\(\int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx\) [3508]

Optimal result

Integrand size = 30, antiderivative size = 22 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=e^x-\frac {2}{x}+e^{e^{1-\frac {60}{e^4}}} x \]

[Out]

exp(x)-2/x+exp(exp(1-60/exp(4)))*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2225} \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=e^{e^{1-\frac {60}{e^4}}} x+e^x-\frac {2}{x} \]

[In]

Int[(2 + E^E^((-60 + E^4)/E^4)*x^2 + E^x*x^2)/x^2,x]

[Out]

E^x - 2/x + E^E^(1 - 60/E^4)*x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^x+\frac {2+e^{e^{1-\frac {60}{e^4}}} x^2}{x^2}\right ) \, dx \\ & = \int e^x \, dx+\int \frac {2+e^{e^{1-\frac {60}{e^4}}} x^2}{x^2} \, dx \\ & = e^x+\int \left (e^{e^{1-\frac {60}{e^4}}}+\frac {2}{x^2}\right ) \, dx \\ & = e^x-\frac {2}{x}+e^{e^{1-\frac {60}{e^4}}} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=e^x-\frac {2}{x}+e^{e^{1-\frac {60}{e^4}}} x \]

[In]

Integrate[(2 + E^E^((-60 + E^4)/E^4)*x^2 + E^x*x^2)/x^2,x]

[Out]

E^x - 2/x + E^E^(1 - 60/E^4)*x

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(x \,{\mathrm e}^{{\mathrm e}^{1-60 \,{\mathrm e}^{-4}}}-\frac {2}{x}+{\mathrm e}^{x}\)	\(19\)
parts	\(x \,{\mathrm e}^{{\mathrm e}^{\left ({\mathrm e}^{4}-60\right ) {\mathrm e}^{-4}}}-\frac {2}{x}+{\mathrm e}^{x}\)	\(20\)
default	\({\mathrm e}^{x}-\frac {2}{x}+x \,{\mathrm e}^{{\mathrm e} \,{\mathrm e}^{-60 \,{\mathrm e}^{-4}}}\)	\(22\)
norman	\(\frac {-2+x^{2} {\mathrm e}^{{\mathrm e} \,{\mathrm e}^{-60 \,{\mathrm e}^{-4}}}+{\mathrm e}^{x} x}{x}\)	\(26\)
parallelrisch	\(\frac {-2+x^{2} {\mathrm e}^{{\mathrm e}^{\left ({\mathrm e}^{4}-60\right ) {\mathrm e}^{-4}}}+{\mathrm e}^{x} x}{x}\)	\(26\)

[In]

int((x^2*exp(exp((exp(4)-60)/exp(4)))+exp(x)*x^2+2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(exp(1-60*exp(-4)))-2/x+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=\frac {x^{2} e^{\left (e^{\left ({\left (e^{4} - 60\right )} e^{\left (-4\right )}\right )}\right )} + x e^{x} - 2}{x} \]

[In]

integrate((x^2*exp(exp((exp(4)-60)/exp(4)))+exp(x)*x^2+2)/x^2,x, algorithm="fricas")

[Out]

(x^2*e^(e^((e^4 - 60)*e^(-4))) + x*e^x - 2)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=x e^{\frac {e}{e^{\frac {60}{e^{4}}}}} + e^{x} - \frac {2}{x} \]

[In]

integrate((x**2*exp(exp((exp(4)-60)/exp(4)))+exp(x)*x**2+2)/x**2,x)

[Out]

x*exp(E*exp(-60*exp(-4))) + exp(x) - 2/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=x e^{\left (e^{\left (-60 \, e^{\left (-4\right )} + 1\right )}\right )} - \frac {2}{x} + e^{x} \]

[In]

integrate((x^2*exp(exp((exp(4)-60)/exp(4)))+exp(x)*x^2+2)/x^2,x, algorithm="maxima")

[Out]

x*e^(e^(-60*e^(-4) + 1)) - 2/x + e^x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx=\frac {x^{2} e^{\left (e^{\left ({\left (e^{4} - 60\right )} e^{\left (-4\right )}\right )}\right )} + x e^{x} - 2}{x} \]

[In]

integrate((x^2*exp(exp((exp(4)-60)/exp(4)))+exp(x)*x^2+2)/x^2,x, algorithm="giac")

[Out]

(x^2*e^(e^((e^4 - 60)*e^(-4))) + x*e^x - 2)/x

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {2+e^{e^{\frac {-60+e^4}{e^4}}} x^2+e^x x^2}{x^2} \, dx={\mathrm {e}}^x+x\,{\mathrm {e}}^{{\mathrm {e}}^{-60\,{\mathrm {e}}^{-4}}\,\mathrm {e}}-\frac {2}{x} \]

[In]

int((x^2*exp(x) + x^2*exp(exp(exp(-4)*(exp(4) - 60))) + 2)/x^2,x)

[Out]

exp(x) + x*exp(exp(-60*exp(-4))*exp(1)) - 2/x