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

Optimal. Leaf size=9 \[ \frac {\log \left (\tanh ^{-1}(a x)\right )}{a} \]

[Out]

ln(arctanh(a*x))/a

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Rubi [A]  time = 0.03, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {5946} \[ \frac {\log \left (\tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)*ArcTanh[a*x]),x]

[Out]

Log[ArcTanh[a*x]]/a

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

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Mathematica [A]  time = 0.03, size = 9, normalized size = 1.00 \[ \frac {\log \left (\tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[ArcTanh[a*x]]/a

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fricas [B]  time = 0.97, size = 20, normalized size = 2.22 \[ \frac {\log \left (\log \left (-\frac {a x + 1}{a x - 1}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(-(a*x + 1)/(a*x - 1)))/a

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giac [B]  time = 0.22, size = 21, normalized size = 2.33 \[ \frac {\log \left ({\left | \log \left (-\frac {a x + 1}{a x - 1}\right ) \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(log(-(a*x + 1)/(a*x - 1))))/a

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maple [A]  time = 0.02, size = 10, normalized size = 1.11 \[ \frac {\ln \left (\arctanh \left (a x \right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(arctanh(a*x))/a

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maxima [B]  time = 0.31, size = 21, normalized size = 2.33 \[ \frac {\log \left (-\log \left (a x + 1\right ) + \log \left (-a x + 1\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-log(a*x + 1) + log(-a*x + 1))/a

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mupad [B]  time = 0.82, size = 9, normalized size = 1.00 \[ \frac {\ln \left (\mathrm {atanh}\left (a\,x\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(atanh(a*x)*(a^2*x^2 - 1)),x)

[Out]

log(atanh(a*x))/a

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sympy [A]  time = 0.64, size = 7, normalized size = 0.78 \[ \frac {\log {\left (\operatorname {atanh}{\left (a x \right )} \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(atanh(a*x))/a

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