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

Optimal. Leaf size=24 \[ \frac {1}{4} e^{-2-2 e^x} x \left (2-2 e^x x^2\right ) \]

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Rubi [F]  time = 0.53, 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} e^{-2-2 e^x} \left (1+2 e^{2 x} x^3+e^x \left (-2 x-3 x^2-x^3\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 23, normalized size = 0.96 \begin {gather*} \frac {1}{2} e^{-2 \left (1+e^x\right )} \left (x-e^x x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x - E^x*x^3)/(2*E^(2*(1 + E^x)))

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fricas [A]  time = 0.55, size = 19, normalized size = 0.79 \begin {gather*} -\frac {1}{2} \, {\left (x^{3} e^{x} - x\right )} e^{\left (-2 \, e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(x^3*e^x - x)*e^(-2*e^x - 2)

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giac [A]  time = 0.15, size = 32, normalized size = 1.33 \begin {gather*} -\frac {1}{2} \, {\left (x^{3} e^{\left (2 \, x - 2 \, e^{x}\right )} - x e^{\left (x - 2 \, e^{x}\right )}\right )} e^{\left (-x - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 19, normalized size = 0.79

method	result	size
risch	\(\frac {\left (-{\mathrm e}^{x} x^{3}+x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{x}-2}}{2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.16, size = 18, normalized size = 0.75 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-2\,{\mathrm {e}}^x-2}\,\left (x^2\,{\mathrm {e}}^x-1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.20, size = 19, normalized size = 0.79 \begin {gather*} \frac {\left (- x^{3} e^{x} + x\right ) e^{- 2 e^{x} - 2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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