3.52.57 \(\int e^{2-4 e-8 x+4 x^2+(-8+8 x) \log (4)+4 \log ^2(4)} (-8+8 x+8 \log (4)) \, dx\) [5157]

Optimal. Leaf size=17 \[ e^{-2-4 e+(2-2 (x+\log (4)))^2} \]

[Out]

exp((2-2*x-4*ln(2))^2-2-4*exp(1))

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Rubi [A]
time = 0.18, antiderivative size = 34, normalized size of antiderivative = 2.00, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2276, 2268} \begin {gather*} \exp \left (4 x^2-8 x (1-\log (4))+2 \left (1-2 e+2 \log ^2(4)-4 \log (4)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2268

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]

Rule 2276

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

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

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Mathematica [A]
time = 0.43, size = 33, normalized size = 1.94 \begin {gather*} 65536^{-1+x} e^{2-4 e-8 x+4 x^2-16 \log ^2(2)+8 \log ^2(4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

65536^(-1 + x)*E^(2 - 4*E - 8*x + 4*x^2 - 16*Log[2]^2 + 8*Log[4]^2)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.35, size = 184, normalized size = 10.82

method	result	size
risch	\(2^{16 x -16} {\mathrm e}^{16 \ln \left (2\right )^{2}+2-4 \,{\mathrm e}+4 x^{2}-8 x}\)	\(30\)
gosper	\({\mathrm e}^{16 \ln \left (2\right )^{2}+16 x \ln \left (2\right )+4 x^{2}-4 \,{\mathrm e}-16 \ln \left (2\right )-8 x +2}\)	\(31\)
derivativedivides	\({\mathrm e}^{16 \ln \left (2\right )^{2}+2 \left (8 x -8\right ) \ln \left (2\right )-4 \,{\mathrm e}+4 x^{2}-8 x +2}\)	\(31\)
norman	\({\mathrm e}^{16 \ln \left (2\right )^{2}+2 \left (8 x -8\right ) \ln \left (2\right )-4 \,{\mathrm e}+4 x^{2}-8 x +2}\)	\(31\)
default	\({\mathrm e}^{4 x^{2}+\left (16 \ln \left (2\right )-8\right ) x +16 \ln \left (2\right )^{2}-4 \,{\mathrm e}-16 \ln \left (2\right )+2}+\frac {i \left (16 \ln \left (2\right )-8\right ) \sqrt {\pi }\, {\mathrm e}^{16 \ln \left (2\right )^{2}-4 \,{\mathrm e}-16 \ln \left (2\right )+2-\frac {\left (16 \ln \left (2\right )-8\right )^{2}}{16}} \erf \left (2 i x +\frac {i \left (16 \ln \left (2\right )-8\right )}{4}\right )}{4}-4 i \ln \left (2\right ) \sqrt {\pi }\, {\mathrm e}^{16 \ln \left (2\right )^{2}-4 \,{\mathrm e}-16 \ln \left (2\right )+2-\frac {\left (16 \ln \left (2\right )-8\right )^{2}}{16}} \erf \left (2 i x +\frac {i \left (16 \ln \left (2\right )-8\right )}{4}\right )+2 i \sqrt {\pi }\, {\mathrm e}^{16 \ln \left (2\right )^{2}-4 \,{\mathrm e}-16 \ln \left (2\right )+2-\frac {\left (16 \ln \left (2\right )-8\right )^{2}}{16}} \erf \left (2 i x +\frac {i \left (16 \ln \left (2\right )-8\right )}{4}\right )\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*ln(2)+8*x-8)*exp(16*ln(2)^2+2*(8*x-8)*ln(2)-4*exp(1)+4*x^2-8*x+2),x,method=_RETURNVERBOSE)

[Out]

exp(4*x^2+(16*ln(2)-8)*x+16*ln(2)^2-4*exp(1)-16*ln(2)+2)+1/4*I*(16*ln(2)-8)*Pi^(1/2)*exp(16*ln(2)^2-4*exp(1)-1
6*ln(2)+2-1/16*(16*ln(2)-8)^2)*erf(2*I*x+1/4*I*(16*ln(2)-8))-4*I*ln(2)*Pi^(1/2)*exp(16*ln(2)^2-4*exp(1)-16*ln(
2)+2-1/16*(16*ln(2)-8)^2)*erf(2*I*x+1/4*I*(16*ln(2)-8))+2*I*Pi^(1/2)*exp(16*ln(2)^2-4*exp(1)-16*ln(2)+2-1/16*(
16*ln(2)-8)^2)*erf(2*I*x+1/4*I*(16*ln(2)-8))

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Maxima [A]
time = 0.27, size = 28, normalized size = 1.65 \begin {gather*} e^{\left (4 \, x^{2} + 16 \, {\left (x - 1\right )} \log \left (2\right ) + 16 \, \log \left (2\right )^{2} - 8 \, x - 4 \, e + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(2)+8*x-8)*exp(16*log(2)^2+2*(8*x-8)*log(2)-4*exp(1)+4*x^2-8*x+2),x, algorithm="maxima")

[Out]

e^(4*x^2 + 16*(x - 1)*log(2) + 16*log(2)^2 - 8*x - 4*e + 2)

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Fricas [A]
time = 0.38, size = 28, normalized size = 1.65 \begin {gather*} e^{\left (4 \, x^{2} + 16 \, {\left (x - 1\right )} \log \left (2\right ) + 16 \, \log \left (2\right )^{2} - 8 \, x - 4 \, e + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(2)+8*x-8)*exp(16*log(2)^2+2*(8*x-8)*log(2)-4*exp(1)+4*x^2-8*x+2),x, algorithm="fricas")

[Out]

e^(4*x^2 + 16*(x - 1)*log(2) + 16*log(2)^2 - 8*x - 4*e + 2)

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Sympy [A]
time = 0.06, size = 31, normalized size = 1.82 \begin {gather*} e^{4 x^{2} - 8 x + \left (16 x - 16\right ) \log {\left (2 \right )} - 4 e + 2 + 16 \log {\left (2 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*ln(2)+8*x-8)*exp(16*ln(2)**2+2*(8*x-8)*ln(2)-4*exp(1)+4*x**2-8*x+2),x)

[Out]

exp(4*x**2 - 8*x + (16*x - 16)*log(2) - 4*E + 2 + 16*log(2)**2)

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Giac [A]
time = 0.42, size = 30, normalized size = 1.76 \begin {gather*} e^{\left (4 \, x^{2} + 16 \, x \log \left (2\right ) + 16 \, \log \left (2\right )^{2} - 8 \, x - 4 \, e - 16 \, \log \left (2\right ) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*log(2)+8*x-8)*exp(16*log(2)^2+2*(8*x-8)*log(2)-4*exp(1)+4*x^2-8*x+2),x, algorithm="giac")

[Out]

e^(4*x^2 + 16*x*log(2) + 16*log(2)^2 - 8*x - 4*e - 16*log(2) + 2)

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Mupad [B]
time = 0.14, size = 31, normalized size = 1.82 \begin {gather*} \frac {2^{16\,x}\,{\mathrm {e}}^{-4\,\mathrm {e}}\,{\mathrm {e}}^{-8\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{16\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{4\,x^2}}{65536} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*log(2)*(8*x - 8) - 4*exp(1) - 8*x + 16*log(2)^2 + 4*x^2 + 2)*(8*x + 16*log(2) - 8),x)

[Out]

(2^(16*x)*exp(-4*exp(1))*exp(-8*x)*exp(2)*exp(16*log(2)^2)*exp(4*x^2))/65536

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