e}^{x^{2}}\)	\(16\)
meijerg	\(-\frac {\left (-2 \ln \relax (3)+8\right ) \left (1-{\mathrm e}^{x^{2}}\right )}{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(3)+8*x)*exp(x^2),x,method=_RETURNVERBOSE)

[Out]

-(-4+ln(3))*exp(x^2)

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maxima [A]  time = 0.35, size = 15, normalized size = 0.79 \begin {gather*} -e^{\left (x^{2}\right )} \log \relax (3) + 4 \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(3)+8*x)*exp(x^2),x, algorithm="maxima")

[Out]

-e^(x^2)*log(3) + 4*e^(x^2)

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mupad [B]  time = 0.05, size = 10, normalized size = 0.53 \begin {gather*} -{\mathrm {e}}^{x^2}\,\left (\ln \relax (3)-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*(8*x - 2*x*log(3)),x)

[Out]

-exp(x^2)*(log(3) - 4)

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sympy [A]  time = 0.10, size = 8, normalized size = 0.42 \begin {gather*} \left (4 - \log {\relax (3 )}\right ) e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(3)+8*x)*exp(x**2),x)

[Out]

(4 - log(3))*exp(x**2)

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