3.3.56 \(\int \frac {1}{(1-a^2 x^2) \tanh ^{-1}(a x)^3} \, dx\) [256]

Optimal. Leaf size=13 \[ -\frac {1}{2 a \tanh ^{-1}(a x)^2} \]

[Out]

-1/2/a/arctanh(a*x)^2

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Rubi [A]
time = 0.02, antiderivative size = 13, 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}{2 a \tanh ^{-1}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)*ArcTanh[a*x]^3),x]

[Out]

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

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)^3} \, dx &=-\frac {1}{2 a \tanh ^{-1}(a x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 13, normalized size = 1.00 \begin {gather*} -\frac {1}{2 a \tanh ^{-1}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)*ArcTanh[a*x]^3),x]

[Out]

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

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Maple [A]
time = 0.32, size = 12, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)/arctanh(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/a/arctanh(a*x)^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (11) = 22\).
time = 0.26, size = 42, normalized size = 3.23 \begin {gather*} -\frac {2}{a \log \left (a x + 1\right )^{2} - 2 \, a \log \left (a x + 1\right ) \log \left (-a x + 1\right ) + a \log \left (-a x + 1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)/arctanh(a*x)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 22, normalized size = 1.69 \begin {gather*} -\frac {2}{a \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)/arctanh(a*x)^3,x, algorithm="fricas")

[Out]

-2/(a*log(-(a*x + 1)/(a*x - 1))^2)

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Sympy [A]
time = 0.90, size = 12, normalized size = 0.92 \begin {gather*} - \frac {1}{2 a \operatorname {atanh}^{2}{\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)**3,x)

[Out]

-1/(2*a*atanh(a*x)**2)

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Giac [A]
time = 0.39, size = 22, normalized size = 1.69 \begin {gather*} -\frac {2}{a \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)/arctanh(a*x)^3,x, algorithm="giac")

[Out]

-2/(a*log(-(a*x + 1)/(a*x - 1))^2)

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Mupad [B]
time = 0.80, size = 23, normalized size = 1.77 \begin {gather*} -\frac {2}{a\,{\left (\ln \left (a\,x+1\right )-\ln \left (1-a\,x\right )\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(atanh(a*x)^3*(a^2*x^2 - 1)),x)

[Out]

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

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