3.7.85 \(\int e^{-1-e^x} (e^{1+e^x}-18 e^x+80 e^{2 x}-32 e^{3 x}) \, dx\)

Optimal. Leaf size=22 \[ 2 e^{-1-e^x} \left (-1+4 e^x\right )^2+x \]

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Rubi [A]  time = 0.13, antiderivative size = 39, normalized size of antiderivative = 1.77, number of steps used = 11, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2282, 14, 2196, 2194, 2176} \begin {gather*} x+2 e^{-e^x-1}-16 e^{x-e^x-1}+32 e^{2 x-e^x-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-1 - E^x)*(E^(1 + E^x) - 18*E^x + 80*E^(2*x) - 32*E^(3*x)),x]

[Out]

2*E^(-1 - E^x) - 16*E^(-1 - E^x + x) + 32*E^(-1 - E^x + 2*x) + x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

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} &=\operatorname {Subst}\left (\int \frac {1-2 e^{-1-x} x \left (9-40 x+16 x^2\right )}{x} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{x}-2 e^{-1-x} (-9+4 x) (-1+4 x)\right ) \, dx,x,e^x\right )\\ &=x-2 \operatorname {Subst}\left (\int e^{-1-x} (-9+4 x) (-1+4 x) \, dx,x,e^x\right )\\ &=x-2 \operatorname {Subst}\left (\int \left (9 e^{-1-x}-40 e^{-1-x} x+16 e^{-1-x} x^2\right ) \, dx,x,e^x\right )\\ &=x-18 \operatorname {Subst}\left (\int e^{-1-x} \, dx,x,e^x\right )-32 \operatorname {Subst}\left (\int e^{-1-x} x^2 \, dx,x,e^x\right )+80 \operatorname {Subst}\left (\int e^{-1-x} x \, dx,x,e^x\right )\\ &=18 e^{-1-e^x}-80 e^{-1-e^x+x}+32 e^{-1-e^x+2 x}+x-64 \operatorname {Subst}\left (\int e^{-1-x} x \, dx,x,e^x\right )+80 \operatorname {Subst}\left (\int e^{-1-x} \, dx,x,e^x\right )\\ &=-62 e^{-1-e^x}-16 e^{-1-e^x+x}+32 e^{-1-e^x+2 x}+x-64 \operatorname {Subst}\left (\int e^{-1-x} \, dx,x,e^x\right )\\ &=2 e^{-1-e^x}-16 e^{-1-e^x+x}+32 e^{-1-e^x+2 x}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 1.45 \begin {gather*} e^{-e^x} \left (\frac {2}{e}-16 e^{-1+x}+32 e^{-1+2 x}\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-1 - E^x)*(E^(1 + E^x) - 18*E^x + 80*E^(2*x) - 32*E^(3*x)),x]

[Out]

(2/E - 16*E^(-1 + x) + 32*E^(-1 + 2*x))/E^E^x + x

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fricas [A]  time = 0.59, size = 27, normalized size = 1.23 \begin {gather*} {\left (x e^{\left (e^{x} + 1\right )} + 32 \, e^{\left (2 \, x\right )} - 16 \, e^{x} + 2\right )} e^{\left (-e^{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp(exp(x)-3)-32*exp(x)^3+80*exp(x)^2-18*exp(x))/exp(4)/exp(exp(x)-3),x, algorithm="fricas")

[Out]

(x*e^(e^x + 1) + 32*e^(2*x) - 16*e^x + 2)*e^(-e^x - 1)

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giac [A]  time = 0.59, size = 35, normalized size = 1.59 \begin {gather*} {\left (x e + 32 \, e^{\left (2 \, x - e^{x}\right )} - 16 \, e^{\left (x - e^{x}\right )} + 2 \, e^{\left (-e^{x}\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp(exp(x)-3)-32*exp(x)^3+80*exp(x)^2-18*exp(x))/exp(4)/exp(exp(x)-3),x, algorithm="giac")

[Out]

(x*e + 32*e^(2*x - e^x) - 16*e^(x - e^x) + 2*e^(-e^x))*e^(-1)

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maple [A]  time = 0.10, size = 23, normalized size = 1.05

method	result	size
risch	\(x +\left (32 \,{\mathrm e}^{2 x}-16 \,{\mathrm e}^{x}+2\right ) {\mathrm e}^{-{\mathrm e}^{x}-1}\)	\(23\)
norman	\(\left (x \,{\mathrm e}^{{\mathrm e}^{x}-3}+2 \,{\mathrm e}^{-4}+32 \,{\mathrm e}^{-4} {\mathrm e}^{2 x}-16 \,{\mathrm e}^{x} {\mathrm e}^{-4}\right ) {\mathrm e}^{-{\mathrm e}^{x}+3}\)	\(41\)
derivativedivides	\({\mathrm e}^{-4} \left ({\mathrm e}^{4} \ln \left ({\mathrm e}^{x}\right )+402 \,{\mathrm e}^{-{\mathrm e}^{x}+3}+208 \left ({\mathrm e}^{x}-5\right ) {\mathrm e}^{-{\mathrm e}^{x}+3}+32 \left (\left ({\mathrm e}^{x}-3\right )^{2}-{\mathrm e}^{x}+11\right ) {\mathrm e}^{-{\mathrm e}^{x}+3}\right )\)	\(56\)
default	\({\mathrm e}^{-4} \left ({\mathrm e}^{4} \ln \left ({\mathrm e}^{x}\right )+402 \,{\mathrm e}^{-{\mathrm e}^{x}+3}+208 \left ({\mathrm e}^{x}-5\right ) {\mathrm e}^{-{\mathrm e}^{x}+3}+32 \left (\left ({\mathrm e}^{x}-3\right )^{2}-{\mathrm e}^{x}+11\right ) {\mathrm e}^{-{\mathrm e}^{x}+3}\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(4)*exp(exp(x)-3)-32*exp(x)^3+80*exp(x)^2-18*exp(x))/exp(4)/exp(exp(x)-3),x,method=_RETURNVERBOSE)

[Out]

x+(32*exp(2*x)-16*exp(x)+2)*exp(-exp(x)-1)

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maxima [B]  time = 0.60, size = 43, normalized size = 1.95 \begin {gather*} 32 \, {\left (e^{\left (2 \, x\right )} + 2 \, e^{x} + 2\right )} e^{\left (-e^{x} - 1\right )} - 80 \, {\left (e^{x} + 1\right )} e^{\left (-e^{x} - 1\right )} + x + 18 \, e^{\left (-e^{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp(exp(x)-3)-32*exp(x)^3+80*exp(x)^2-18*exp(x))/exp(4)/exp(exp(x)-3),x, algorithm="maxima")

[Out]

32*(e^(2*x) + 2*e^x + 2)*e^(-e^x - 1) - 80*(e^x + 1)*e^(-e^x - 1) + x + 18*e^(-e^x - 1)

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mupad [B]  time = 0.58, size = 33, normalized size = 1.50 \begin {gather*} x+2\,{\mathrm {e}}^{-{\mathrm {e}}^x-1}-16\,{\mathrm {e}}^{x-{\mathrm {e}}^x-1}+32\,{\mathrm {e}}^{2\,x-{\mathrm {e}}^x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(3 - exp(x))*exp(-4)*(80*exp(2*x) - 32*exp(3*x) - 18*exp(x) + exp(4)*exp(exp(x) - 3)),x)

[Out]

x + 2*exp(- exp(x) - 1) - 16*exp(x - exp(x) - 1) + 32*exp(2*x - exp(x) - 1)

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sympy [A]  time = 0.21, size = 24, normalized size = 1.09 \begin {gather*} x + \frac {\left (32 e^{2 x} - 16 e^{x} + 2\right ) e^{3 - e^{x}}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(4)*exp(exp(x)-3)-32*exp(x)**3+80*exp(x)**2-18*exp(x))/exp(4)/exp(exp(x)-3),x)

[Out]

x + (32*exp(2*x) - 16*exp(x) + 2)*exp(-4)*exp(3 - exp(x))

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