3.32 \(\int \frac{1}{x (x-x^3)} \, dx\)

Optimal. Leaf size=8 \[ \tanh ^{-1}(x)-\frac{1}{x} \]

[Out]

-x^(-1) + ArcTanh[x]

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Rubi [A]  time = 0.007941, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1584, 325, 206} \[ \tanh ^{-1}(x)-\frac{1}{x} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(x - x^3)),x]

[Out]

-x^(-1) + ArcTanh[x]

Rule 1584

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]

Rule 325

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 206

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

Rubi steps

\begin{align*} \int \frac{1}{x \left (x-x^3\right )} \, dx &=\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]  time = 0.0033792, size = 24, normalized size = 3. \[ -\frac{1}{x}-\frac{1}{2} \log (1-x)+\frac{1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(x - x^3)),x]

[Out]

-x^(-1) - Log[1 - x]/2 + Log[1 + x]/2

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Maple [B]  time = 0.005, size = 19, normalized size = 2.4 \begin{align*}{\frac{\ln \left ( 1+x \right ) }{2}}-{\frac{\ln \left ( -1+x \right ) }{2}}-{x}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(-x^3+x),x)

[Out]

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

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Maxima [B]  time = 1.0997, size = 24, normalized size = 3. \begin{align*} -\frac{1}{x} + \frac{1}{2} \, \log \left (x + 1\right ) - \frac{1}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.43392, size = 55, normalized size = 6.88 \begin{align*} \frac{x \log \left (x + 1\right ) - x \log \left (x - 1\right ) - 2}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.093039, size = 15, normalized size = 1.88 \begin{align*} - \frac{\log{\left (x - 1 \right )}}{2} + \frac{\log{\left (x + 1 \right )}}{2} - \frac{1}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x - 1)/2 + log(x + 1)/2 - 1/x

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Giac [B]  time = 1.19293, size = 27, normalized size = 3.38 \begin{align*} -\frac{1}{x} + \frac{1}{2} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))