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

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

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Rubi [A]  time = 0.01, antiderivative size = 8, normalized size of antiderivative = 1.00, 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} \begin {gather*} \tanh ^{-1}(x)-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(x - x^3)),x]

[Out]

-x^(-1) + ArcTanh[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])

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 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]

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*}

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Mathematica [B]  time = 0.00, size = 24, normalized size = 3.00 \begin {gather*} -\frac {1}{x}-\frac {1}{2} \log (1-x)+\frac {1}{2} \log (x+1) \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (x-x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x*(x - x^3)),x]

[Out]

IntegrateAlgebraic[1/(x*(x - x^3)), x]

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fricas [B]  time = 0.41, size = 20, normalized size = 2.50 \begin {gather*} \frac {x \log \left (x + 1\right ) - x \log \left (x - 1\right ) - 2}{2 \, x} \end {gather*}

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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giac [B]  time = 0.15, size = 20, normalized size = 2.50 \begin {gather*} -\frac {1}{x} + \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

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))

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maple [B]  time = 0.05, size = 19, normalized size = 2.38 \begin {gather*} -\frac {\ln \left (x -1\right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.33, size = 18, normalized size = 2.25 \begin {gather*} -\frac {1}{x} + \frac {1}{2} \, \log \left (x + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) \end {gather*}

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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mupad [B]  time = 0.03, size = 8, normalized size = 1.00 \begin {gather*} \mathrm {atanh}\relax (x)-\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(x - x^3)),x)

[Out]

atanh(x) - 1/x

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sympy [B]  time = 0.13, size = 15, normalized size = 1.88 \begin {gather*} - \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x + 1 \right )}}{2} - \frac {1}{x} \end {gather*}

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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