\(\int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx\) [4331]

Optimal result

Integrand size = 21, antiderivative size = 23 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=-16+e^3+e^{3-\frac {e^5+x}{x}}+\log ^2(2) \]

[Out]

exp(3)-16+ln(2)^2+exp(3-(exp(5)+x)/x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2257, 2240} \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{2-\frac {e^5}{x}} \]

[In]

Int[E^(5 + (-E^5 + 2*x)/x)/x^2,x]

[Out]

E^(2 - E^5/x)

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2257

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{7-\frac {e^5}{x}}}{x^2} \, dx \\ & = e^{2-\frac {e^5}{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{2-\frac {e^5}{x}} \]

[In]

Integrate[E^(5 + (-E^5 + 2*x)/x)/x^2,x]

[Out]

E^(2 - E^5/x)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57

method	result	size
gosper	\({\mathrm e}^{-\frac {{\mathrm e}^{5}-2 x}{x}}\)	\(13\)
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{5}-2 x}{x}}\)	\(13\)
parallelrisch	\({\mathrm e}^{-\frac {{\mathrm e}^{5}-2 x}{x}}\)	\(13\)
derivativedivides	\({\mathrm e}^{\frac {-{\mathrm e}^{5}+2 x}{x}}\)	\(14\)
default	\({\mathrm e}^{\frac {-{\mathrm e}^{5}+2 x}{x}}\)	\(14\)
norman	\({\mathrm e}^{\frac {-{\mathrm e}^{5}+2 x}{x}}\)	\(14\)
meijerg	\(-{\mathrm e}^{2} \left (1-{\mathrm e}^{-\frac {{\mathrm e}^{5}}{x}}\right )\)	\(17\)

[In]

int(exp(5)*exp((-exp(5)+2*x)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(-(exp(5)-2*x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{\left (\frac {7 \, x - e^{5}}{x} - 5\right )} \]

[In]

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

[Out]

e^((7*x - e^5)/x - 5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.35 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{\frac {2 x - e^{5}}{x}} \]

[In]

integrate(exp(5)*exp((-exp(5)+2*x)/x)/x**2,x)

[Out]

exp((2*x - exp(5))/x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.43 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{\left (-\frac {e^{5}}{x} + 2\right )} \]

[In]

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

[Out]

e^(-e^5/x + 2)

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.43 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx=e^{\left (-\frac {e^{5}}{x} + 2\right )} \]

[In]

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

[Out]

e^(-e^5/x + 2)

Mupad [B] (verification not implemented)

Time = 11.42 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.48 \[ \int \frac {e^{5+\frac {-e^5+2 x}{x}}}{x^2} \, dx={\mathrm {e}}^{-\frac {{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^2 \]

[In]

int((exp(5)*exp((2*x - exp(5))/x))/x^2,x)

[Out]

exp(-exp(5)/x)*exp(2)