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

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

[Out]

-Log[1 - 2*ArcTanh[x]]/(2*a*b)

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Rubi [A]  time = 0.0444294, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {5946} \[ -\frac{\log \left (1-2 \tanh ^{-1}(x)\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - 2*ArcTanh[x]]/(2*a*b)

Rule 5946

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*}

Mathematica [A]  time = 0.0546647, size = 17, normalized size = 1. \[ -\frac{\log \left (2 \tanh ^{-1}(x)-1\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[-1 + 2*ArcTanh[x]]/(2*a*b)

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Maple [A]  time = 0.059, size = 19, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( 2\,b{\it Artanh} \left ( x \right ) -b \right ) }{2\,ab}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970007, size = 31, normalized size = 1.82 \begin{align*} -\frac{\log \left (-\log \left (x + 1\right ) + \log \left (-x + 1\right ) + 1\right )}{2 \, a b} \end{align*}

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 = 1.91186, size = 58, normalized size = 3.41 \begin{align*} -\frac{\log \left (\log \left (-\frac{x + 1}{x - 1}\right ) - 1\right )}{2 \, a b} \end{align*}

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 = 1.57957, size = 14, normalized size = 0.82 \begin{align*} - \frac{\log{\left (\operatorname{atanh}{\left (x \right )} - \frac{1}{2} \right )}}{2 a b} \end{align*}

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]  time = 1.16613, size = 107, normalized size = 6.29 \begin{align*} -\frac{\log \left (\frac{1}{4} \,{\left (\pi{\left (\mathrm{sgn}\left (x + 1\right ) - 1\right )} - \pi{\left (\mathrm{sgn}\left (x - 1\right ) + 1\right )} + 4 \, \pi \left \lfloor -\frac{\pi{\left (\mathrm{sgn}\left (x + 1\right ) - 1\right )} - \pi{\left (\mathrm{sgn}\left (x - 1\right ) + 1\right )}}{4 \, \pi } + \frac{1}{2} \right \rfloor \right )}^{2} +{\left (\log \left (\frac{{\left | x + 1 \right |}}{{\left | -x + 1 \right |}}\right ) - 1\right )}^{2}\right )}{4 \, a b} \end{align*}

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