\(\int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} (-4+12 x-3 x^2-\log (\log (4))) \, dx\) [1594]

Optimal result

Integrand size = 42, antiderivative size = 23 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{(-5+x) x \left (-x+\frac {1+x-\log (\log (4))}{x}\right )} \]

[Out]

exp(x*(-5+x)*((1+x-ln(2*ln(2)))/x-x))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6838} \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{-x^3+6 x^2-4 x-5} \log ^{5-x}(4) \]

[In]

Int[E^(-5 - 4*x + 6*x^2 - x^3 + (5 - x)*Log[Log[4]])*(-4 + 12*x - 3*x^2 - Log[Log[4]]),x]

[Out]

E^(-5 - 4*x + 6*x^2 - x^3)*Log[4]^(5 - x)

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = e^{-5-4 x+6 x^2-x^3} \log ^{5-x}(4) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{-5-4 x+6 x^2-x^3} \log ^{5-x}(4) \]

[In]

Integrate[E^(-5 - 4*x + 6*x^2 - x^3 + (5 - x)*Log[Log[4]])*(-4 + 12*x - 3*x^2 - Log[Log[4]]),x]

[Out]

E^(-5 - 4*x + 6*x^2 - x^3)*Log[4]^(5 - x)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\({\mathrm e}^{-\left (-5+x \right ) \left (x^{2}+\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )-x -1\right )}\)	\(20\)
derivativedivides	\({\mathrm e}^{\left (5-x \right ) \ln \left (2 \ln \left (2\right )\right )-x^{3}+6 x^{2}-4 x -5}\)	\(28\)
default	\({\mathrm e}^{\left (5-x \right ) \ln \left (2 \ln \left (2\right )\right )-x^{3}+6 x^{2}-4 x -5}\)	\(28\)
norman	\({\mathrm e}^{\left (5-x \right ) \ln \left (2 \ln \left (2\right )\right )-x^{3}+6 x^{2}-4 x -5}\)	\(28\)
parallelrisch	\({\mathrm e}^{\left (5-x \right ) \ln \left (2 \ln \left (2\right )\right )-x^{3}+6 x^{2}-4 x -5}\)	\(28\)
gosper	\({\mathrm e}^{-x^{3}-x \ln \left (2 \ln \left (2\right )\right )+6 x^{2}+5 \ln \left (2 \ln \left (2\right )\right )-4 x -5}\)	\(32\)

[In]

int((-ln(2*ln(2))-3*x^2+12*x-4)*exp((5-x)*ln(2*ln(2))-x^3+6*x^2-4*x-5),x,method=_RETURNVERBOSE)

[Out]

exp(-(-5+x)*(x^2+ln(2)+ln(ln(2))-x-1))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{\left (-x^{3} + 6 \, x^{2} - {\left (x - 5\right )} \log \left (2 \, \log \left (2\right )\right ) - 4 \, x - 5\right )} \]

[In]

integrate((-log(2*log(2))-3*x^2+12*x-4)*exp((5-x)*log(2*log(2))-x^3+6*x^2-4*x-5),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{- x^{3} + 6 x^{2} - 4 x + \left (5 - x\right ) \log {\left (2 \log {\left (2 \right )} \right )} - 5} \]

[In]

integrate((-ln(2*ln(2))-3*x**2+12*x-4)*exp((5-x)*ln(2*ln(2))-x**3+6*x**2-4*x-5),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{\left (-x^{3} + 6 \, x^{2} - {\left (x - 5\right )} \log \left (2 \, \log \left (2\right )\right ) - 4 \, x - 5\right )} \]

[In]

integrate((-log(2*log(2))-3*x^2+12*x-4)*exp((5-x)*log(2*log(2))-x^3+6*x^2-4*x-5),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx=e^{\left (-x^{3} + 6 \, x^{2} - x \log \left (2 \, \log \left (2\right )\right ) - 4 \, x + 5 \, \log \left (2 \, \log \left (2\right )\right ) - 5\right )} \]

[In]

integrate((-log(2*log(2))-3*x^2+12*x-4)*exp((5-x)*log(2*log(2))-x^3+6*x^2-4*x-5),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int e^{-5-4 x+6 x^2-x^3+(5-x) \log (\log (4))} \left (-4+12 x-3 x^2-\log (\log (4))\right ) \, dx={\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{6\,x^2}\,{\left (2\,\ln \left (2\right )\right )}^{5-x} \]

[In]

int(-exp(6*x^2 - log(2*log(2))*(x - 5) - 4*x - x^3 - 5)*(log(2*log(2)) - 12*x + 3*x^2 + 4),x)

[Out]

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