3.93.80 \(\int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx\)

Optimal. Leaf size=16 \[ \frac {e^2-20 e^{\frac {1}{x}} x}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2209} \begin {gather*} \frac {e^2}{x}-20 e^{\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^2 + 20*E^x^(-1))/x^2,x]

[Out]

-20*E^x^(-1) + E^2/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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^2}{x^2}+\frac {20 e^{\frac {1}{x}}}{x^2}\right ) \, dx\\ &=\frac {e^2}{x}+20 \int \frac {e^{\frac {1}{x}}}{x^2} \, dx\\ &=-20 e^{\frac {1}{x}}+\frac {e^2}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.94 \begin {gather*} -20 e^{\frac {1}{x}}+\frac {e^2}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^2 + 20*E^x^(-1))/x^2,x]

[Out]

-20*E^x^(-1) + E^2/x

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fricas [A]  time = 0.92, size = 17, normalized size = 1.06 \begin {gather*} -\frac {20 \, x e^{\frac {1}{x}} - e^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*exp(1/x)-exp(2))/x^2,x, algorithm="fricas")

[Out]

-(20*x*e^(1/x) - e^2)/x

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giac [A]  time = 0.21, size = 13, normalized size = 0.81 \begin {gather*} \frac {e^{2}}{x} - 20 \, e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*exp(1/x)-exp(2))/x^2,x, algorithm="giac")

[Out]

e^2/x - 20*e^(1/x)

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maple [A]  time = 0.04, size = 14, normalized size = 0.88

method	result	size
derivativedivides	\(-20 \,{\mathrm e}^{\frac {1}{x}}+\frac {{\mathrm e}^{2}}{x}\)	\(14\)
default	\(-20 \,{\mathrm e}^{\frac {1}{x}}+\frac {{\mathrm e}^{2}}{x}\)	\(14\)
risch	\(-20 \,{\mathrm e}^{\frac {1}{x}}+\frac {{\mathrm e}^{2}}{x}\)	\(14\)
norman	\(\frac {{\mathrm e}^{2}-20 x \,{\mathrm e}^{\frac {1}{x}}}{x}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*exp(1/x)-exp(2))/x^2,x,method=_RETURNVERBOSE)

[Out]

-20*exp(1/x)+exp(2)/x

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maxima [A]  time = 0.35, size = 13, normalized size = 0.81 \begin {gather*} \frac {e^{2}}{x} - 20 \, e^{\frac {1}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*exp(1/x)-exp(2))/x^2,x, algorithm="maxima")

[Out]

e^2/x - 20*e^(1/x)

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mupad [B]  time = 7.47, size = 13, normalized size = 0.81 \begin {gather*} \frac {{\mathrm {e}}^2}{x}-20\,{\mathrm {e}}^{1/x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*exp(1/x) - exp(2))/x^2,x)

[Out]

exp(2)/x - 20*exp(1/x)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.62 \begin {gather*} - 20 e^{\frac {1}{x}} + \frac {e^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*exp(1/x)-exp(2))/x**2,x)

[Out]

-20*exp(1/x) + exp(2)/x

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