3.104.12 \(\int \frac {1}{8} (8+e^{-2-e^x+x} (-8+8 e^x)+x) \, dx\)

Optimal. Leaf size=22 \[ -4-e^{-2-e^x+x}+x+\frac {x^2}{16} \]

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Rubi [A]  time = 0.05, antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {12, 2282, 2176, 2194} \begin {gather*} \frac {x^2}{16}+x-e^{-e^x-2}+e^{-e^x-2} \left (1-e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(-2 - E^x) + E^(-2 - E^x)*(1 - E^x) + x + x^2/16

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 21, normalized size = 0.95 \begin {gather*} -e^{-2-e^x+x}+x+\frac {x^2}{16} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(-2 - E^x + x) + x + x^2/16

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fricas [A]  time = 0.97, size = 17, normalized size = 0.77 \begin {gather*} \frac {1}{16} \, x^{2} + x - e^{\left (x - e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^2 + x - e^(x - e^x - 2)

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giac [A]  time = 0.13, size = 17, normalized size = 0.77 \begin {gather*} \frac {1}{16} \, x^{2} + x - e^{\left (x - e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^2 + x - e^(x - e^x - 2)

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maple [A]  time = 0.04, size = 18, normalized size = 0.82

method	result	size
norman	\(x +\frac {x^{2}}{16}-{\mathrm e}^{-{\mathrm e}^{x}+x -2}\)	\(18\)
risch	\(x +\frac {x^{2}}{16}-{\mathrm e}^{-{\mathrm e}^{x}+x -2}\)	\(18\)
default	\(x +\frac {x^{2}}{16}+{\mathrm e}^{-2} \left (-{\mathrm e}^{-{\mathrm e}^{x}} {\mathrm e}^{x}-{\mathrm e}^{-{\mathrm e}^{x}}\right )+{\mathrm e}^{-2} {\mathrm e}^{-{\mathrm e}^{x}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 17, normalized size = 0.77 \begin {gather*} \frac {1}{16} \, x^{2} + x - e^{\left (x - e^{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*x^2 + x - e^(x - e^x - 2)

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mupad [B]  time = 0.08, size = 18, normalized size = 0.82 \begin {gather*} x+\frac {x^2}{16}-{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/8 + (exp(x - exp(x) - 2)*(8*exp(x) - 8))/8 + 1,x)

[Out]

x + x^2/16 - exp(-2)*exp(-exp(x))*exp(x)

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sympy [A]  time = 0.14, size = 14, normalized size = 0.64 \begin {gather*} \frac {x^{2}}{16} + x - e^{x - e^{x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/16 + x - exp(x - exp(x) - 2)

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