\(\int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx\) [1180]

Optimal result

Integrand size = 37, antiderivative size = 36 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=5 \left (e^4-e^{5 e^5}-e^{\frac {-e^{2/x}+x}{x}}\right )+x \]

[Out]

x-5*exp((-exp(2/x)+x)/x)-5*exp(5*exp(5))+5*exp(4)

Rubi [F]

\[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=\int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx \]

[In]

Int[(E^(2/x + (-E^(2/x) + x)/x)*(-10 - 5*x) + x^3)/x^3,x]

[Out]

x - 10*Defer[Int][E^(1 + 2/x - E^(2/x)/x)/x^3, x] + 5*Defer[Subst][Defer[Int][E^(1 + 2*x - E^(2*x)*x), x], x,
x^(-1)]

Rubi steps \begin{align*} \text {integral}& = \int \left (1-\frac {5 e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}} (2+x)}{x^3}\right ) \, dx \\ & = x-5 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}} (2+x)}{x^3} \, dx \\ & = x-5 \int \left (\frac {2 e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3}+\frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^2}\right ) \, dx \\ & = x-5 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^2} \, dx-10 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3} \, dx \\ & = x+5 \text {Subst}\left (\int e^{1+2 x-e^{2 x} x} \, dx,x,\frac {1}{x}\right )-10 \int \frac {e^{1+\frac {2}{x}-\frac {e^{2/x}}{x}}}{x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=-5 e^{1-\frac {e^{2/x}}{x}}+x \]

[In]

Integrate[(E^(2/x + (-E^(2/x) + x)/x)*(-10 - 5*x) + x^3)/x^3,x]

[Out]

-5*E^(1 - E^(2/x)/x) + x

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.56

method	result	size
risch	\(x -5 \,{\mathrm e}^{\frac {-{\mathrm e}^{\frac {2}{x}}+x}{x}}\)	\(20\)
parts	\(x -5 \,{\mathrm e}^{\frac {-{\mathrm e}^{\frac {2}{x}}+x}{x}}\)	\(20\)
parallelrisch	\(x -5 \,{\mathrm e}^{-\frac {{\mathrm e}^{\frac {2}{x}}-x}{x}}\)	\(21\)
norman	\(\frac {x^{3}-5 x^{2} {\mathrm e}^{\frac {-{\mathrm e}^{\frac {2}{x}}+x}{x}}}{x^{2}}\)	\(29\)

[In]

int(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx={\left (x e^{\frac {2}{x}} - 5 \, e^{\left (\frac {x - e^{\frac {2}{x}} + 2}{x}\right )}\right )} e^{\left (-\frac {2}{x}\right )} \]

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="fricas")

[Out]

(x*e^(2/x) - 5*e^((x - e^(2/x) + 2)/x))*e^(-2/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.33 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=x - 5 e^{\frac {x - e^{\frac {2}{x}}}{x}} \]

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x**3)/x**3,x)

[Out]

x - 5*exp((x - exp(2/x))/x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=x - 5 \, e^{\left (-\frac {e^{\frac {2}{x}}}{x} + 1\right )} \]

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="maxima")

[Out]

x - 5*e^(-e^(2/x)/x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx={\left (x e^{\frac {2}{x}} - 5 \, e^{\left (\frac {x - e^{\frac {2}{x}} + 2}{x}\right )}\right )} e^{\left (-\frac {2}{x}\right )} \]

[In]

integrate(((-5*x-10)*exp(2/x)*exp((-exp(2/x)+x)/x)+x^3)/x^3,x, algorithm="giac")

[Out]

(x*e^(2/x) - 5*e^((x - e^(2/x) + 2)/x))*e^(-2/x)

Mupad [B] (verification not implemented)

Time = 8.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.50 \[ \int \frac {e^{\frac {2}{x}+\frac {-e^{2/x}+x}{x}} (-10-5 x)+x^3}{x^3} \, dx=x-5\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2/x}}{x}}\,\mathrm {e} \]

[In]

int((x^3 - exp((x - exp(2/x))/x)*exp(2/x)*(5*x + 10))/x^3,x)

[Out]

x - 5*exp(-exp(2/x)/x)*exp(1)