\(\int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx\) [9280]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=\frac {e^2-20 e^{\frac {1}{x}} x}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {14, 2240} \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=\frac {e^2}{x}-20 e^{\frac {1}{x}} \]

[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 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]

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=-20 e^{\frac {1}{x}}+\frac {e^2}{x} \]

[In]

Integrate[(-E^2 + 20*E^x^(-1))/x^2,x]

[Out]

-20*E^x^(-1) + E^2/x

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 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	\(-20 \,{\mathrm e}^{\frac {1}{x}}+\frac {{\mathrm e}^{2}}{x}\)	\(14\)
parts	\(-20 \,{\mathrm e}^{\frac {1}{x}}+\frac {{\mathrm e}^{2}}{x}\)	\(14\)
norman	\(\frac {{\mathrm e}^{2}-20 x \,{\mathrm e}^{\frac {1}{x}}}{x}\)	\(15\)
parallelrisch	\(\frac {{\mathrm e}^{2}-20 x \,{\mathrm e}^{\frac {1}{x}}}{x}\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=-\frac {20 \, x e^{\frac {1}{x}} - e^{2}}{x} \]

[In]

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

[Out]

-(20*x*e^(1/x) - e^2)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=- 20 e^{\frac {1}{x}} + \frac {e^{2}}{x} \]

[In]

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

[Out]

-20*exp(1/x) + exp(2)/x

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=\frac {e^{2}}{x} - 20 \, e^{\frac {1}{x}} \]

[In]

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

[Out]

e^2/x - 20*e^(1/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=\frac {e^{2}}{x} - 20 \, e^{\frac {1}{x}} \]

[In]

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

[Out]

e^2/x - 20*e^(1/x)

Mupad [B] (verification not implemented)

Time = 12.51 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {-e^2+20 e^{\frac {1}{x}}}{x^2} \, dx=\frac {{\mathrm {e}}^2}{x}-20\,{\mathrm {e}}^{1/x} \]

[In]

int((20*exp(1/x) - exp(2))/x^2,x)

[Out]

exp(2)/x - 20*exp(1/x)