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

Optimal. Leaf size=23 \[ e^{(-5+x) x \left (-x+\frac {1+x-\log (\log (4))}{x}\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 26, normalized size of antiderivative = 1.13, number of steps used = 1, number of rules used = 1, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6706} \begin {gather*} e^{-x^3+6 x^2-4 x-5} \log ^{5-x}(4) \end {gather*}

Antiderivative was successfully verified.

[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 6706

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

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Mathematica [A]  time = 0.36, size = 26, normalized size = 1.13 \begin {gather*} e^{-5-4 x+6 x^2-x^3} \log ^{5-x}(4) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.83, size = 26, normalized size = 1.13 \begin {gather*} e^{\left (-x^{3} + 6 \, x^{2} - {\left (x - 5\right )} \log \left (2 \, \log \relax (2)\right ) - 4 \, x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.48, size = 31, normalized size = 1.35 \begin {gather*} e^{\left (-x^{3} + 6 \, x^{2} - x \log \left (2 \, \log \relax (2)\right ) - 4 \, x + 5 \, \log \left (2 \, \log \relax (2)\right ) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 20, normalized size = 0.87

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.43, size = 26, normalized size = 1.13 \begin {gather*} e^{\left (-x^{3} + 6 \, x^{2} - {\left (x - 5\right )} \log \left (2 \, \log \relax (2)\right ) - 4 \, x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.44, size = 29, normalized size = 1.26 \begin {gather*} {\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{6\,x^2}\,{\left (2\,\ln \relax (2)\right )}^{5-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.15, size = 24, normalized size = 1.04 \begin {gather*} e^{- x^{3} + 6 x^{2} - 4 x + \left (5 - x\right ) \log {\left (2 \log {\relax (2 )} \right )} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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