3.3.53 \(\int \frac {1}{(1-a^2 x^2) \tanh ^{-1}(a x)^2} \, dx\) [253]

Optimal. Leaf size=11 \[ -\frac {1}{a \tanh ^{-1}(a x)} \]

[Out]

-1/a/arctanh(a*x)

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Rubi [A]
time = 0.02, antiderivative size = 11, 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 = {6095} \begin {gather*} -\frac {1}{a \tanh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*ArcTanh[a*x]))

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{a \tanh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*ArcTanh[a*x]))

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Maple [A]
time = 0.28, size = 12, normalized size = 1.09

method	result	size
derivativedivides	\(-\frac {1}{a \arctanh \left (a x \right )}\)	\(12\)
default	\(-\frac {1}{a \arctanh \left (a x \right )}\)	\(12\)
risch	\(\frac {2}{a \left (-\ln \left (a x +1\right )+\ln \left (-a x +1\right )\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/arctanh(a*x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (11) = 22\).
time = 0.26, size = 23, normalized size = 2.09 \begin {gather*} -\frac {2}{a \log \left (a x + 1\right ) - a \log \left (-a x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*log(a*x + 1) - a*log(-a*x + 1))

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Fricas [A]
time = 0.36, size = 22, normalized size = 2.00 \begin {gather*} -\frac {2}{a \log \left (-\frac {a x + 1}{a x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*log(-(a*x + 1)/(a*x - 1)))

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Sympy [A]
time = 0.47, size = 8, normalized size = 0.73 \begin {gather*} - \frac {1}{a \operatorname {atanh}{\left (a x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a*atanh(a*x))

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Giac [A]
time = 0.39, size = 22, normalized size = 2.00 \begin {gather*} -\frac {2}{a \log \left (-\frac {a x + 1}{a x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*log(-(a*x + 1)/(a*x - 1)))

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Mupad [B]
time = 0.78, size = 23, normalized size = 2.09 \begin {gather*} -\frac {2}{a\,\left (\ln \left (a\,x+1\right )-\ln \left (1-a\,x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(a*(log(a*x + 1) - log(1 - a*x)))

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