3.46.82 \(\int e^{-24+2 x} (8+8 e^{12} \log (4)+2 e^{24} \log ^2(4)) \, dx\)

Optimal. Leaf size=18 \[ e^{2 x} \left (-\frac {2}{e^{12}}-\log (4)\right )^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2194} \begin {gather*} e^{2 x-24} \left (2+e^{12} \log (4)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-24 + 2*x)*(8 + 8*E^12*Log[4] + 2*E^24*Log[4]^2),x]

[Out]

E^(-24 + 2*x)*(2 + E^12*Log[4])^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 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} &=\left (2 \left (2+e^{12} \log (4)\right )^2\right ) \int e^{-24+2 x} \, dx\\ &=e^{-24+2 x} \left (2+e^{12} \log (4)\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} e^{-24+2 x} \left (2+e^{12} \log (4)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-24 + 2*x)*(8 + 8*E^12*Log[4] + 2*E^24*Log[4]^2),x]

[Out]

E^(-24 + 2*x)*(2 + E^12*Log[4])^2

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fricas [A]  time = 0.58, size = 23, normalized size = 1.28 \begin {gather*} 4 \, {\left (e^{24} \log \relax (2)^{2} + 2 \, e^{12} \log \relax (2) + 1\right )} e^{\left (2 \, x - 24\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="fricas")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

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giac [A]  time = 0.12, size = 23, normalized size = 1.28 \begin {gather*} 4 \, {\left (e^{24} \log \relax (2)^{2} + 2 \, e^{12} \log \relax (2) + 1\right )} e^{\left (2 \, x - 24\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="giac")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

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maple [A]  time = 0.03, size = 28, normalized size = 1.56

method	result	size
gosper	\(4 \left ({\mathrm e}^{24} \ln \relax (2)^{2}+2 \,{\mathrm e}^{12} \ln \relax (2)+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
default	\(4 \left ({\mathrm e}^{24} \ln \relax (2)^{2}+2 \,{\mathrm e}^{12} \ln \relax (2)+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
norman	\(4 \left ({\mathrm e}^{24} \ln \relax (2)^{2}+2 \,{\mathrm e}^{12} \ln \relax (2)+1\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}\)	\(28\)
derivativedivides	\(\frac {\left (8 \,{\mathrm e}^{24} \ln \relax (2)^{2}+16 \,{\mathrm e}^{12} \ln \relax (2)+8\right ) {\mathrm e}^{2 x} {\mathrm e}^{-24}}{2}\)	\(29\)
risch	\(4 \ln \relax (2)^{2} {\mathrm e}^{24} {\mathrm e}^{2 x -24}+8 \,{\mathrm e}^{12} \ln \relax (2) {\mathrm e}^{2 x -24}+4 \,{\mathrm e}^{2 x -24}\)	\(36\)
meijerg	\(-4 \ln \relax (2)^{2} \left (1-{\mathrm e}^{2 x}\right )-8 \,{\mathrm e}^{-12} \ln \relax (2) \left (1-{\mathrm e}^{2 x}\right )-4 \,{\mathrm e}^{-24} \left (1-{\mathrm e}^{2 x}\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(12)^2*ln(2)^2+16*exp(12)*ln(2)+8)*exp(x)^2/exp(12)^2,x,method=_RETURNVERBOSE)

[Out]

4*(exp(12)^2*ln(2)^2+2*exp(12)*ln(2)+1)*exp(x)^2/exp(12)^2

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maxima [A]  time = 0.35, size = 23, normalized size = 1.28 \begin {gather*} 4 \, {\left (e^{24} \log \relax (2)^{2} + 2 \, e^{12} \log \relax (2) + 1\right )} e^{\left (2 \, x - 24\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(12)^2*log(2)^2+16*exp(12)*log(2)+8)*exp(x)^2/exp(12)^2,x, algorithm="maxima")

[Out]

4*(e^24*log(2)^2 + 2*e^12*log(2) + 1)*e^(2*x - 24)

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mupad [B]  time = 0.08, size = 23, normalized size = 1.28 \begin {gather*} {\mathrm {e}}^{2\,x-24}\,\left (8\,{\mathrm {e}}^{12}\,\ln \relax (2)+4\,{\mathrm {e}}^{24}\,{\ln \relax (2)}^2+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*exp(-24)*(16*exp(12)*log(2) + 8*exp(24)*log(2)^2 + 8),x)

[Out]

exp(2*x - 24)*(8*exp(12)*log(2) + 4*exp(24)*log(2)^2 + 4)

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sympy [A]  time = 0.10, size = 27, normalized size = 1.50 \begin {gather*} \frac {\left (4 + 8 e^{12} \log {\relax (2 )} + 4 e^{24} \log {\relax (2 )}^{2}\right ) e^{2 x}}{e^{24}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*exp(12)**2*ln(2)**2+16*exp(12)*ln(2)+8)*exp(x)**2/exp(12)**2,x)

[Out]

(4 + 8*exp(12)*log(2) + 4*exp(24)*log(2)**2)*exp(-24)*exp(2*x)

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