3.95.69 \(\int \frac {1}{3} (-6 e^{8-2 x}+6 e^{4-x}+(-2 e^{8-2 x}+2 e^{4-x}) \log (\frac {4}{e^4})) \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{3} \left (-1+e^{4-x}\right )^2 \left (3+\log \left (\frac {4}{e^4}\right )\right ) \]

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Rubi [B]  time = 0.03, antiderivative size = 51, normalized size of antiderivative = 2.22, number of steps used = 7, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 2194} \begin {gather*} e^{8-2 x}-2 e^{4-x}-\frac {1}{3} e^{8-2 x} (4-\log (4))+\frac {2}{3} e^{4-x} (4-\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*E^(8 - 2*x) + 6*E^(4 - x) + (-2*E^(8 - 2*x) + 2*E^(4 - x))*Log[4/E^4])/3,x]

[Out]

E^(8 - 2*x) - 2*E^(4 - x) - (E^(8 - 2*x)*(4 - Log[4]))/3 + (2*E^(4 - x)*(4 - Log[4]))/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \left (-6 e^{8-2 x}+6 e^{4-x}+\left (-2 e^{8-2 x}+2 e^{4-x}\right ) \log \left (\frac {4}{e^4}\right )\right ) \, dx\\ &=-\left (2 \int e^{8-2 x} \, dx\right )+2 \int e^{4-x} \, dx+\frac {1}{3} (-4+\log (4)) \int \left (-2 e^{8-2 x}+2 e^{4-x}\right ) \, dx\\ &=e^{8-2 x}-2 e^{4-x}+\frac {1}{3} (2 (4-\log (4))) \int e^{8-2 x} \, dx-\frac {1}{3} (2 (4-\log (4))) \int e^{4-x} \, dx\\ &=e^{8-2 x}-2 e^{4-x}-\frac {1}{3} e^{8-2 x} (4-\log (4))+\frac {2}{3} e^{4-x} (4-\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 1.26 \begin {gather*} \frac {2}{3} \left (\frac {1}{2} e^{8-2 x}-e^{4-x}\right ) (-1+\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*E^(8 - 2*x) + 6*E^(4 - x) + (-2*E^(8 - 2*x) + 2*E^(4 - x))*Log[4/E^4])/3,x]

[Out]

(2*(E^(8 - 2*x)/2 - E^(4 - x))*(-1 + Log[4]))/3

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fricas [A]  time = 0.51, size = 29, normalized size = 1.26 \begin {gather*} -\frac {2}{3} \, {\left (2 \, \log \relax (2) - 1\right )} e^{\left (-x + 4\right )} + \frac {1}{3} \, {\left (2 \, \log \relax (2) - 1\right )} e^{\left (-2 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(-x+4)^2+2*exp(-x+4))*log(4/exp(4))-2*exp(-x+4)^2+2*exp(-x+4),x, algorithm="fricas")

[Out]

-2/3*(2*log(2) - 1)*e^(-x + 4) + 1/3*(2*log(2) - 1)*e^(-2*x + 8)

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giac [B]  time = 0.20, size = 39, normalized size = 1.70 \begin {gather*} -\frac {1}{3} \, {\left (2 \, e^{\left (-x + 4\right )} - e^{\left (-2 \, x + 8\right )}\right )} \log \left (4 \, e^{\left (-4\right )}\right ) - 2 \, e^{\left (-x + 4\right )} + e^{\left (-2 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(-x+4)^2+2*exp(-x+4))*log(4/exp(4))-2*exp(-x+4)^2+2*exp(-x+4),x, algorithm="giac")

[Out]

-1/3*(2*e^(-x + 4) - e^(-2*x + 8))*log(4*e^(-4)) - 2*e^(-x + 4) + e^(-2*x + 8)

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maple [A]  time = 0.06, size = 30, normalized size = 1.30

method	result	size
norman	\(\left (-\frac {4 \ln \relax (2)}{3}+\frac {2}{3}\right ) {\mathrm e}^{-x +4}+\left (\frac {2 \ln \relax (2)}{3}-\frac {1}{3}\right ) {\mathrm e}^{-2 x +8}\)	\(30\)
derivativedivides	\(-\frac {\left (2 \ln \left (4 \,{\mathrm e}^{-4}\right )+6\right ) \left (-\frac {{\mathrm e}^{-2 x +8}}{2}+{\mathrm e}^{-x +4}\right )}{3}\)	\(31\)
risch	\(\frac {2 \,{\mathrm e}^{-2 x +8} \ln \relax (2)}{3}-\frac {{\mathrm e}^{-2 x +8}}{3}-\frac {4 \,{\mathrm e}^{-x +4} \ln \relax (2)}{3}+\frac {2 \,{\mathrm e}^{-x +4}}{3}\)	\(38\)
default	\({\mathrm e}^{-2 x +8}+\frac {\ln \left (4 \,{\mathrm e}^{-4}\right ) {\mathrm e}^{-2 x +8}}{3}-\frac {2 \ln \left (4 \,{\mathrm e}^{-4}\right ) {\mathrm e}^{-x +4}}{3}-2 \,{\mathrm e}^{-x +4}\)	\(50\)
meijerg	\(-\frac {{\mathrm e}^{-2 x +2 x \,{\mathrm e}^{8}} \left (\ln \left (4 \,{\mathrm e}^{-4}\right )+3\right ) \left (1-{\mathrm e}^{-2 x \,{\mathrm e}^{8}}\right )}{3}+\left (\frac {2 \ln \left (4 \,{\mathrm e}^{-4}\right ) {\mathrm e}^{4}}{3}+2 \,{\mathrm e}^{4}\right ) \left (1-{\mathrm e}^{-x}\right )\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-2*exp(-x+4)^2+2*exp(-x+4))*ln(4/exp(4))-2*exp(-x+4)^2+2*exp(-x+4),x,method=_RETURNVERBOSE)

[Out]

(-4/3*ln(2)+2/3)*exp(-x+4)+(2/3*ln(2)-1/3)*exp(-x+4)^2

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maxima [B]  time = 0.36, size = 39, normalized size = 1.70 \begin {gather*} -\frac {1}{3} \, {\left (2 \, e^{\left (-x + 4\right )} - e^{\left (-2 \, x + 8\right )}\right )} \log \left (4 \, e^{\left (-4\right )}\right ) - 2 \, e^{\left (-x + 4\right )} + e^{\left (-2 \, x + 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(-x+4)^2+2*exp(-x+4))*log(4/exp(4))-2*exp(-x+4)^2+2*exp(-x+4),x, algorithm="maxima")

[Out]

-1/3*(2*e^(-x + 4) - e^(-2*x + 8))*log(4*e^(-4)) - 2*e^(-x + 4) + e^(-2*x + 8)

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mupad [B]  time = 5.41, size = 28, normalized size = 1.22 \begin {gather*} {\mathrm {e}}^{8-2\,x}\,\left (\frac {\ln \relax (4)}{3}-\frac {1}{3}\right )-{\mathrm {e}}^{4-x}\,\left (\frac {\ln \left (16\right )}{3}-\frac {2}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(4 - x) - 2*exp(8 - 2*x) + (log(4*exp(-4))*(2*exp(4 - x) - 2*exp(8 - 2*x)))/3,x)

[Out]

exp(8 - 2*x)*(log(4)/3 - 1/3) - exp(4 - x)*(log(16)/3 - 2/3)

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sympy [A]  time = 0.13, size = 27, normalized size = 1.17 \begin {gather*} \frac {\left (6 - 12 \log {\relax (2 )}\right ) e^{4 - x}}{9} + \frac {\left (-3 + 6 \log {\relax (2 )}\right ) e^{8 - 2 x}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-2*exp(-x+4)**2+2*exp(-x+4))*ln(4/exp(4))-2*exp(-x+4)**2+2*exp(-x+4),x)

[Out]

(6 - 12*log(2))*exp(4 - x)/9 + (-3 + 6*log(2))*exp(8 - 2*x)/9

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