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

Optimal. Leaf size=23 \[ -16+e^3+e^{3-\frac {e^5+x}{x}}+\log ^2(2) \]

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Rubi [A]  time = 0.04, antiderivative size = 12, normalized size of antiderivative = 0.52, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2225, 2209} \begin {gather*} e^{2-\frac {e^5}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(5 + (-E^5 + 2*x)/x)/x^2,x]

[Out]

E^(2 - E^5/x)

Rule 2209

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 2225

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

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.52 \begin {gather*} e^{2-\frac {e^5}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 - E^5/x)

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fricas [A]  time = 0.50, size = 15, normalized size = 0.65 \begin {gather*} e^{\left (\frac {7 \, x - e^{5}}{x} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^((7*x - e^5)/x - 5)

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giac [A]  time = 0.12, size = 10, normalized size = 0.43 \begin {gather*} e^{\left (-\frac {e^{5}}{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-e^5/x + 2)

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maple [A]  time = 0.04, size = 11, normalized size = 0.48

method	result	size
derivativedivides	\({\mathrm e}^{2-\frac {{\mathrm e}^{5}}{x}}\)	\(11\)
default	\({\mathrm e}^{2-\frac {{\mathrm e}^{5}}{x}}\)	\(11\)
gosper	\({\mathrm e}^{-\frac {{\mathrm e}^{5}-2 x}{x}}\)	\(13\)
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{5}-2 x}{x}}\)	\(13\)
norman	\({\mathrm e}^{\frac {-{\mathrm e}^{5}+2 x}{x}}\)	\(14\)
meijerg	\(-{\mathrm e}^{2} \left (1-{\mathrm e}^{-\frac {{\mathrm e}^{5}}{x}}\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2-exp(5)/x)

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maxima [A]  time = 0.36, size = 10, normalized size = 0.43 \begin {gather*} e^{\left (-\frac {e^{5}}{x} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-e^5/x + 2)

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mupad [B]  time = 3.04, size = 11, normalized size = 0.48 \begin {gather*} {\mathrm {e}}^{-\frac {{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.11, size = 8, normalized size = 0.35 \begin {gather*} e^{\frac {2 x - e^{5}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((2*x - exp(5))/x)

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