3.45.75 \(\int \frac {1}{10} e^{-e^{1-x}} (1+e^{1-x} (1+x)) \, dx\)

Optimal. Leaf size=29 \[ \frac {-x+5 e^{-e^{1-x}} \left (x+x^2\right )}{50 x} \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 2288} \begin {gather*} \frac {1}{10} e^{-e^{1-x}} (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^(1 - x)*(1 + x))/(10*E^E^(1 - x)),x]

[Out]

(1 + x)/(10*E^E^(1 - 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} &=\frac {1}{10} \int e^{-e^{1-x}} \left (1+e^{1-x} (1+x)\right ) \, dx\\ &=\frac {1}{10} e^{-e^{1-x}} (1+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 18, normalized size = 0.62 \begin {gather*} \frac {1}{10} e^{-e^{1-x}} (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(1 - x)*(1 + x))/(10*E^E^(1 - x)),x]

[Out]

(1 + x)/(10*E^E^(1 - x))

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fricas [A]  time = 0.51, size = 14, normalized size = 0.48 \begin {gather*} \frac {1}{10} \, {\left (x + 1\right )} e^{\left (-e^{\left (-x + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x+1)*exp(-x+1)+1)/exp(exp(-x+1)),x, algorithm="fricas")

[Out]

1/10*(x + 1)*e^(-e^(-x + 1))

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giac [A]  time = 0.24, size = 37, normalized size = 1.28 \begin {gather*} \frac {1}{10} \, {\left (x e^{\left (-x - e^{\left (-x + 1\right )} + 1\right )} + e^{\left (-x - e^{\left (-x + 1\right )} + 1\right )}\right )} e^{\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x+1)*exp(-x+1)+1)/exp(exp(-x+1)),x, algorithm="giac")

[Out]

1/10*(x*e^(-x - e^(-x + 1) + 1) + e^(-x - e^(-x + 1) + 1))*e^(x - 1)

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maple [A]  time = 0.05, size = 15, normalized size = 0.52

method	result	size
risch	\(\frac {\left (x +1\right ) {\mathrm e}^{-{\mathrm e}^{1-x}}}{10}\)	\(15\)
norman	\(\left (\frac {x}{10}+\frac {1}{10}\right ) {\mathrm e}^{-{\mathrm e}^{1-x}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((x+1)*exp(1-x)+1)/exp(exp(1-x)),x,method=_RETURNVERBOSE)

[Out]

1/10*(x+1)*exp(-exp(1-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{10} \, e^{\left (-e^{\left (-x + 1\right )}\right )} + \frac {1}{10} \, \int {\left (x e + e^{x}\right )} e^{\left (-x - e^{\left (-x + 1\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x+1)*exp(-x+1)+1)/exp(exp(-x+1)),x, algorithm="maxima")

[Out]

1/10*e^(-e^(-x + 1)) + 1/10*integrate((x*e + e^x)*e^(-x - e^(-x + 1)), x)

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mupad [B]  time = 0.06, size = 14, normalized size = 0.48 \begin {gather*} \frac {{\mathrm {e}}^{-{\mathrm {e}}^{-x}\,\mathrm {e}}\,\left (x+1\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-exp(1 - x))*((exp(1 - x)*(x + 1))/10 + 1/10),x)

[Out]

(exp(-exp(-x)*exp(1))*(x + 1))/10

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sympy [A]  time = 0.15, size = 10, normalized size = 0.34 \begin {gather*} \frac {\left (x + 1\right ) e^{- e^{1 - x}}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((x+1)*exp(-x+1)+1)/exp(exp(-x+1)),x)

[Out]

(x + 1)*exp(-exp(1 - x))/10

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