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

Optimal. Leaf size=28 \[ 3+e^{-3+3 \left (1+e^{\frac {x^2}{x+x^2}}\right )-x} x \]

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Rubi [F]  time = 3.30, 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 {e^{3 e^{\frac {x}{1+x}}-x} \left (1+x+3 e^{\frac {x}{1+x}} x-x^2-x^3\right )}{1+2 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.64, size = 21, normalized size = 0.75 \begin {gather*} e^{3 e^{1-\frac {1}{1+x}}-x} x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(3*E^(1 - (1 + x)^(-1)) - x)*x

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fricas [A]  time = 0.79, size = 17, normalized size = 0.61 \begin {gather*} x e^{\left (-x + 3 \, e^{\left (\frac {x}{x + 1}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-x + 3*e^(x/(x + 1)))

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giac [A]  time = 0.21, size = 44, normalized size = 1.57 \begin {gather*} x e^{\left (-\frac {x^{2} - 3 \, x e^{\left (\frac {x}{x + 1}\right )} - 3 \, e^{\left (\frac {x}{x + 1}\right )}}{x + 1} - \frac {x}{x + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.32, size = 18, normalized size = 0.64

method	result	size
risch	\(x \,{\mathrm e}^{3 \,{\mathrm e}^{\frac {x}{x +1}}-x}\)	\(18\)
norman	\(\frac {\left (x^{2}+x \right ) {\mathrm e}^{3 \,{\mathrm e}^{\frac {x}{x +1}}-x}}{x +1}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 19, normalized size = 0.68 \begin {gather*} x e^{\left (-x + 3 \, e^{\left (-\frac {1}{x + 1} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-x + 3*e^(-1/(x + 1) + 1))

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mupad [B]  time = 1.26, size = 17, normalized size = 0.61 \begin {gather*} x\,{\mathrm {e}}^{3\,{\mathrm {e}}^{\frac {x}{x+1}}}\,{\mathrm {e}}^{-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(3*exp(x/(x + 1)))*exp(-x)

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sympy [A]  time = 6.85, size = 12, normalized size = 0.43 \begin {gather*} x e^{- x + 3 e^{\frac {x}{x + 1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-x + 3*exp(x/(x + 1)))

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