\(\int \frac {e^{-2/x} (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x))}{x^3} \, dx\) [4674]

Optimal result

Integrand size = 39, antiderivative size = 28 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=\frac {16 (-3+x)}{x}+\frac {16 \left (16-e^{-1/x} x\right )^2}{x^2} \]

[Out]

16*(16-x/exp(1/x))^2/x^2+16*(-3+x)/x

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6874, 2326, 2240, 37} \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=\frac {(512-3 x)^2}{64 x^2}+16 e^{-2/x}-\frac {512 e^{-1/x}}{x} \]

[In]

Int[(32*x + E^(2/x)*(-8192 + 48*x) + E^x^(-1)*(-512 + 512*x))/(E^(2/x)*x^3),x]

[Out]

16/E^(2/x) + (512 - 3*x)^2/(64*x^2) - 512/(E^x^(-1)*x)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {512 e^{-1/x} (-1+x)}{x^3}+\frac {32 e^{-2/x}}{x^2}+\frac {16 (-512+3 x)}{x^3}\right ) \, dx \\ & = 16 \int \frac {-512+3 x}{x^3} \, dx+32 \int \frac {e^{-2/x}}{x^2} \, dx+512 \int \frac {e^{-1/x} (-1+x)}{x^3} \, dx \\ & = 16 e^{-2/x}+\frac {(512-3 x)^2}{64 x^2}-\frac {512 e^{-1/x}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=16 \left (e^{-2/x}+\frac {256}{x^2}-\frac {3}{x}-\frac {32 e^{-1/x}}{x}\right ) \]

[In]

Integrate[(32*x + E^(2/x)*(-8192 + 48*x) + E^x^(-1)*(-512 + 512*x))/(E^(2/x)*x^3),x]

[Out]

16*(E^(-2/x) + 256/x^2 - 3/x - 32/(E^x^(-1)*x))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07

method	result	size
risch	\(\frac {-48 x +4096}{x^{2}}-\frac {512 \,{\mathrm e}^{-\frac {1}{x}}}{x}+16 \,{\mathrm e}^{-\frac {2}{x}}\)	\(30\)
derivativedivides	\(\frac {4096}{x^{2}}-\frac {48}{x}+16 \,{\mathrm e}^{-\frac {2}{x}}-\frac {512 \,{\mathrm e}^{-\frac {1}{x}}}{x}\)	\(31\)
default	\(\frac {4096}{x^{2}}-\frac {48}{x}+16 \,{\mathrm e}^{-\frac {2}{x}}-\frac {512 \,{\mathrm e}^{-\frac {1}{x}}}{x}\)	\(31\)
parts	\(\frac {4096}{x^{2}}-\frac {48}{x}+16 \,{\mathrm e}^{-\frac {2}{x}}-\frac {512 \,{\mathrm e}^{-\frac {1}{x}}}{x}\)	\(31\)
norman	\(\frac {\left (16 x^{2}+4096 \,{\mathrm e}^{\frac {2}{x}}-512 x \,{\mathrm e}^{\frac {1}{x}}-48 x \,{\mathrm e}^{\frac {2}{x}}\right ) {\mathrm e}^{-\frac {2}{x}}}{x^{2}}\)	\(41\)
parallelrisch	\(-\frac {\left (48 \ln \left ({\mathrm e}^{\frac {1}{x}}\right ) x^{2} {\mathrm e}^{\frac {2}{x}}-16 x^{2}+512 x \,{\mathrm e}^{\frac {1}{x}}-4096 \,{\mathrm e}^{\frac {2}{x}}\right ) {\mathrm e}^{-\frac {2}{x}}}{x^{2}}\)	\(49\)

[In]

int(((48*x-8192)*exp(1/x)^2+(512*x-512)*exp(1/x)+32*x)/x^3/exp(1/x)^2,x,method=_RETURNVERBOSE)

[Out]

(-48*x+4096)/x^2-512/x*exp(-1/x)+16*exp(-2/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=\frac {16 \, {\left (x^{2} - {\left (3 \, x - 256\right )} e^{\frac {2}{x}} - 32 \, x e^{\frac {1}{x}}\right )} e^{\left (-\frac {2}{x}\right )}}{x^{2}} \]

[In]

integrate(((48*x-8192)*exp(1/x)^2+(512*x-512)*exp(1/x)+32*x)/x^3/exp(1/x)^2,x, algorithm="fricas")

[Out]

16*(x^2 - (3*x - 256)*e^(2/x) - 32*x*e^(1/x))*e^(-2/x)/x^2

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=\frac {16 x e^{- \frac {2}{x}} - 512 e^{- \frac {1}{x}}}{x} + \frac {4096 - 48 x}{x^{2}} \]

[In]

integrate(((48*x-8192)*exp(1/x)**2+(512*x-512)*exp(1/x)+32*x)/x**3/exp(1/x)**2,x)

[Out]

(16*x*exp(-2/x) - 512*exp(-1/x))/x + (4096 - 48*x)/x**2

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=-\frac {48}{x} + \frac {4096}{x^{2}} + 512 \, e^{\left (-\frac {1}{x}\right )} + 16 \, e^{\left (-\frac {2}{x}\right )} - 512 \, \Gamma \left (2, \frac {1}{x}\right ) \]

[In]

integrate(((48*x-8192)*exp(1/x)^2+(512*x-512)*exp(1/x)+32*x)/x^3/exp(1/x)^2,x, algorithm="maxima")

[Out]

-48/x + 4096/x^2 + 512*e^(-1/x) + 16*e^(-2/x) - 512*gamma(2, 1/x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=-\frac {512 \, e^{\left (-\frac {1}{x}\right )}}{x} - \frac {48}{x} + \frac {4096}{x^{2}} + 16 \, e^{\left (-\frac {2}{x}\right )} \]

[In]

integrate(((48*x-8192)*exp(1/x)^2+(512*x-512)*exp(1/x)+32*x)/x^3/exp(1/x)^2,x, algorithm="giac")

[Out]

-512*e^(-1/x)/x - 48/x + 4096/x^2 + 16*e^(-2/x)

Mupad [B] (verification not implemented)

Time = 10.58 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2/x} \left (32 x+e^{2/x} (-8192+48 x)+e^{\frac {1}{x}} (-512+512 x)\right )}{x^3} \, dx=16\,{\mathrm {e}}^{-\frac {2}{x}}-\frac {48\,x-4096}{x^2}-\frac {512\,{\mathrm {e}}^{-\frac {1}{x}}}{x} \]

[In]

int((exp(-2/x)*(32*x + exp(2/x)*(48*x - 8192) + exp(1/x)*(512*x - 512)))/x^3,x)

[Out]

16*exp(-2/x) - (48*x - 4096)/x^2 - (512*exp(-1/x))/x