\(\int e^{e^{2016 x}+6048 x} \, dx\) [153]

Optimal result

Integrand size = 11, antiderivative size = 42 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {e^{e^{2016 x}}}{1008}-\frac {e^{e^{2016 x}+2016 x}}{1008}+\frac {e^{e^{2016 x}+4032 x}}{2016} \]

[Out]

1/1008*exp(exp(1)^(2016*x))-1/1008*exp(exp(1)^(2016*x)+2016*x)+1/2016*exp(exp(1)^(2016*x)+4032*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2320, 2207, 2225} \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {e^{e^{2016 x}}}{1008}-\frac {e^{2016 x+e^{2016 x}}}{1008}+\frac {e^{4032 x+e^{2016 x}}}{2016} \]

[In]

Int[E^(E^(2016*x) + 6048*x),x]

[Out]

E^E^(2016*x)/1008 - E^(E^(2016*x) + 2016*x)/1008 + E^(E^(2016*x) + 4032*x)/2016

Rule 2207

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] &&  !TrueQ[$UseGamma]

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]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^x x^2 \, dx,x,e^{2016 x}\right )}{2016} \\ & = \frac {e^{e^{2016 x}+4032 x}}{2016}-\frac {\text {Subst}\left (\int e^x x \, dx,x,e^{2016 x}\right )}{1008} \\ & = -\frac {e^{e^{2016 x}+2016 x}}{1008}+\frac {e^{e^{2016 x}+4032 x}}{2016}+\frac {\text {Subst}\left (\int e^x \, dx,x,e^{2016 x}\right )}{1008} \\ & = \frac {e^{e^{2016 x}}}{1008}-\frac {e^{e^{2016 x}+2016 x}}{1008}+\frac {e^{e^{2016 x}+4032 x}}{2016} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.60 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {e^{e^{2016 x}} \left (2-2 e^{2016 x}+e^{4032 x}\right )}{2016} \]

[In]

Integrate[E^(E^(2016*x) + 6048*x),x]

[Out]

(E^E^(2016*x)*(2 - 2*E^(2016*x) + E^(4032*x)))/2016

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.48

method	result	size
risch	\(\frac {\left ({\mathrm e}^{4032 x}-2 \,{\mathrm e}^{2016 x}+2\right ) {\mathrm e}^{{\mathrm e}^{2016 x}}}{2016}\)	\(20\)

[In]

int(exp(exp(2016*x)+6048*x),x,method=_RETURNVERBOSE)

[Out]

1/2016*(exp(4032*x)-2*exp(2016*x)+2)*exp(exp(2016*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.45 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {1}{2016} \, {\left (e^{\left (4032 \, x\right )} - 2 \, e^{\left (2016 \, x\right )} + 2\right )} e^{\left (e^{\left (2016 \, x\right )}\right )} \]

[In]

integrate(exp(exp(2016*x)+6048*x),x, algorithm="fricas")

[Out]

1/2016*(e^(4032*x) - 2*e^(2016*x) + 2)*e^(e^(2016*x))

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {e^{4032 x} e^{e^{2016 x}}}{2016} - \frac {e^{2016 x} e^{e^{2016 x}}}{1008} + \frac {e^{e^{2016 x}}}{1008} \]

[In]

integrate(exp(exp(2016*x)+6048*x),x)

[Out]

exp(4032*x)*exp(exp(2016*x))/2016 - exp(2016*x)*exp(exp(2016*x))/1008 + exp(exp(2016*x))/1008

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.45 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {1}{2016} \, {\left (e^{\left (4032 \, x\right )} - 2 \, e^{\left (2016 \, x\right )} + 2\right )} e^{\left (e^{\left (2016 \, x\right )}\right )} \]

[In]

integrate(exp(exp(2016*x)+6048*x),x, algorithm="maxima")

[Out]

1/2016*(e^(4032*x) - 2*e^(2016*x) + 2)*e^(e^(2016*x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {1}{2016} \, {\left (e^{\left (10080 \, x + e^{\left (2016 \, x\right )}\right )} - 2 \, e^{\left (8064 \, x + e^{\left (2016 \, x\right )}\right )} + 2 \, e^{\left (6048 \, x + e^{\left (2016 \, x\right )}\right )}\right )} e^{\left (-6048 \, x\right )} \]

[In]

integrate(exp(exp(2016*x)+6048*x),x, algorithm="giac")

[Out]

1/2016*(e^(10080*x + e^(2016*x)) - 2*e^(8064*x + e^(2016*x)) + 2*e^(6048*x + e^(2016*x)))*e^(-6048*x)

Mupad [B] (verification not implemented)

Time = 15.59 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int e^{e^{2016 x}+6048 x} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{2016\,x}}}{1008}-\frac {{\mathrm {e}}^{2016\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{2016\,x}}}{1008}+\frac {{\mathrm {e}}^{4032\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{2016\,x}}}{2016} \]

[In]

int(exp(6048*x + exp(2016*x)),x)

[Out]

exp(exp(2016*x))/1008 - (exp(2016*x)*exp(exp(2016*x)))/1008 + (exp(4032*x)*exp(exp(2016*x)))/2016