\(\int \frac {1}{x (x-x^3)} \, dx\) [32]

Optimal result

Integrand size = 13, antiderivative size = 8 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=-\frac {1}{x}+\text {arctanh}(x) \]

[Out]

-1/x+arctanh(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1598, 331, 212} \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=\text {arctanh}(x)-\frac {1}{x} \]

[In]

Int[1/(x*(x - x^3)),x]

[Out]

-x^(-1) + ArcTanh[x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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^2 \left (1-x^2\right )} \, dx \\ & = -\frac {1}{x}+\int \frac {1}{1-x^2} \, dx \\ & = -\frac {1}{x}+\tanh ^{-1}(x) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(24\) vs. \(2(8)=16\).

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 3.00 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=-\frac {1}{x}-\frac {1}{2} \log (1-x)+\frac {1}{2} \log (1+x) \]

[In]

Integrate[1/(x*(x - x^3)),x]

[Out]

-x^(-1) - Log[1 - x]/2 + Log[1 + x]/2

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 2.00

method	result	size
meijerg	\(\frac {i \left (\frac {2 i}{x}-2 i \operatorname {arctanh}\left (x \right )\right )}{2}\)	\(16\)
default	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (1+x \right )}{2}-\frac {1}{x}\)	\(19\)
norman	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (1+x \right )}{2}-\frac {1}{x}\)	\(19\)
risch	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (1+x \right )}{2}-\frac {1}{x}\)	\(19\)
parallelrisch	\(\frac {\ln \left (1+x \right ) x -\ln \left (-1+x \right ) x -2}{2 x}\)	\(21\)

[In]

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

[Out]

1/2*I*(2*I/x-2*I*arctanh(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (8) = 16\).

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.50 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=\frac {x \log \left (x + 1\right ) - x \log \left (x - 1\right ) - 2}{2 \, x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (5) = 10\).

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.88 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=- \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x + 1 \right )}}{2} - \frac {1}{x} \]

[In]

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

[Out]

-log(x - 1)/2 + log(x + 1)/2 - 1/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (8) = 16\).

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=-\frac {1}{x} + \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (8) = 16\).

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 2.50 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=-\frac {1}{x} + \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (x-x^3\right )} \, dx=\mathrm {atanh}\left (x\right )-\frac {1}{x} \]

[In]

int(1/(x*(x - x^3)),x)

[Out]

atanh(x) - 1/x