3.97.52 \(\int \frac {1}{3} e^{-x^2} (e^{x^2} (9+9 x^2)+8 x \log (4)) \, dx\)

Optimal. Leaf size=23 \[ x \left (3+x^2-\frac {4 e^{-x^2} \log (4)}{3 x}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 6688, 2209} \begin {gather*} x^3-\frac {4}{3} e^{-x^2} \log (4)+3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(9 + 9*x^2) + 8*x*Log[4])/(3*E^x^2),x]

[Out]

3*x + x^3 - (4*Log[4])/(3*E^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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{-x^2} \left (e^{x^2} \left (9+9 x^2\right )+8 x \log (4)\right ) \, dx\\ &=\frac {1}{3} \int \left (9+9 x^2+8 e^{-x^2} x \log (4)\right ) \, dx\\ &=3 x+x^3+\frac {1}{3} (8 \log (4)) \int e^{-x^2} x \, dx\\ &=3 x+x^3-\frac {4}{3} e^{-x^2} \log (4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.87 \begin {gather*} 3 x+x^3-\frac {4}{3} e^{-x^2} \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(9 + 9*x^2) + 8*x*Log[4])/(3*E^x^2),x]

[Out]

3*x + x^3 - (4*Log[4])/(3*E^x^2)

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fricas [A]  time = 0.66, size = 26, normalized size = 1.13 \begin {gather*} \frac {1}{3} \, {\left (3 \, {\left (x^{3} + 3 \, x\right )} e^{\left (x^{2}\right )} - 8 \, \log \relax (2)\right )} e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^2+9)*exp(x^2)+16*x*log(2))/exp(x^2),x, algorithm="fricas")

[Out]

1/3*(3*(x^3 + 3*x)*e^(x^2) - 8*log(2))*e^(-x^2)

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giac [A]  time = 0.12, size = 17, normalized size = 0.74 \begin {gather*} x^{3} - \frac {8}{3} \, e^{\left (-x^{2}\right )} \log \relax (2) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^2+9)*exp(x^2)+16*x*log(2))/exp(x^2),x, algorithm="giac")

[Out]

x^3 - 8/3*e^(-x^2)*log(2) + 3*x

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maple [A]  time = 0.03, size = 18, normalized size = 0.78

method	result	size
default	\(x^{3}+3 x -\frac {8 \ln \relax (2) {\mathrm e}^{-x^{2}}}{3}\)	\(18\)
risch	\(x^{3}+3 x -\frac {8 \ln \relax (2) {\mathrm e}^{-x^{2}}}{3}\)	\(18\)
norman	\(\left (x^{3} {\mathrm e}^{x^{2}}+3 \,{\mathrm e}^{x^{2}} x -\frac {8 \ln \relax (2)}{3}\right ) {\mathrm e}^{-x^{2}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((9*x^2+9)*exp(x^2)+16*x*ln(2))/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

x^3+3*x-8/3*ln(2)/exp(x^2)

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maxima [A]  time = 0.35, size = 17, normalized size = 0.74 \begin {gather*} x^{3} - \frac {8}{3} \, e^{\left (-x^{2}\right )} \log \relax (2) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x^2+9)*exp(x^2)+16*x*log(2))/exp(x^2),x, algorithm="maxima")

[Out]

x^3 - 8/3*e^(-x^2)*log(2) + 3*x

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mupad [B]  time = 0.08, size = 17, normalized size = 0.74 \begin {gather*} 3\,x-\frac {8\,{\mathrm {e}}^{-x^2}\,\ln \relax (2)}{3}+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x^2)*((exp(x^2)*(9*x^2 + 9))/3 + (16*x*log(2))/3),x)

[Out]

3*x - (8*exp(-x^2)*log(2))/3 + x^3

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sympy [A]  time = 0.10, size = 17, normalized size = 0.74 \begin {gather*} x^{3} + 3 x - \frac {8 e^{- x^{2}} \log {\relax (2 )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((9*x**2+9)*exp(x**2)+16*x*ln(2))/exp(x**2),x)

[Out]

x**3 + 3*x - 8*exp(-x**2)*log(2)/3

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