\(\int \frac {1}{5} e^{e^{-e^4} (1+2 e^{e^4})+x} (10+4 x+2 x^2) \, dx\) [5337]

Optimal result

Integrand size = 35, antiderivative size = 21 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{2+e^{-e^4}+x} \left (5+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 2227, 2225, 2207} \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{x+e^{-e^4}+2} x^2+2 e^{x+e^{-e^4}+2} \]

[In]

Int[(E^((1 + 2*E^E^4)/E^E^4 + x)*(10 + 4*x + 2*x^2))/5,x]

[Out]

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

Rule 12

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

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 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx \\ & = \frac {1}{5} \int \left (10 e^{2+e^{-e^4}+x}+4 e^{2+e^{-e^4}+x} x+2 e^{2+e^{-e^4}+x} x^2\right ) \, dx \\ & = \frac {2}{5} \int e^{2+e^{-e^4}+x} x^2 \, dx+\frac {4}{5} \int e^{2+e^{-e^4}+x} x \, dx+2 \int e^{2+e^{-e^4}+x} \, dx \\ & = 2 e^{2+e^{-e^4}+x}+\frac {4}{5} e^{2+e^{-e^4}+x} x+\frac {2}{5} e^{2+e^{-e^4}+x} x^2-\frac {4}{5} \int e^{2+e^{-e^4}+x} \, dx-\frac {4}{5} \int e^{2+e^{-e^4}+x} x \, dx \\ & = \frac {6}{5} e^{2+e^{-e^4}+x}+\frac {2}{5} e^{2+e^{-e^4}+x} x^2+\frac {4}{5} \int e^{2+e^{-e^4}+x} \, dx \\ & = 2 e^{2+e^{-e^4}+x}+\frac {2}{5} e^{2+e^{-e^4}+x} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} e^{2+e^{-e^4}+x} \left (5+x^2\right ) \]

[In]

Integrate[(E^((1 + 2*E^E^4)/E^E^4 + x)*(10 + 4*x + 2*x^2))/5,x]

[Out]

(2*E^(2 + E^(-E^4) + x)*(5 + x^2))/5

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\left (2 x^{2}+10\right ) {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}+2+x}}{5}\)	\(19\)
gosper	\(\frac {2 \,{\mathrm e}^{x +\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (x^{2}+5\right )}{5}\)	\(24\)
default	\(\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left ({\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{x}\right )}{5}\)	\(28\)
norman	\(2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}} {\mathrm e}^{x}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{-{\mathrm e}^{4}}} x^{2} {\mathrm e}^{x}}{5}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (2 \,{\mathrm e}^{x} x^{2}+10 \,{\mathrm e}^{x}\right )}{5}\)	\(29\)
parts	\(\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} {\mathrm e}^{x} x^{2}}{5}+2 \,{\mathrm e}^{x} {\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}}\)	\(41\)
meijerg	\(-2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (1-{\mathrm e}^{x}\right )-\frac {2 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (2-\frac {\left (3 x^{2}-6 x +6\right ) {\mathrm e}^{x}}{3}\right )}{5}+\frac {4 \,{\mathrm e}^{\left (2 \,{\mathrm e}^{{\mathrm e}^{4}}+1\right ) {\mathrm e}^{-{\mathrm e}^{4}}} \left (1-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}\right )}{5}\)	\(83\)

[In]

int(1/5*(2*x^2+4*x+10)*exp(x)*exp((2*exp(exp(4))+1)/exp(exp(4))),x,method=_RETURNVERBOSE)

[Out]

1/5*(2*x^2+10)*exp(exp(-exp(4))+2+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} + 5\right )} e^{\left ({\left ({\left (x + 2\right )} e^{\left (e^{4}\right )} + 1\right )} e^{\left (-e^{4}\right )}\right )} \]

[In]

integrate(1/5*(2*x^2+4*x+10)*exp(x)*exp((2*exp(exp(4))+1)/exp(exp(4))),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {\left (2 x^{2} e^{2} e^{e^{- e^{4}}} + 10 e^{2} e^{e^{- e^{4}}}\right ) e^{x}}{5} \]

[In]

integrate(1/5*(2*x**2+4*x+10)*exp(x)*exp((2*exp(exp(4))+1)/exp(exp(4))),x)

[Out]

(2*x**2*exp(2)*exp(exp(-exp(4))) + 10*exp(2)*exp(exp(-exp(4))))*exp(x)/5

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (16) = 32\).

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.57 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} e^{\left (e^{\left (-e^{4}\right )} + 2\right )} - 2 \, x e^{\left (e^{\left (-e^{4}\right )} + 2\right )} + 2 \, e^{\left (e^{\left (-e^{4}\right )} + 2\right )}\right )} e^{x} + \frac {4}{5} \, {\left (x e^{\left (e^{\left (-e^{4}\right )} + 2\right )} - e^{\left (e^{\left (-e^{4}\right )} + 2\right )}\right )} e^{x} + 2 \, e^{\left (x + e^{\left (-e^{4}\right )} + 2\right )} \]

[In]

integrate(1/5*(2*x^2+4*x+10)*exp(x)*exp((2*exp(exp(4))+1)/exp(exp(4))),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2}{5} \, {\left (x^{2} + 5\right )} e^{\left (x + e^{\left (-e^{4}\right )} + 2\right )} \]

[In]

integrate(1/5*(2*x^2+4*x+10)*exp(x)*exp((2*exp(exp(4))+1)/exp(exp(4))),x, algorithm="giac")

[Out]

2/5*(x^2 + 5)*e^(x + e^(-e^4) + 2)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} e^{e^{-e^4} \left (1+2 e^{e^4}\right )+x} \left (10+4 x+2 x^2\right ) \, dx=\frac {2\,{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^4}}\,{\mathrm {e}}^x\,\left (x^2+5\right )}{5} \]

[In]

int((exp(exp(-exp(4))*(2*exp(exp(4)) + 1))*exp(x)*(4*x + 2*x^2 + 10))/5,x)

[Out]

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