3.92.16 \(\int 4 e^{-8+\frac {4 (25 x-4 x^3-x^3 \log (4))}{e^8}} (25-12 x^2-3 x^2 \log (4)) \, dx\)

Optimal. Leaf size=19 \[ e^{\frac {4 x \left (25-x^2 (4+\log (4))\right )}{e^8}} \]

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Rubi [A]  time = 0.12, antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {6, 12, 6706} \begin {gather*} 4^{-\frac {4 x^3}{e^8}} e^{\frac {4 \left (25 x-4 x^3\right )}{e^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[4*E^(-8 + (4*(25*x - 4*x^3 - x^3*Log[4]))/E^8)*(25 - 12*x^2 - 3*x^2*Log[4]),x]

[Out]

E^((4*(25*x - 4*x^3))/E^8)/4^((4*x^3)/E^8)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\int 4 e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25+x^2 (-12-3 \log (4))\right ) \, dx\\ &=4 \int e^{-8+\frac {4 \left (25 x-4 x^3-x^3 \log (4)\right )}{e^8}} \left (25+x^2 (-12-3 \log (4))\right ) \, dx\\ &=4^{-\frac {4 x^3}{e^8}} e^{\frac {4 \left (25 x-4 x^3\right )}{e^8}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 18, normalized size = 0.95 \begin {gather*} e^{-\frac {4 x \left (-25+x^2 (4+\log (4))\right )}{e^8}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[4*E^(-8 + (4*(25*x - 4*x^3 - x^3*Log[4]))/E^8)*(25 - 12*x^2 - 3*x^2*Log[4]),x]

[Out]

E^((-4*x*(-25 + x^2*(4 + Log[4])))/E^8)

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fricas [A]  time = 1.05, size = 26, normalized size = 1.37 \begin {gather*} e^{\left (-{\left (2 \, x^{3} \log \relax (2) + 4 \, x^{3} - 25 \, x\right )} e^{\left (2 \, \log \relax (2) - 8\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^2*log(2)-12*x^2+25)*exp(2*log(2)-8)*exp((-2*x^3*log(2)-4*x^3+25*x)*exp(2*log(2)-8)),x, algorit
hm="fricas")

[Out]

e^(-(2*x^3*log(2) + 4*x^3 - 25*x)*e^(2*log(2) - 8))

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giac [A]  time = 0.15, size = 44, normalized size = 2.32 \begin {gather*} \frac {1}{4} \, e^{\left (-2 \, x^{3} e^{\left (2 \, \log \relax (2) - 8\right )} \log \relax (2) - 4 \, x^{3} e^{\left (2 \, \log \relax (2) - 8\right )} + 25 \, x e^{\left (2 \, \log \relax (2) - 8\right )} + 2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^2*log(2)-12*x^2+25)*exp(2*log(2)-8)*exp((-2*x^3*log(2)-4*x^3+25*x)*exp(2*log(2)-8)),x, algorit
hm="giac")

[Out]

1/4*e^(-2*x^3*e^(2*log(2) - 8)*log(2) - 4*x^3*e^(2*log(2) - 8) + 25*x*e^(2*log(2) - 8) + 2*log(2))

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maple [A]  time = 0.06, size = 21, normalized size = 1.11

method	result	size
risch	\({\mathrm e}^{-4 x \left (2 x^{2} \ln \relax (2)+4 x^{2}-25\right ) {\mathrm e}^{-8}}\)	\(21\)
gosper	\({\mathrm e}^{-x \left (2 x^{2} \ln \relax (2)+4 x^{2}-25\right ) {\mathrm e}^{2 \ln \relax (2)-8}}\)	\(26\)
derivativedivides	\({\mathrm e}^{\left (-2 x^{3} \ln \relax (2)-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \relax (2)-8}}\)	\(26\)
default	\({\mathrm e}^{\left (-2 x^{3} \ln \relax (2)-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \relax (2)-8}}\)	\(26\)
norman	\({\mathrm e}^{-8} {\mathrm e}^{8} {\mathrm e}^{\left (-2 x^{3} \ln \relax (2)-4 x^{3}+25 x \right ) {\mathrm e}^{2 \ln \relax (2)-8}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*x^2*ln(2)-12*x^2+25)*exp(2*ln(2)-8)*exp((-2*x^3*ln(2)-4*x^3+25*x)*exp(2*ln(2)-8)),x,method=_RETURNVERB
OSE)

[Out]

exp(-4*x*(2*x^2*ln(2)+4*x^2-25)*exp(-8))

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maxima [A]  time = 0.35, size = 21, normalized size = 1.11 \begin {gather*} e^{\left (-4 \, {\left (2 \, x^{3} \log \relax (2) + 4 \, x^{3} - 25 \, x\right )} e^{\left (-8\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x^2*log(2)-12*x^2+25)*exp(2*log(2)-8)*exp((-2*x^3*log(2)-4*x^3+25*x)*exp(2*log(2)-8)),x, algorit
hm="maxima")

[Out]

e^(-4*(2*x^3*log(2) + 4*x^3 - 25*x)*e^(-8))

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mupad [B]  time = 0.25, size = 26, normalized size = 1.37 \begin {gather*} \frac {{\mathrm {e}}^{-16\,x^3\,{\mathrm {e}}^{-8}}\,{\mathrm {e}}^{100\,x\,{\mathrm {e}}^{-8}}}{2^{8\,x^3\,{\mathrm {e}}^{-8}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-exp(2*log(2) - 8)*(2*x^3*log(2) - 25*x + 4*x^3))*exp(2*log(2) - 8)*(6*x^2*log(2) + 12*x^2 - 25),x)

[Out]

(exp(-16*x^3*exp(-8))*exp(100*x*exp(-8)))/2^(8*x^3*exp(-8))

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sympy [A]  time = 0.14, size = 20, normalized size = 1.05 \begin {gather*} e^{\frac {4 \left (- 4 x^{3} - 2 x^{3} \log {\relax (2 )} + 25 x\right )}{e^{8}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x**2*ln(2)-12*x**2+25)*exp(2*ln(2)-8)*exp((-2*x**3*ln(2)-4*x**3+25*x)*exp(2*ln(2)-8)),x)

[Out]

exp(4*(-4*x**3 - 2*x**3*log(2) + 25*x)*exp(-8))

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