3.72.58 \(\int (e^{4-x} (-12-3 x)+e^{4-x+x^2} (-6+12 x) \log (\log (4))) \, dx\)

Optimal. Leaf size=22 \[ 3 e^{4-x} \left (5+x+2 e^{x^2} \log (\log (4))\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2176, 2194, 2236} \begin {gather*} 6 e^{x^2-x+4} \log (\log (4))+3 e^{4-x} (x+4)+3 e^{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4 - x)*(-12 - 3*x) + E^(4 - x + x^2)*(-6 + 12*x)*Log[Log[4]],x]

[Out]

3*E^(4 - x) + 3*E^(4 - x)*(4 + x) + 6*E^(4 - x + x^2)*Log[Log[4]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (4)) \int e^{4-x+x^2} (-6+12 x) \, dx+\int e^{4-x} (-12-3 x) \, dx\\ &=3 e^{4-x} (4+x)+6 e^{4-x+x^2} \log (\log (4))-3 \int e^{4-x} \, dx\\ &=3 e^{4-x}+3 e^{4-x} (4+x)+6 e^{4-x+x^2} \log (\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 22, normalized size = 1.00 \begin {gather*} 3 e^{4-x} \left (5+x+2 e^{x^2} \log (\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4 - x)*(-12 - 3*x) + E^(4 - x + x^2)*(-6 + 12*x)*Log[Log[4]],x]

[Out]

3*E^(4 - x)*(5 + x + 2*E^x^2*Log[Log[4]])

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fricas [A]  time = 0.62, size = 28, normalized size = 1.27 \begin {gather*} 3 \, {\left (x + 5\right )} e^{\left (-x + 4\right )} + 6 \, e^{\left (x^{2} - x + 4\right )} \log \left (2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x-6)*exp(-x+4)*exp(x^2)*log(2*log(2))+(-3*x-12)*exp(-x+4),x, algorithm="fricas")

[Out]

3*(x + 5)*e^(-x + 4) + 6*e^(x^2 - x + 4)*log(2*log(2))

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giac [A]  time = 0.17, size = 28, normalized size = 1.27 \begin {gather*} 3 \, {\left (x + 5\right )} e^{\left (-x + 4\right )} + 6 \, e^{\left (x^{2} - x + 4\right )} \log \left (2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x-6)*exp(-x+4)*exp(x^2)*log(2*log(2))+(-3*x-12)*exp(-x+4),x, algorithm="giac")

[Out]

3*(x + 5)*e^(-x + 4) + 6*e^(x^2 - x + 4)*log(2*log(2))

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maple [A]  time = 0.04, size = 39, normalized size = 1.77

method	result	size
default	\(-3 \,{\mathrm e}^{-x +4} \left (-x +4\right )+27 \,{\mathrm e}^{-x +4}+6 \ln \left (2 \ln \relax (2)\right ) {\mathrm e}^{x^{2}-x +4}\)	\(39\)
norman	\(\left (6 \ln \relax (2)+6 \ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{x^{2}} {\mathrm e}^{-x +4}+3 x \,{\mathrm e}^{-x +4}+15 \,{\mathrm e}^{-x +4}\)	\(40\)
risch	\(6 \,{\mathrm e}^{x^{2}-x +4} \ln \relax (2)+6 \,{\mathrm e}^{x^{2}-x +4} \ln \left (\ln \relax (2)\right )+\left (15+3 x \right ) {\mathrm e}^{-x +4}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x-6)*exp(-x+4)*exp(x^2)*ln(2*ln(2))+(-3*x-12)*exp(-x+4),x,method=_RETURNVERBOSE)

[Out]

-3*exp(-x+4)*(-x+4)+27*exp(-x+4)+6*ln(2*ln(2))*exp(x^2-x+4)

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maxima [A]  time = 0.41, size = 38, normalized size = 1.73 \begin {gather*} 3 \, {\left (x e^{4} + e^{4}\right )} e^{\left (-x\right )} + 6 \, e^{\left (x^{2} - x + 4\right )} \log \left (2 \, \log \relax (2)\right ) + 12 \, e^{\left (-x + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x-6)*exp(-x+4)*exp(x^2)*log(2*log(2))+(-3*x-12)*exp(-x+4),x, algorithm="maxima")

[Out]

3*(x*e^4 + e^4)*e^(-x) + 6*e^(x^2 - x + 4)*log(2*log(2)) + 12*e^(-x + 4)

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mupad [B]  time = 0.11, size = 33, normalized size = 1.50 \begin {gather*} 15\,{\mathrm {e}}^{4-x}+\ln \left ({\ln \relax (4)}^6\right )\,{\mathrm {e}}^{x^2-x+4}+3\,x\,{\mathrm {e}}^{4-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(2*log(2))*exp(x^2)*exp(4 - x)*(12*x - 6) - exp(4 - x)*(3*x + 12),x)

[Out]

15*exp(4 - x) + log(log(4)^6)*exp(x^2 - x + 4) + 3*x*exp(4 - x)

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sympy [A]  time = 1.02, size = 31, normalized size = 1.41 \begin {gather*} \left (3 x + 6 e^{x^{2}} \log {\left (\log {\relax (2 )} \right )} + 6 e^{x^{2}} \log {\relax (2 )} + 15\right ) e^{4 - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x-6)*exp(-x+4)*exp(x**2)*ln(2*ln(2))+(-3*x-12)*exp(-x+4),x)

[Out]

(3*x + 6*exp(x**2)*log(log(2)) + 6*exp(x**2)*log(2) + 15)*exp(4 - x)

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