3.30.98 \(\int \frac {1}{4} e^{\frac {1}{4} (-4+4 x+64 x^2-64 x^3+(1-x) \log (9))} (4+4 x+128 x^2-192 x^3-x \log (9)) \, dx\)

Optimal. Leaf size=21 \[ e^{(-1+x) \left (1-16 x^2-\frac {\log (9)}{4}\right )} x \]

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Rubi [B]  time = 0.15, antiderivative size = 62, normalized size of antiderivative = 2.95, number of steps used = 3, number of rules used = 3, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6, 12, 2288} \begin {gather*} \frac {3^{\frac {1-x}{2}} e^{-16 x^3+16 x^2+x-1} \left (-192 x^3+128 x^2+x (4-\log (9))\right )}{-192 x^2+128 x+4-\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-4 + 4*x + 64*x^2 - 64*x^3 + (1 - x)*Log[9])/4)*(4 + 4*x + 128*x^2 - 192*x^3 - x*Log[9]))/4,x]

[Out]

(3^((1 - x)/2)*E^(-1 + x + 16*x^2 - 16*x^3)*(128*x^2 - 192*x^3 + x*(4 - Log[9])))/(4 + 128*x - 192*x^2 - Log[9
])

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{4} \exp \left (\frac {1}{4} \left (-4+4 x+64 x^2-64 x^3+(1-x) \log (9)\right )\right ) \left (4+128 x^2-192 x^3+x (4-\log (9))\right ) \, dx\\ &=\frac {1}{4} \int \exp \left (\frac {1}{4} \left (-4+4 x+64 x^2-64 x^3+(1-x) \log (9)\right )\right ) \left (4+128 x^2-192 x^3+x (4-\log (9))\right ) \, dx\\ &=\frac {3^{\frac {1-x}{2}} e^{-1+x+16 x^2-16 x^3} \left (128 x^2-192 x^3+x (4-\log (9))\right )}{4+128 x-192 x^2-\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.29, size = 56, normalized size = 2.67 \begin {gather*} \frac {1}{4} \int e^{\frac {1}{4} \left (-4+4 x+64 x^2-64 x^3+(1-x) \log (9)\right )} \left (4+4 x+128 x^2-192 x^3-x \log (9)\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-4 + 4*x + 64*x^2 - 64*x^3 + (1 - x)*Log[9])/4)*(4 + 4*x + 128*x^2 - 192*x^3 - x*Log[9]))/4,x]

[Out]

Integrate[E^((-4 + 4*x + 64*x^2 - 64*x^3 + (1 - x)*Log[9])/4)*(4 + 4*x + 128*x^2 - 192*x^3 - x*Log[9]), x]/4

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fricas [A]  time = 0.67, size = 23, normalized size = 1.10 \begin {gather*} x e^{\left (-16 \, x^{3} + 16 \, x^{2} - \frac {1}{2} \, {\left (x - 1\right )} \log \relax (3) + x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x*log(3)-192*x^3+128*x^2+4*x+4)*exp(1/2*(-x+1)*log(3)-16*x^3+16*x^2+x-1),x, algorithm="frica
s")

[Out]

x*e^(-16*x^3 + 16*x^2 - 1/2*(x - 1)*log(3) + x - 1)

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giac [A]  time = 0.30, size = 24, normalized size = 1.14 \begin {gather*} \sqrt {3} x e^{\left (-16 \, x^{3} + 16 \, x^{2} - \frac {1}{2} \, x \log \relax (3) + x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x*log(3)-192*x^3+128*x^2+4*x+4)*exp(1/2*(-x+1)*log(3)-16*x^3+16*x^2+x-1),x, algorithm="giac"
)

[Out]

sqrt(3)*x*e^(-16*x^3 + 16*x^2 - 1/2*x*log(3) + x - 1)

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maple [A]  time = 0.06, size = 26, normalized size = 1.24

method	result	size
gosper	\({\mathrm e}^{-\frac {x \ln \relax (3)}{2}+\frac {\ln \relax (3)}{2}-16 x^{3}+16 x^{2}+x -1} x\)	\(26\)
norman	\(x \,{\mathrm e}^{\frac {\left (1-x \right ) \ln \relax (3)}{2}-16 x^{3}+16 x^{2}+x -1}\)	\(26\)
risch	\(x 3^{-\frac {x}{2}+\frac {1}{2}} {\mathrm e}^{-\left (x -1\right ) \left (4 x -1\right ) \left (4 x +1\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-2*x*ln(3)-192*x^3+128*x^2+4*x+4)*exp(1/2*(1-x)*ln(3)-16*x^3+16*x^2+x-1),x,method=_RETURNVERBOSE)

[Out]

exp(-1/2*x*ln(3)+1/2*ln(3)-16*x^3+16*x^2+x-1)*x

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maxima [A]  time = 1.03, size = 24, normalized size = 1.14 \begin {gather*} \sqrt {3} x e^{\left (-16 \, x^{3} + 16 \, x^{2} - \frac {1}{2} \, x \log \relax (3) + x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x*log(3)-192*x^3+128*x^2+4*x+4)*exp(1/2*(-x+1)*log(3)-16*x^3+16*x^2+x-1),x, algorithm="maxim
a")

[Out]

sqrt(3)*x*e^(-16*x^3 + 16*x^2 - 1/2*x*log(3) + x - 1)

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mupad [B]  time = 0.11, size = 28, normalized size = 1.33 \begin {gather*} \frac {\sqrt {3}\,x\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{-16\,x^3}\,{\mathrm {e}}^x}{3^{x/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - (log(3)*(x - 1))/2 + 16*x^2 - 16*x^3 - 1)*(4*x - 2*x*log(3) + 128*x^2 - 192*x^3 + 4))/4,x)

[Out]

(3^(1/2)*x*exp(-1)*exp(16*x^2)*exp(-16*x^3)*exp(x))/3^(x/2)

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sympy [A]  time = 0.15, size = 26, normalized size = 1.24 \begin {gather*} x e^{- 16 x^{3} + 16 x^{2} + x + \left (\frac {1}{2} - \frac {x}{2}\right ) \log {\relax (3 )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x*ln(3)-192*x**3+128*x**2+4*x+4)*exp(1/2*(-x+1)*ln(3)-16*x**3+16*x**2+x-1),x)

[Out]

x*exp(-16*x**3 + 16*x**2 + x + (1/2 - x/2)*log(3) - 1)

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