\(\int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx\) [6775]

Optimal result

Integrand size = 19, antiderivative size = 14 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=36-e^x+\frac {1}{16 x^2} \]

[Out]

1/16/x^2+36-exp(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 14, 2225} \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=\frac {1}{16 x^2}-e^x \]

[In]

Int[(-x^(-2) - 8*E^x*x)/(8*x),x]

[Out]

-E^x + 1/(16*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \frac {-\frac {1}{x^2}-8 e^x x}{x} \, dx \\ & = \frac {1}{8} \int \left (-8 e^x-\frac {1}{x^3}\right ) \, dx \\ & = \frac {1}{16 x^2}-\int e^x \, dx \\ & = -e^x+\frac {1}{16 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=-e^x+\frac {1}{16 x^2} \]

[In]

Integrate[(-x^(-2) - 8*E^x*x)/(8*x),x]

[Out]

-E^x + 1/(16*x^2)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {1}{16 x^{2}}-{\mathrm e}^{x}\)	\(11\)
risch	\(\frac {1}{16 x^{2}}-{\mathrm e}^{x}\)	\(11\)
parts	\(\frac {1}{16 x^{2}}-{\mathrm e}^{x}\)	\(11\)
norman	\(\frac {\frac {1}{16}-{\mathrm e}^{x} x^{2}}{x^{2}}\)	\(14\)
parallelrisch	\(-\frac {16 \,{\mathrm e}^{x} x^{2}-1}{16 x^{2}}\)	\(15\)

[In]

int(1/8*(-1/x^2-8*exp(x)*x)/x,x,method=_RETURNVERBOSE)

[Out]

1/16/x^2-exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=-\frac {16 \, x^{2} e^{x} - 1}{16 \, x^{2}} \]

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="fricas")

[Out]

-1/16*(16*x^2*e^x - 1)/x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.57 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=- e^{x} + \frac {1}{16 x^{2}} \]

[In]

integrate(1/8*(-1/x**2-8*exp(x)*x)/x,x)

[Out]

-exp(x) + 1/(16*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=\frac {1}{16 \, x^{2}} - e^{x} \]

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="maxima")

[Out]

1/16/x^2 - e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=-\frac {16 \, x^{2} e^{x} - 1}{16 \, x^{2}} \]

[In]

integrate(1/8*(-1/x^2-8*exp(x)*x)/x,x, algorithm="giac")

[Out]

-1/16*(16*x^2*e^x - 1)/x^2

Mupad [B] (verification not implemented)

Time = 12.34 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {-\frac {1}{x^2}-8 e^x x}{8 x} \, dx=\frac {1}{16\,x^2}-{\mathrm {e}}^x \]

[In]

int(-(x*exp(x) + 1/(8*x^2))/x,x)

[Out]

1/(16*x^2) - exp(x)