3.72.32 \(\int \frac {1}{64} e^{-x^2+\frac {1}{4} e^{-x^2} (4 x+3 e^{x^2} x)} (-4-3 e^{x^2}+8 x^2) \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{16} \left (e^5-e^{\frac {3 x}{4}+e^{-x^2} x}\right ) \]

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Rubi [F]  time = 0.59, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{64} e^{-x^2+\frac {1}{4} e^{-x^2} \left (4 x+3 e^{x^2} x\right )} \left (-4-3 e^{x^2}+8 x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x^2 + (4*x + 3*E^x^2*x)/(4*E^x^2))*(-4 - 3*E^x^2 + 8*x^2))/64,x]

[Out]

-1/16*Defer[Int][E^(((3 + 4/E^x^2 - 4*x)*x)/4), x] - (3*Defer[Int][E^((3*x)/4 + x/E^x^2), x])/64 + Defer[Int][
E^(((3 + 4/E^x^2 - 4*x)*x)/4)*x^2, x]/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int e^{-x^2+\frac {1}{4} e^{-x^2} \left (4 x+3 e^{x^2} x\right )} \left (-4-3 e^{x^2}+8 x^2\right ) \, dx\\ &=\frac {1}{64} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} \left (-4-3 e^{x^2}+8 x^2\right ) \, dx\\ &=\frac {1}{64} \int \left (-4 e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x}-3 e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x+x^2}+8 e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} x^2\right ) \, dx\\ &=-\left (\frac {3}{64} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x+x^2} \, dx\right )-\frac {1}{16} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} \, dx+\frac {1}{8} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} x^2 \, dx\\ &=-\left (\frac {3}{64} \int e^{\frac {3 x}{4}+e^{-x^2} x} \, dx\right )-\frac {1}{16} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} \, dx+\frac {1}{8} \int e^{\frac {1}{4} \left (3+4 e^{-x^2}-4 x\right ) x} x^2 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 21, normalized size = 0.78 \begin {gather*} -\frac {1}{16} e^{\frac {3 x}{4}+e^{-x^2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x^2 + (4*x + 3*E^x^2*x)/(4*E^x^2))*(-4 - 3*E^x^2 + 8*x^2))/64,x]

[Out]

-1/16*E^((3*x)/4 + x/E^x^2)

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fricas [A]  time = 0.95, size = 33, normalized size = 1.22 \begin {gather*} -\frac {1}{16} \, e^{\left (x^{2} - \frac {1}{4} \, {\left ({\left (4 \, x^{2} - 3 \, x\right )} e^{\left (x^{2}\right )} - 4 \, x\right )} e^{\left (-x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-3*exp(x^2)+8*x^2-4)*exp(1/4*(3*exp(x^2)*x+4*x)/exp(x^2))/exp(x^2),x, algorithm="fricas")

[Out]

-1/16*e^(x^2 - 1/4*((4*x^2 - 3*x)*e^(x^2) - 4*x)*e^(-x^2))

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giac [A]  time = 0.20, size = 15, normalized size = 0.56 \begin {gather*} -\frac {1}{16} \, e^{\left (x e^{\left (-x^{2}\right )} + \frac {3}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-3*exp(x^2)+8*x^2-4)*exp(1/4*(3*exp(x^2)*x+4*x)/exp(x^2))/exp(x^2),x, algorithm="giac")

[Out]

-1/16*e^(x*e^(-x^2) + 3/4*x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.78

method	result	size
risch	\(-\frac {{\mathrm e}^{\frac {x \left (3 \,{\mathrm e}^{x^{2}}+4\right ) {\mathrm e}^{-x^{2}}}{4}}}{16}\)	\(21\)
norman	\(-\frac {{\mathrm e}^{\frac {\left (3 \,{\mathrm e}^{x^{2}} x +4 x \right ) {\mathrm e}^{-x^{2}}}{4}}}{16}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*(-3*exp(x^2)+8*x^2-4)*exp(1/4*(3*exp(x^2)*x+4*x)/exp(x^2))/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

-1/16*exp(1/4*x*(3*exp(x^2)+4)*exp(-x^2))

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maxima [A]  time = 0.40, size = 15, normalized size = 0.56 \begin {gather*} -\frac {1}{16} \, e^{\left (x e^{\left (-x^{2}\right )} + \frac {3}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-3*exp(x^2)+8*x^2-4)*exp(1/4*(3*exp(x^2)*x+4*x)/exp(x^2))/exp(x^2),x, algorithm="maxima")

[Out]

-1/16*e^(x*e^(-x^2) + 3/4*x)

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mupad [B]  time = 4.19, size = 15, normalized size = 0.56 \begin {gather*} -\frac {{\mathrm {e}}^{\frac {3\,x}{4}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-x^2}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(-x^2)*(x + (3*x*exp(x^2))/4))*exp(-x^2)*(3*exp(x^2) - 8*x^2 + 4))/64,x)

[Out]

-(exp((3*x)/4)*exp(x*exp(-x^2)))/16

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sympy [A]  time = 0.37, size = 20, normalized size = 0.74 \begin {gather*} - \frac {e^{\left (\frac {3 x e^{x^{2}}}{4} + x\right ) e^{- x^{2}}}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-3*exp(x**2)+8*x**2-4)*exp(1/4*(3*exp(x**2)*x+4*x)/exp(x**2))/exp(x**2),x)

[Out]

-exp((3*x*exp(x**2)/4 + x)*exp(-x**2))/16

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