\(\int e^{-e^x} (5 e^x+e^{e^x} (8 x^3+10 x^4)) \, dx\) [7927]

Optimal result

Integrand size = 31, antiderivative size = 18 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=-5 e^{-e^x}+2 x^4 (1+x) \]

[Out]

2*x^4*(1+x)-5/exp(exp(x))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6820, 2320, 2225, 45} \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=2 x^5+2 x^4-5 e^{-e^x} \]

[In]

Int[(5*E^x + E^E^x*(8*x^3 + 10*x^4))/E^E^x,x]

[Out]

-5/E^E^x + 2*x^4 + 2*x^5

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2225

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 2320

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]]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (5 e^{-e^x+x}+2 x^3 (4+5 x)\right ) \, dx \\ & = 2 \int x^3 (4+5 x) \, dx+5 \int e^{-e^x+x} \, dx \\ & = 2 \int \left (4 x^3+5 x^4\right ) \, dx+5 \text {Subst}\left (\int e^{-x} \, dx,x,e^x\right ) \\ & = -5 e^{-e^x}+2 x^4+2 x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=-5 e^{-e^x}+2 x^4+2 x^5 \]

[In]

Integrate[(5*E^x + E^E^x*(8*x^3 + 10*x^4))/E^E^x,x]

[Out]

-5/E^E^x + 2*x^4 + 2*x^5

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
default	\(2 x^{5}+2 x^{4}-5 \,{\mathrm e}^{-{\mathrm e}^{x}}\)	\(19\)
risch	\(2 x^{5}+2 x^{4}-5 \,{\mathrm e}^{-{\mathrm e}^{x}}\)	\(19\)
parts	\(2 x^{5}+2 x^{4}-5 \,{\mathrm e}^{-{\mathrm e}^{x}}\)	\(19\)
norman	\(\left (-5+2 x^{4} {\mathrm e}^{{\mathrm e}^{x}}+2 x^{5} {\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-{\mathrm e}^{x}}\)	\(25\)
parallelrisch	\(\left (-5+2 x^{4} {\mathrm e}^{{\mathrm e}^{x}}+2 x^{5} {\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-{\mathrm e}^{x}}\)	\(25\)

[In]

int(((10*x^4+8*x^3)*exp(exp(x))+5*exp(x))/exp(exp(x)),x,method=_RETURNVERBOSE)

[Out]

2*x^5+2*x^4-5/exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx={\left (2 \, {\left (x^{5} + x^{4}\right )} e^{\left (e^{x}\right )} - 5\right )} e^{\left (-e^{x}\right )} \]

[In]

integrate(((10*x^4+8*x^3)*exp(exp(x))+5*exp(x))/exp(exp(x)),x, algorithm="fricas")

[Out]

(2*(x^5 + x^4)*e^(e^x) - 5)*e^(-e^x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=2 x^{5} + 2 x^{4} - 5 e^{- e^{x}} \]

[In]

integrate(((10*x**4+8*x**3)*exp(exp(x))+5*exp(x))/exp(exp(x)),x)

[Out]

2*x**5 + 2*x**4 - 5*exp(-exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=2 \, x^{5} + 2 \, x^{4} - 5 \, e^{\left (-e^{x}\right )} \]

[In]

integrate(((10*x^4+8*x^3)*exp(exp(x))+5*exp(x))/exp(exp(x)),x, algorithm="maxima")

[Out]

2*x^5 + 2*x^4 - 5*e^(-e^x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=2 \, x^{5} + 2 \, x^{4} - 5 \, e^{\left (-e^{x}\right )} \]

[In]

integrate(((10*x^4+8*x^3)*exp(exp(x))+5*exp(x))/exp(exp(x)),x, algorithm="giac")

[Out]

2*x^5 + 2*x^4 - 5*e^(-e^x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-e^x} \left (5 e^x+e^{e^x} \left (8 x^3+10 x^4\right )\right ) \, dx=2\,x^4-5\,{\mathrm {e}}^{-{\mathrm {e}}^x}+2\,x^5 \]

[In]

int(exp(-exp(x))*(5*exp(x) + exp(exp(x))*(8*x^3 + 10*x^4)),x)

[Out]

2*x^4 - 5*exp(-exp(x)) + 2*x^5