\(\int \frac {1}{135} e^{\frac {1}{27} (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4)} (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)) \, dx\) [6525]

Optimal result

Integrand size = 67, antiderivative size = 29 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{-2+e^{e^{1+\frac {x}{5}} x}+3 x \left (4+\frac {x^3}{81}\right )} \]

[Out]

exp(3*(4+1/81*x^3)*x-2+exp(x*exp(1+1/5*x)))

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12, 6838} \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{\frac {1}{27} \left (x^4+324 x+27 e^{e^{\frac {x+5}{5}} x}-54\right )} \]

[In]

Int[(E^((-54 + 27*E^(E^((5 + x)/5)*x) + 324*x + x^4)/27)*(1620 + 20*x^3 + E^(E^((5 + x)/5)*x + (5 + x)/5)*(135
+ 27*x)))/135,x]

[Out]

E^((-54 + 27*E^(E^((5 + x)/5)*x) + 324*x + x^4)/27)

Rule 12

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{135} \int e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx \\ & = e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{-2+e^{e^{1+\frac {x}{5}} x}+12 x+\frac {x^4}{27}} \]

[In]

Integrate[(E^((-54 + 27*E^(E^((5 + x)/5)*x) + 324*x + x^4)/27)*(1620 + 20*x^3 + E^(E^((5 + x)/5)*x + (5 + x)/5
)*(135 + 27*x)))/135,x]

[Out]

E^(-2 + E^(E^(1 + x/5)*x) + 12*x + x^4/27)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{1+\frac {x}{5}}}+\frac {x^{4}}{27}+12 x -2}\)	\(21\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{1+\frac {x}{5}}}+\frac {x^{4}}{27}+12 x -2}\)	\(21\)

[In]

int(1/135*((27*x+135)*exp(1+1/5*x)*exp(x*exp(1+1/5*x))+20*x^3+1620)*exp(exp(x*exp(1+1/5*x))+1/27*x^4+12*x-2),x
,method=_RETURNVERBOSE)

[Out]

exp(exp(x*exp(1+1/5*x))+1/27*x^4+12*x-2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{\left (\frac {1}{27} \, {\left ({\left (x^{4} + 324 \, x - 54\right )} e^{\left (\frac {1}{5} \, x + 1\right )} + 27 \, e^{\left (x e^{\left (\frac {1}{5} \, x + 1\right )} + \frac {1}{5} \, x + 1\right )}\right )} e^{\left (-\frac {1}{5} \, x - 1\right )}\right )} \]

[In]

integrate(1/135*((27*x+135)*exp(1+1/5*x)*exp(x*exp(1+1/5*x))+20*x^3+1620)*exp(exp(x*exp(1+1/5*x))+1/27*x^4+12*
x-2),x, algorithm="fricas")

[Out]

e^(1/27*((x^4 + 324*x - 54)*e^(1/5*x + 1) + 27*e^(x*e^(1/5*x + 1) + 1/5*x + 1))*e^(-1/5*x - 1))

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{\frac {x^{4}}{27} + 12 x + e^{x e^{\frac {x}{5} + 1}} - 2} \]

[In]

integrate(1/135*((27*x+135)*exp(1+1/5*x)*exp(x*exp(1+1/5*x))+20*x**3+1620)*exp(exp(x*exp(1+1/5*x))+1/27*x**4+1
2*x-2),x)

[Out]

exp(x**4/27 + 12*x + exp(x*exp(x/5 + 1)) - 2)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=e^{\left (\frac {1}{27} \, x^{4} + 12 \, x + e^{\left (x e^{\left (\frac {1}{5} \, x + 1\right )}\right )} - 2\right )} \]

[In]

integrate(1/135*((27*x+135)*exp(1+1/5*x)*exp(x*exp(1+1/5*x))+20*x^3+1620)*exp(exp(x*exp(1+1/5*x))+1/27*x^4+12*
x-2),x, algorithm="maxima")

[Out]

e^(1/27*x^4 + 12*x + e^(x*e^(1/5*x + 1)) - 2)

Giac [F]

\[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx=\int { \frac {1}{135} \, {\left (20 \, x^{3} + 27 \, {\left (x + 5\right )} e^{\left (x e^{\left (\frac {1}{5} \, x + 1\right )} + \frac {1}{5} \, x + 1\right )} + 1620\right )} e^{\left (\frac {1}{27} \, x^{4} + 12 \, x + e^{\left (x e^{\left (\frac {1}{5} \, x + 1\right )}\right )} - 2\right )} \,d x } \]

[In]

integrate(1/135*((27*x+135)*exp(1+1/5*x)*exp(x*exp(1+1/5*x))+20*x^3+1620)*exp(exp(x*exp(1+1/5*x))+1/27*x^4+12*
x-2),x, algorithm="giac")

[Out]

integrate(1/135*(20*x^3 + 27*(x + 5)*e^(x*e^(1/5*x + 1) + 1/5*x + 1) + 1620)*e^(1/27*x^4 + 12*x + e^(x*e^(1/5*
x + 1)) - 2), x)

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {1}{135} e^{\frac {1}{27} \left (-54+27 e^{e^{\frac {5+x}{5}} x}+324 x+x^4\right )} \left (1620+20 x^3+e^{e^{\frac {5+x}{5}} x+\frac {5+x}{5}} (135+27 x)\right ) \, dx={\mathrm {e}}^{12\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{x/5}\,\mathrm {e}}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{\frac {x^4}{27}} \]

[In]

int((exp(12*x + exp(x*exp(x/5 + 1)) + x^4/27 - 2)*(20*x^3 + exp(x/5 + 1)*exp(x*exp(x/5 + 1))*(27*x + 135) + 16
20))/135,x)

[Out]

exp(12*x)*exp(exp(x*exp(x/5)*exp(1)))*exp(-2)*exp(x^4/27)