3.6.5 \(\int \frac {1}{(a-a x^2) (b-2 b \tanh ^{-1}(x))} \, dx\) [505]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {6093} \begin {gather*} -\frac {\log \left (1-2 \tanh ^{-1}(x)\right )}{2 a b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Log[1 - 2*ArcTanh[x]]/(a*b)

Rule 6093

Int[1/(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[Log[RemoveContent[a + b*A
rcTanh[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 \tanh ^{-1}(x)\right )} \, dx &=-\frac {\log \left (1-2 \tanh ^{-1}(x)\right )}{2 a b}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 17, normalized size = 1.00 \begin {gather*} -\frac {\log \left (-1+2 \tanh ^{-1}(x)\right )}{2 a b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Log[-1 + 2*ArcTanh[x]]/(a*b)

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Maple [A]
time = 1.08, size = 19, normalized size = 1.12

method result size
default \(-\frac {\ln \left (2 b \arctanh \left (x \right )-b \right )}{2 a b}\) \(19\)
risch \(-\frac {\ln \left (-\ln \left (1-x \right )+\ln \left (1+x \right )-1\right )}{2 a b}\) \(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a*x^2+a)/(b-2*b*arctanh(x)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 23, normalized size = 1.35 \begin {gather*} -\frac {\log \left (-\log \left (x + 1\right ) + \log \left (-x + 1\right ) + 1\right )}{2 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 22, normalized size = 1.29 \begin {gather*} -\frac {\log \left (\log \left (-\frac {x + 1}{x - 1}\right ) - 1\right )}{2 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.24, size = 14, normalized size = 0.82 \begin {gather*} - \frac {\log {\left (\operatorname {atanh}{\left (x \right )} - \frac {1}{2} \right )}}{2 a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(atanh(x) - 1/2)/(2*a*b)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (15) = 30\).
time = 0.41, size = 48, normalized size = 2.82 \begin {gather*} -\frac {\log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\left (x - 1\right ) \mathrm {sgn}\left (-x - 1\right ) - 1\right )}^{2} + {\left (\log \left (\frac {{\left | -x - 1 \right |}}{{\left | x - 1 \right |}}\right ) - 1\right )}^{2}\right )}{4 \, a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(1/4*pi^2*(sgn(x - 1)*sgn(-x - 1) - 1)^2 + (log(abs(-x - 1)/abs(x - 1)) - 1)^2)/(a*b)

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Mupad [B]
time = 0.98, size = 15, normalized size = 0.88 \begin {gather*} -\frac {\ln \left (2\,\mathrm {atanh}\left (x\right )-1\right )}{2\,a\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(2*atanh(x) - 1)/(2*a*b)

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