3.292 \(\int \frac {1}{(a-a x^2) (b-2 b \coth ^{-1}(x))} \, dx\)

Optimal. Leaf size=17 \[ -\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b} \]

[Out]

-1/2*ln(1-2*arccoth(x))/a/b

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {5947} \[ -\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - a*x^2)*(b - 2*b*ArcCoth[x])),x]

[Out]

-Log[1 - 2*ArcCoth[x]]/(2*a*b)

Rule 5947

Int[1/(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[Log[RemoveContent[a + b*A
rcCoth[c*x], x]]/(b*c*d), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx &=-\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 17, normalized size = 1.00 \[ -\frac {\log \left (2 \coth ^{-1}(x)-1\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a - a*x^2)*(b - 2*b*ArcCoth[x])),x]

[Out]

-1/2*Log[-1 + 2*ArcCoth[x]]/(a*b)

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fricas [A]  time = 0.73, size = 21, normalized size = 1.24 \[ -\frac {\log \left (\log \left (\frac {x + 1}{x - 1}\right ) - 1\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a*x^2+a)/(b-2*b*arccoth(x)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a x^{2} - a\right )} {\left (2 \, b \operatorname {arcoth}\relax (x) - b\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a*x^2+a)/(b-2*b*arccoth(x)),x, algorithm="giac")

[Out]

integrate(1/((a*x^2 - a)*(2*b*arccoth(x) - b)), x)

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maple [A]  time = 0.07, size = 19, normalized size = 1.12 \[ -\frac {\ln \left (2 b \,\mathrm {arccoth}\relax (x )-b \right )}{2 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a*x^2+a)/(b-2*b*arccoth(x)),x)

[Out]

-1/2/a*ln(2*b*arccoth(x)-b)/b

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maxima [A]  time = 0.33, size = 21, normalized size = 1.24 \[ -\frac {\log \left (\log \left (x + 1\right ) - \log \left (x - 1\right ) - 1\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a*x^2+a)/(b-2*b*arccoth(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.40, size = 15, normalized size = 0.88 \[ -\frac {\ln \left (2\,\mathrm {acoth}\relax (x)-1\right )}{2\,a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - a*x^2)*(b - 2*b*acoth(x))),x)

[Out]

-log(2*acoth(x) - 1)/(2*a*b)

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sympy [A]  time = 0.57, size = 14, normalized size = 0.82 \[ - \frac {\log {\left (\operatorname {acoth}{\relax (x )} - \frac {1}{2} \right )}}{2 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a*x**2+a)/(b-2*b*acoth(x)),x)

[Out]

-log(acoth(x) - 1/2)/(2*a*b)

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