\(\int \frac {e^{-x/2}}{x^3} \, dx\) [536]

Optimal result

Integrand size = 11, antiderivative size = 39 \[ \int \frac {e^{-x/2}}{x^3} \, dx=-\frac {e^{-x/2}}{2 x^2}+\frac {e^{-x/2}}{4 x}+\frac {\operatorname {ExpIntegralEi}\left (-\frac {x}{2}\right )}{8} \]

[Out]

-1/2/exp(1/2*x)/x^2+1/4/exp(1/2*x)/x+1/8*Ei(-1/2*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2208, 2209} \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {\operatorname {ExpIntegralEi}\left (-\frac {x}{2}\right )}{8}-\frac {e^{-x/2}}{2 x^2}+\frac {e^{-x/2}}{4 x} \]

[In]

Int[1/(E^(x/2)*x^3),x]

[Out]

-1/2*1/(E^(x/2)*x^2) + 1/(4*E^(x/2)*x) + ExpIntegralEi[-1/2*x]/8

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = -\frac {e^{-x/2}}{2 x^2}-\frac {1}{4} \int \frac {e^{-x/2}}{x^2} \, dx \\ & = -\frac {e^{-x/2}}{2 x^2}+\frac {e^{-x/2}}{4 x}+\frac {1}{8} \int \frac {e^{-x/2}}{x} \, dx \\ & = -\frac {e^{-x/2}}{2 x^2}+\frac {e^{-x/2}}{4 x}+\frac {\operatorname {ExpIntegralEi}\left (-\frac {x}{2}\right )}{8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {1}{8} \left (\frac {2 e^{-x/2} (-2+x)}{x^2}+\operatorname {ExpIntegralEi}\left (-\frac {x}{2}\right )\right ) \]

[In]

Integrate[1/(E^(x/2)*x^3),x]

[Out]

((2*(-2 + x))/(E^(x/2)*x^2) + ExpIntegralEi[-1/2*x])/8

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69

method	result	size
risch	\(-\frac {{\mathrm e}^{-\frac {x}{2}}}{2 x^{2}}+\frac {{\mathrm e}^{-\frac {x}{2}}}{4 x}-\frac {\operatorname {Ei}_{1}\left (\frac {x}{2}\right )}{8}\)	\(27\)
derivativedivides	\(-\frac {{\mathrm e}^{-\frac {x}{2}}}{2 x^{2}}+\frac {{\mathrm e}^{-\frac {x}{2}}}{4 x}-\frac {\operatorname {Ei}_{1}\left (\frac {x}{2}\right )}{8}\)	\(31\)
default	\(-\frac {{\mathrm e}^{-\frac {x}{2}}}{2 x^{2}}+\frac {{\mathrm e}^{-\frac {x}{2}}}{4 x}-\frac {\operatorname {Ei}_{1}\left (\frac {x}{2}\right )}{8}\)	\(31\)
meijerg	\(-\frac {1}{2 x^{2}}+\frac {1}{2 x}-\frac {3}{16}+\frac {\ln \left (x \right )}{8}-\frac {\ln \left (2\right )}{8}+\frac {\frac {9}{4} x^{2}-6 x +6}{12 x^{2}}-\frac {\left (-\frac {3 x}{2}+3\right ) {\mathrm e}^{-\frac {x}{2}}}{6 x^{2}}-\frac {\ln \left (\frac {x}{2}\right )}{8}-\frac {\operatorname {Ei}_{1}\left (\frac {x}{2}\right )}{8}\)	\(63\)

[In]

int(1/exp(1/2*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*exp(-1/2*x)/x^2+1/4*exp(-1/2*x)/x-1/8*Ei(1,1/2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {x^{2} {\rm Ei}\left (-\frac {1}{2} \, x\right ) + 2 \, {\left (x - 2\right )} e^{\left (-\frac {1}{2} \, x\right )}}{8 \, x^{2}} \]

[In]

integrate(1/exp(1/2*x)/x^3,x, algorithm="fricas")

[Out]

1/8*(x^2*Ei(-1/2*x) + 2*(x - 2)*e^(-1/2*x))/x^2

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {\operatorname {Ei}{\left (\frac {x e^{i \pi }}{2} \right )}}{8} + \frac {e^{- \frac {x}{2}}}{4 x} - \frac {e^{- \frac {x}{2}}}{2 x^{2}} \]

[In]

integrate(1/exp(1/2*x)/x**3,x)

[Out]

Ei(x*exp_polar(I*pi)/2)/8 + exp(-x/2)/(4*x) - exp(-x/2)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.18 \[ \int \frac {e^{-x/2}}{x^3} \, dx=-\frac {1}{4} \, \Gamma \left (-2, \frac {1}{2} \, x\right ) \]

[In]

integrate(1/exp(1/2*x)/x^3,x, algorithm="maxima")

[Out]

-1/4*gamma(-2, 1/2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {x^{2} {\rm Ei}\left (-\frac {1}{2} \, x\right ) + 2 \, x e^{\left (-\frac {1}{2} \, x\right )} - 4 \, e^{\left (-\frac {1}{2} \, x\right )}}{8 \, x^{2}} \]

[In]

integrate(1/exp(1/2*x)/x^3,x, algorithm="giac")

[Out]

1/8*(x^2*Ei(-1/2*x) + 2*x*e^(-1/2*x) - 4*e^(-1/2*x))/x^2

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-x/2}}{x^3} \, dx=\frac {{\mathrm {e}}^{-\frac {x}{2}}\,\left (\frac {1}{x}-\frac {2}{x^2}\right )}{4}-\frac {\mathrm {expint}\left (\frac {x}{2}\right )}{8} \]

[In]

int(exp(-x/2)/x^3,x)

[Out]

(exp(-x/2)*(1/x - 2/x^2))/4 - expint(x/2)/8