\(\int \frac {1}{8} (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} (4+e^{5+x} (48+4 x))) \, dx\) [6867]

Optimal result

Integrand size = 42, antiderivative size = 18 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\left (-3-e^{e^{5+x}}-\frac {x}{4}\right )^2 \]

[Out]

(-1/4*x-exp(exp(5+x))-3)^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2320, 2225, 2326} \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {x^2}{16}+\frac {3 x}{2}+e^{2 e^{x+5}}+\frac {1}{2} e^{e^{x+5}} (x+12) \]

[In]

Int[(12 + 16*E^(5 + 2*E^(5 + x) + x) + x + E^E^(5 + x)*(4 + E^(5 + x)*(48 + 4*x)))/8,x]

[Out]

E^(2*E^(5 + x)) + (3*x)/2 + x^2/16 + (E^E^(5 + x)*(12 + x))/2

Rule 12

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

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx \\ & = \frac {3 x}{2}+\frac {x^2}{16}+\frac {1}{8} \int e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right ) \, dx+2 \int e^{5+2 e^{5+x}+x} \, dx \\ & = \frac {3 x}{2}+\frac {x^2}{16}+\frac {1}{2} e^{e^{5+x}} (12+x)+2 \text {Subst}\left (\int e^{5+2 e^5 x} \, dx,x,e^x\right ) \\ & = e^{2 e^{5+x}}+\frac {3 x}{2}+\frac {x^2}{16}+\frac {1}{2} e^{e^{5+x}} (12+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {1}{16} \left (12+4 e^{e^{5+x}}+x\right )^2 \]

[In]

Integrate[(12 + 16*E^(5 + 2*E^(5 + x) + x) + x + E^E^(5 + x)*(4 + E^(5 + x)*(48 + 4*x)))/8,x]

[Out]

(12 + 4*E^E^(5 + x) + x)^2/16

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61

method	result	size
risch	\({\mathrm e}^{2 \,{\mathrm e}^{5+x}}+\frac {\left (4 x +48\right ) {\mathrm e}^{{\mathrm e}^{5+x}}}{8}+\frac {x^{2}}{16}+\frac {3 x}{2}\)	\(29\)
default	\({\mathrm e}^{2 \,{\mathrm e}^{5+x}}+\frac {x \,{\mathrm e}^{{\mathrm e}^{5+x}}}{2}+6 \,{\mathrm e}^{{\mathrm e}^{5+x}}+\frac {x^{2}}{16}+\frac {3 x}{2}\)	\(32\)
norman	\({\mathrm e}^{2 \,{\mathrm e}^{5+x}}+\frac {x \,{\mathrm e}^{{\mathrm e}^{5+x}}}{2}+6 \,{\mathrm e}^{{\mathrm e}^{5+x}}+\frac {x^{2}}{16}+\frac {3 x}{2}\)	\(32\)
parallelrisch	\({\mathrm e}^{2 \,{\mathrm e}^{5+x}}+\frac {x \,{\mathrm e}^{{\mathrm e}^{5+x}}}{2}+6 \,{\mathrm e}^{{\mathrm e}^{5+x}}+\frac {x^{2}}{16}+\frac {3 x}{2}\)	\(32\)

[In]

int(2*exp(5+x)*exp(exp(5+x))^2+1/8*((4*x+48)*exp(5+x)+4)*exp(exp(5+x))+1/8*x+3/2,x,method=_RETURNVERBOSE)

[Out]

exp(2*exp(5+x))+1/8*(4*x+48)*exp(exp(5+x))+1/16*x^2+3/2*x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {1}{16} \, x^{2} + \frac {1}{2} \, {\left (x + 12\right )} e^{\left (e^{\left (x + 5\right )}\right )} + \frac {3}{2} \, x + e^{\left (2 \, e^{\left (x + 5\right )}\right )} \]

[In]

integrate(2*exp(5+x)*exp(exp(5+x))^2+1/8*((4*x+48)*exp(5+x)+4)*exp(exp(5+x))+1/8*x+3/2,x, algorithm="fricas")

[Out]

1/16*x^2 + 1/2*(x + 12)*e^(e^(x + 5)) + 3/2*x + e^(2*e^(x + 5))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (14) = 28\).

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {x^{2}}{16} + \frac {3 x}{2} + \frac {\left (x + 12\right ) e^{e^{x + 5}}}{2} + e^{2 e^{x + 5}} \]

[In]

integrate(2*exp(5+x)*exp(exp(5+x))**2+1/8*((4*x+48)*exp(5+x)+4)*exp(exp(5+x))+1/8*x+3/2,x)

[Out]

x**2/16 + 3*x/2 + (x + 12)*exp(exp(x + 5))/2 + exp(2*exp(x + 5))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {1}{16} \, x^{2} + \frac {1}{2} \, {\left (x + 12\right )} e^{\left (e^{\left (x + 5\right )}\right )} + \frac {3}{2} \, x + e^{\left (2 \, e^{\left (x + 5\right )}\right )} \]

[In]

integrate(2*exp(5+x)*exp(exp(5+x))^2+1/8*((4*x+48)*exp(5+x)+4)*exp(exp(5+x))+1/8*x+3/2,x, algorithm="maxima")

[Out]

1/16*x^2 + 1/2*(x + 12)*e^(e^(x + 5)) + 3/2*x + e^(2*e^(x + 5))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.50 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {1}{16} \, x^{2} + \frac {1}{2} \, {\left (x e^{\left (x + e^{\left (x + 5\right )} + 5\right )} + 12 \, e^{\left (x + e^{\left (x + 5\right )} + 5\right )}\right )} e^{\left (-x - 5\right )} + \frac {3}{2} \, x + e^{\left (2 \, e^{\left (x + 5\right )}\right )} \]

[In]

integrate(2*exp(5+x)*exp(exp(5+x))^2+1/8*((4*x+48)*exp(5+x)+4)*exp(exp(5+x))+1/8*x+3/2,x, algorithm="giac")

[Out]

1/16*x^2 + 1/2*(x*e^(x + e^(x + 5) + 5) + 12*e^(x + e^(x + 5) + 5))*e^(-x - 5) + 3/2*x + e^(2*e^(x + 5))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {1}{8} \left (12+16 e^{5+2 e^{5+x}+x}+x+e^{e^{5+x}} \left (4+e^{5+x} (48+4 x)\right )\right ) \, dx=\frac {3\,x}{2}+6\,{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}+{\mathrm {e}}^{2\,{\mathrm {e}}^5\,{\mathrm {e}}^x}+\frac {x^2}{16}+\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^5\,{\mathrm {e}}^x}}{2} \]

[In]

int(x/8 + (exp(exp(x + 5))*(exp(x + 5)*(4*x + 48) + 4))/8 + 2*exp(2*exp(x + 5))*exp(x + 5) + 3/2,x)

[Out]

(3*x)/2 + 6*exp(exp(5)*exp(x)) + exp(2*exp(5)*exp(x)) + x^2/16 + (x*exp(exp(5)*exp(x)))/2