\(\int (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}) \, dx\) [1904]

Optimal result

Integrand size = 10, antiderivative size = 15 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=-\frac {1}{2 x^2}-\frac {1}{x}+\log (x) \]

[Out]

-1/2/x^2-1/x+ln(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=-\frac {1}{2 x^2}-\frac {1}{x}+\log (x) \]

[In]

Int[x^(-3) + x^(-2) + x^(-1),x]

[Out]

-1/2*1/x^2 - x^(-1) + Log[x]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 x^2}-\frac {1}{x}+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=-\frac {1}{2 x^2}-\frac {1}{x}+\log (x) \]

[In]

Integrate[x^(-3) + x^(-2) + x^(-1),x]

[Out]

-1/2*1/x^2 - x^(-1) + Log[x]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {-\frac {1}{2}-x}{x^{2}}+\ln \left (x \right )\)	\(13\)
default	\(-\frac {1}{2 x^{2}}-\frac {1}{x}+\ln \left (x \right )\)	\(14\)
risch	\(-\frac {1}{2 x^{2}}-\frac {1}{x}+\ln \left (x \right )\)	\(14\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2}-2 x -1}{2 x^{2}}\)	\(18\)

[In]

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

[Out]

(-1/2-x)/x^2+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=\frac {2 \, x^{2} \log \left (x\right ) - 2 \, x - 1}{2 \, x^{2}} \]

[In]

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

[Out]

1/2*(2*x^2*log(x) - 2*x - 1)/x^2

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=\log {\left (x \right )} + \frac {- 2 x - 1}{2 x^{2}} \]

[In]

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

[Out]

log(x) + (-2*x - 1)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=-\frac {1}{x} - \frac {1}{2 \, x^{2}} + \log \left (x\right ) \]

[In]

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

[Out]

-1/x - 1/2/x^2 + log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=-\frac {1}{x} - \frac {1}{2 \, x^{2}} + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

-1/x - 1/2/x^2 + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \left (\frac {1}{x^3}+\frac {1}{x^2}+\frac {1}{x}\right ) \, dx=\ln \left (x\right )-\frac {x+\frac {1}{2}}{x^2} \]

[In]

int(1/x + 1/x^2 + 1/x^3,x)

[Out]

log(x) - (x + 1/2)/x^2