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

Optimal. Leaf size=22 \[ e^{-\frac {1}{2} e^{-2 x} (1-x)-2 x} x \]

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Rubi [F]  time = 0.65, 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}{2} \exp \left (-2 x-\frac {1}{2} e^{-2 x} \left (1-x+4 e^{2 x} x\right )\right ) \left (e^{2 x} (2-4 x)+3 x-2 x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 24, normalized size = 1.09 \begin {gather*} e^{\frac {1}{2} e^{-2 x} \left (-1+x-4 e^{2 x} x\right )} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2*x - (1 - x + 4*E^(2*x)*x)/(2*E^(2*x)))*(E^(2*x)*(2 - 4*x) + 3*x - 2*x^2))/2,x]

[Out]

E^((-1 + x - 4*E^(2*x)*x)/(2*E^(2*x)))*x

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fricas [A]  time = 0.60, size = 25, normalized size = 1.14 \begin {gather*} x e^{\left (-\frac {1}{2} \, {\left (8 \, x e^{\left (2 \, x\right )} - x + 1\right )} e^{\left (-2 \, x\right )} + 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-4*x+2)*exp(x)^2-2*x^2+3*x)/exp(x)^2/exp(1/4*(4*x*exp(x)^2-x+1)/exp(x)^2)^2,x, algorithm="fric
as")

[Out]

x*e^(-1/2*(8*x*e^(2*x) - x + 1)*e^(-2*x) + 2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{2} \, {\left (2 \, x^{2} + 2 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 3 \, x\right )} e^{\left (-\frac {1}{2} \, {\left (4 \, x e^{\left (2 \, x\right )} - x + 1\right )} e^{\left (-2 \, x\right )} - 2 \, x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-1/2*(2*x^2 + 2*(2*x - 1)*e^(2*x) - 3*x)*e^(-1/2*(4*x*e^(2*x) - x + 1)*e^(-2*x) - 2*x), x)

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maple [A]  time = 0.06, size = 22, normalized size = 1.00

method	result	size
risch	\(x \,{\mathrm e}^{-\frac {\left (4 x \,{\mathrm e}^{2 x}-x +1\right ) {\mathrm e}^{-2 x}}{2}}\)	\(22\)
norman	\(x \,{\mathrm e}^{-\frac {\left (4 x \,{\mathrm e}^{2 x}-x +1\right ) {\mathrm e}^{-2 x}}{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-1/2*(4*x*exp(2*x)-x+1)*exp(-2*x))

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maxima [A]  time = 0.45, size = 20, normalized size = 0.91 \begin {gather*} x e^{\left (\frac {1}{2} \, x e^{\left (-2 \, x\right )} - 2 \, x - \frac {1}{2} \, e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-4*x+2)*exp(x)^2-2*x^2+3*x)/exp(x)^2/exp(1/4*(4*x*exp(x)^2-x+1)/exp(x)^2)^2,x, algorithm="maxi
ma")

[Out]

x*e^(1/2*x*e^(-2*x) - 2*x - 1/2*e^(-2*x))

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mupad [B]  time = 4.18, size = 21, normalized size = 0.95 \begin {gather*} x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-2\,x}}{2}}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2\,x}}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-exp(-2*x)/2)*exp(-2*x)*exp((x*exp(-2*x))/2)

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sympy [A]  time = 0.21, size = 22, normalized size = 1.00 \begin {gather*} x e^{- 2 \left (x e^{2 x} - \frac {x}{4} + \frac {1}{4}\right ) e^{- 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-2*(x*exp(2*x) - x/4 + 1/4)*exp(-2*x))

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