\(\int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} (91125 e^x+6750 e^{2 x}+125 e^{3 x})}{5832+432 e^x+8 e^{2 x}} \, dx\) [8668]

Optimal result

Integrand size = 69, antiderivative size = 26 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=\frac {25}{16} e^{2 e^{5 e^x}}-\frac {1}{27+e^x}+x \]

[Out]

x-1/(27+exp(x))+25/16*exp(exp(5*exp(x)))^2

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.072, Rules used = {2320, 12, 6874, 2225, 907} \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=x+\frac {25}{16} e^{2 e^{5 e^x}}-\frac {1}{e^x+27} \]

[In]

Int[(5832 + 440*E^x + 8*E^(2*x) + E^(2*E^(5*E^x) + 5*E^x)*(91125*E^x + 6750*E^(2*x) + 125*E^(3*x)))/(5832 + 43
2*E^x + 8*E^(2*x)),x]

[Out]

(25*E^(2*E^(5*E^x)))/16 - (27 + E^x)^(-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 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 6874

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

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {5832+440 x+8 x^2+125 e^{2 e^{5 x}+5 x} x (27+x)^2}{8 x (27+x)^2} \, dx,x,e^x\right ) \\ & = \frac {1}{8} \text {Subst}\left (\int \frac {5832+440 x+8 x^2+125 e^{2 e^{5 x}+5 x} x (27+x)^2}{x (27+x)^2} \, dx,x,e^x\right ) \\ & = \frac {1}{8} \text {Subst}\left (\int \left (125 e^{2 e^{5 x}+5 x}+\frac {8 \left (729+55 x+x^2\right )}{x (27+x)^2}\right ) \, dx,x,e^x\right ) \\ & = \frac {125}{8} \text {Subst}\left (\int e^{2 e^{5 x}+5 x} \, dx,x,e^x\right )+\text {Subst}\left (\int \frac {729+55 x+x^2}{x (27+x)^2} \, dx,x,e^x\right ) \\ & = \frac {25}{8} \text {Subst}\left (\int e^{2 x} \, dx,x,e^{5 e^x}\right )+\text {Subst}\left (\int \left (\frac {1}{x}+\frac {1}{(27+x)^2}\right ) \, dx,x,e^x\right ) \\ & = \frac {25}{16} e^{2 e^{5 e^x}}-\frac {1}{27+e^x}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=\frac {1}{8} \left (\frac {25}{2} e^{2 e^{5 e^x}}-\frac {8}{27+e^x}+8 x\right ) \]

[In]

Integrate[(5832 + 440*E^x + 8*E^(2*x) + E^(2*E^(5*E^x) + 5*E^x)*(91125*E^x + 6750*E^(2*x) + 125*E^(3*x)))/(583
2 + 432*E^x + 8*E^(2*x)),x]

[Out]

((25*E^(2*E^(5*E^x)))/2 - 8/(27 + E^x) + 8*x)/8

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
risch	\(x -\frac {1}{27+{\mathrm e}^{x}}+\frac {25 \,{\mathrm e}^{2 \,{\mathrm e}^{5 \,{\mathrm e}^{x}}}}{16}\)	\(21\)
parallelrisch	\(\frac {25 \,{\mathrm e}^{2 \,{\mathrm e}^{5 \,{\mathrm e}^{x}}} {\mathrm e}^{x}+16 \,{\mathrm e}^{x} x +675 \,{\mathrm e}^{2 \,{\mathrm e}^{5 \,{\mathrm e}^{x}}}+432 x -16}{432+16 \,{\mathrm e}^{x}}\)	\(41\)

[In]

int(((125*exp(x)^3+6750*exp(x)^2+91125*exp(x))*exp(5*exp(x))*exp(exp(5*exp(x)))^2+8*exp(x)^2+440*exp(x)+5832)/
(8*exp(x)^2+432*exp(x)+5832),x,method=_RETURNVERBOSE)

[Out]

x-1/(27+exp(x))+25/16*exp(2*exp(5*exp(x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (20) = 40\).

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=\frac {{\left (16 \, {\left (x e^{x} + 27 \, x - 1\right )} e^{\left (5 \, e^{x}\right )} + 25 \, {\left (e^{x} + 27\right )} e^{\left (5 \, e^{x} + 2 \, e^{\left (5 \, e^{x}\right )}\right )}\right )} e^{\left (-5 \, e^{x}\right )}}{16 \, {\left (e^{x} + 27\right )}} \]

[In]

integrate(((125*exp(x)^3+6750*exp(x)^2+91125*exp(x))*exp(5*exp(x))*exp(exp(5*exp(x)))^2+8*exp(x)^2+440*exp(x)+
5832)/(8*exp(x)^2+432*exp(x)+5832),x, algorithm="fricas")

[Out]

1/16*(16*(x*e^x + 27*x - 1)*e^(5*e^x) + 25*(e^x + 27)*e^(5*e^x + 2*e^(5*e^x)))*e^(-5*e^x)/(e^x + 27)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=x + \frac {25 e^{2 e^{5 e^{x}}}}{16} - \frac {1}{e^{x} + 27} \]

[In]

integrate(((125*exp(x)**3+6750*exp(x)**2+91125*exp(x))*exp(5*exp(x))*exp(exp(5*exp(x)))**2+8*exp(x)**2+440*exp
(x)+5832)/(8*exp(x)**2+432*exp(x)+5832),x)

[Out]

x + 25*exp(2*exp(5*exp(x)))/16 - 1/(exp(x) + 27)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=x - \frac {1}{e^{x} + 27} + \frac {25}{16} \, e^{\left (2 \, e^{\left (5 \, e^{x}\right )}\right )} \]

[In]

integrate(((125*exp(x)^3+6750*exp(x)^2+91125*exp(x))*exp(5*exp(x))*exp(exp(5*exp(x)))^2+8*exp(x)^2+440*exp(x)+
5832)/(8*exp(x)^2+432*exp(x)+5832),x, algorithm="maxima")

[Out]

x - 1/(e^x + 27) + 25/16*e^(2*e^(5*e^x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.92 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=\frac {16 \, x e^{\left (x + 5 \, e^{x}\right )} + 432 \, x e^{\left (5 \, e^{x}\right )} + 25 \, e^{\left (x + 5 \, e^{x} + 2 \, e^{\left (5 \, e^{x}\right )}\right )} - 16 \, e^{\left (5 \, e^{x}\right )} + 675 \, e^{\left (5 \, e^{x} + 2 \, e^{\left (5 \, e^{x}\right )}\right )}}{16 \, {\left (e^{\left (x + 5 \, e^{x}\right )} + 27 \, e^{\left (5 \, e^{x}\right )}\right )}} \]

[In]

integrate(((125*exp(x)^3+6750*exp(x)^2+91125*exp(x))*exp(5*exp(x))*exp(exp(5*exp(x)))^2+8*exp(x)^2+440*exp(x)+
5832)/(8*exp(x)^2+432*exp(x)+5832),x, algorithm="giac")

[Out]

1/16*(16*x*e^(x + 5*e^x) + 432*x*e^(5*e^x) + 25*e^(x + 5*e^x + 2*e^(5*e^x)) - 16*e^(5*e^x) + 675*e^(5*e^x + 2*
e^(5*e^x)))/(e^(x + 5*e^x) + 27*e^(5*e^x))

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {5832+440 e^x+8 e^{2 x}+e^{2 e^{5 e^x}+5 e^x} \left (91125 e^x+6750 e^{2 x}+125 e^{3 x}\right )}{5832+432 e^x+8 e^{2 x}} \, dx=x+\frac {25\,{\mathrm {e}}^{2\,{\mathrm {e}}^{5\,{\mathrm {e}}^x}}}{16}-\frac {1}{{\mathrm {e}}^x+27} \]

[In]

int((8*exp(2*x) + 440*exp(x) + exp(2*exp(5*exp(x)))*exp(5*exp(x))*(6750*exp(2*x) + 125*exp(3*x) + 91125*exp(x)
) + 5832)/(8*exp(2*x) + 432*exp(x) + 5832),x)

[Out]

x + (25*exp(2*exp(5*exp(x))))/16 - 1/(exp(x) + 27)