\(\int \frac {1}{(a-a x^2) (b-2 b \coth ^{-1}(x))} \, dx\) [292]

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6094} \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b} \]

[In]

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

[Out]

-1/2*Log[1 - 2*ArcCoth[x]]/(a*b)

Rule 6094

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*} \text {integral}& = -\frac {\log \left (1-2 \coth ^{-1}(x)\right )}{2 a b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (-1+2 \coth ^{-1}(x)\right )}{2 a b} \]

[In]

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

[Out]

-1/2*Log[-1 + 2*ArcCoth[x]]/(a*b)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(-\frac {\ln \left (\operatorname {arccoth}\left (x \right )-\frac {1}{2}\right )}{2 a b}\)	\(14\)
default	\(-\frac {\ln \left (2 b \,\operatorname {arccoth}\left (x \right )-b \right )}{2 a b}\)	\(19\)
risch	\(-\frac {\ln \left (1-\ln \left (1+x \right )+\ln \left (x -1\right )\right )}{2 a b}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (\log \left (\frac {x + 1}{x - 1}\right ) - 1\right )}{2 \, a b} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=- \frac {\log {\left (\operatorname {acoth}{\left (x \right )} - \frac {1}{2} \right )}}{2 a b} \]

[In]

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

[Out]

-log(acoth(x) - 1/2)/(2*a*b)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (\log \left (x + 1\right ) - \log \left (x - 1\right ) - 1\right )}{2 \, a b} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (15) = 30\).

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.59 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\left (x + 1\right ) \mathrm {sgn}\left (x - 1\right ) - 1\right )}^{2} + {\left (\log \left (\frac {{\left | x + 1 \right |}}{{\left | x - 1 \right |}}\right ) - 1\right )}^{2}\right )}{4 \, a b} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a-a x^2\right ) \left (b-2 b \coth ^{-1}(x)\right )} \, dx=-\frac {\ln \left (2\,\mathrm {acoth}\left (x\right )-1\right )}{2\,a\,b} \]

[In]

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

[Out]

-log(2*acoth(x) - 1)/(2*a*b)