\(\int \frac {1}{x^2 (x-x^3)} \, dx\) [33]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1-x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1598, 272, 46} \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{2} \log \left (1-x^2\right )+\log (x) \]

[In]

Int[1/(x^2*(x - x^3)),x]

[Out]

-1/2*1/x^2 + Log[x] - Log[1 - x^2]/2

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1-x^2\right ) \]

[In]

Integrate[1/(x^2*(x - x^3)),x]

[Out]

-1/2*1/x^2 + Log[x] - Log[1 - x^2]/2

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (x^{2}-1\right )}{2}\)	\(17\)
default	\(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\)	\(21\)
norman	\(-\frac {1}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (-1+x \right )}{2}-\frac {\ln \left (1+x \right )}{2}\)	\(21\)
meijerg	\(-\frac {1}{2 x^{2}}+\ln \left (x \right )+\frac {i \pi }{2}-\frac {\ln \left (-x^{2}+1\right )}{2}\)	\(23\)
parallelrisch	\(\frac {2 \ln \left (x \right ) x^{2}-\ln \left (1+x \right ) x^{2}-\ln \left (-1+x \right ) x^{2}-1}{2 x^{2}}\)	\(33\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.75 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {x^{2} \log \left (x^{2} - 1\right ) - 2 \, x^{2} \log \left (x\right ) + 1}{2 \, x^{2}} \]

[In]

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

[Out]

-1/2*(x^2*log(x^2 - 1) - 2*x^2*log(x) + 1)/x^2

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{2} - 1 \right )}}{2} - \frac {1}{2 x^{2}} \]

[In]

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

[Out]

log(x) - log(x**2 - 1)/2 - 1/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {1}{2 \, x^{2}} - \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) + \log \left (x\right ) \]

[In]

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

[Out]

-1/2/x^2 - 1/2*log(x + 1) - 1/2*log(x - 1) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=-\frac {x^{2} + 1}{2 \, x^{2}} + \frac {1}{2} \, \log \left (x^{2}\right ) - \frac {1}{2} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/2*(x^2 + 1)/x^2 + 1/2*log(x^2) - 1/2*log(abs(x^2 - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^2 \left (x-x^3\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2-1\right )}{2}-\frac {1}{2\,x^2} \]

[In]

int(1/(x^2*(x - x^3)),x)

[Out]

log(x) - log(x^2 - 1)/2 - 1/(2*x^2)