\(\int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx\) [8675]

Optimal result

Integrand size = 39, antiderivative size = 25 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 e^{6-2 x+2 (-x+\log (2))^2-\log ^2(4)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2276, 2268} \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 \exp \left (2 x^2-2 x (1+\log (4))+6-\log ^2(4)+2 \log ^2(2)\right ) \]

[In]

Int[E^(6 - 2*x + 2*x^2 - 4*x*Log[2] + 2*Log[2]^2 - Log[4]^2)*(-8 + 16*x - 16*Log[2]),x]

[Out]

4*E^(6 + 2*x^2 + 2*Log[2]^2 - Log[4]^2 - 2*x*(1 + Log[4]))

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{align*} \text {integral}& = \int \exp \left (6+2 x^2+2 \log ^2(2)-\log ^2(4)-2 x (1+\log (4))\right ) (16 x-8 (1+\log (4))) \, dx \\ & = 4 \exp \left (6+2 x^2+2 \log ^2(2)-\log ^2(4)-2 x (1+\log (4))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=e^{\frac {1}{2} \left (11-2 \log ^2(4)+(1-2 x+\log (4))^2\right )} \]

[In]

Integrate[E^(6 - 2*x + 2*x^2 - 4*x*Log[2] + 2*Log[2]^2 - Log[4]^2)*(-8 + 16*x - 16*Log[2]),x]

[Out]

E^((11 - 2*Log[4]^2 + (1 - 2*x + Log[4])^2)/2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
gosper	\(4 \,{\mathrm e}^{-2 \ln \left (2\right )^{2}-4 x \ln \left (2\right )+2 x^{2}-2 x +6}\)	\(25\)
default	\(4 \,{\mathrm e}^{-2 \ln \left (2\right )^{2}-4 x \ln \left (2\right )+2 x^{2}-2 x +6}\)	\(25\)
norman	\(4 \,{\mathrm e}^{-2 \ln \left (2\right )^{2}-4 x \ln \left (2\right )+2 x^{2}-2 x +6}\)	\(25\)
risch	\(4 \,4^{-2 x} {\mathrm e}^{-2 \ln \left (2\right )^{2}+6+2 x^{2}-2 x}\)	\(25\)
parallelrisch	\(4 \,{\mathrm e}^{-2 \ln \left (2\right )^{2}-4 x \ln \left (2\right )+2 x^{2}-2 x +6}\)	\(25\)

[In]

int((-16*ln(2)+16*x-8)*exp(-ln(2)^2-2*x*ln(2)+x^2-x+3)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 \, e^{\left (2 \, x^{2} - 4 \, x \log \left (2\right ) - 2 \, \log \left (2\right )^{2} - 2 \, x + 6\right )} \]

[In]

integrate((-16*log(2)+16*x-8)*exp(-log(2)^2-2*x*log(2)+x^2-x+3)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 e^{2 x^{2} - 4 x \log {\left (2 \right )} - 2 x - 2 \log {\left (2 \right )}^{2} + 6} \]

[In]

integrate((-16*ln(2)+16*x-8)*exp(-ln(2)**2-2*x*ln(2)+x**2-x+3)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 \, e^{\left (2 \, x^{2} - 4 \, x \log \left (2\right ) - 2 \, \log \left (2\right )^{2} - 2 \, x + 6\right )} \]

[In]

integrate((-16*log(2)+16*x-8)*exp(-log(2)^2-2*x*log(2)+x^2-x+3)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=4 \, e^{\left (2 \, x^{2} - 4 \, x \log \left (2\right ) - 2 \, \log \left (2\right )^{2} - 2 \, x + 6\right )} \]

[In]

integrate((-16*log(2)+16*x-8)*exp(-log(2)^2-2*x*log(2)+x^2-x+3)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int e^{6-2 x+2 x^2-4 x \log (2)+2 \log ^2(2)-\log ^2(4)} (-8+16 x-16 \log (2)) \, dx=\frac {4\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^6\,{\mathrm {e}}^{-2\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{2\,x^2}}{2^{4\,x}} \]

[In]

int(-exp(2*x^2 - 4*x*log(2) - 2*log(2)^2 - 2*x + 6)*(16*log(2) - 16*x + 8),x)

[Out]

(4*exp(-2*x)*exp(6)*exp(-2*log(2)^2)*exp(2*x^2))/2^(4*x)