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

Optimal. Leaf size=13 \[ \frac {\tanh ^{-1}(a x)^3}{3 a} \]

[Out]

1/3*arctanh(a*x)^3/a

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Rubi [A]  time = 0.03, 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 = {5948} \[ \frac {\tanh ^{-1}(a x)^3}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/(1 - a^2*x^2),x]

[Out]

ArcTanh[a*x]^3/(3*a)

Rule 5948

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

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {\tanh ^{-1}(a x)^3}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[a*x]^3/(3*a)

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fricas [A]  time = 0.54, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{3}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*log(-(a*x + 1)/(a*x - 1))^3/a

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giac [A]  time = 0.19, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{3}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*log(-(a*x + 1)/(a*x - 1))^3/a

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maple [A]  time = 0.02, size = 12, normalized size = 0.92 \[ \frac {\arctanh \left (a x \right )^{3}}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*arctanh(a*x)^3/a

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maxima [B]  time = 0.33, size = 127, normalized size = 9.77 \[ \frac {1}{2} \, {\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{2} - \frac {{\left (\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2}\right )} \operatorname {artanh}\left (a x\right )}{4 \, a} + \frac {\log \left (a x + 1\right )^{3} - 3 \, \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 3 \, \log \left (a x + 1\right ) \log \left (a x - 1\right )^{2} - \log \left (a x - 1\right )^{3}}{24 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(a*x + 1)/a - log(a*x - 1)/a)*arctanh(a*x)^2 - 1/4*(log(a*x + 1)^2 - 2*log(a*x + 1)*log(a*x - 1) + log
(a*x - 1)^2)*arctanh(a*x)/a + 1/24*(log(a*x + 1)^3 - 3*log(a*x + 1)^2*log(a*x - 1) + 3*log(a*x + 1)*log(a*x -
1)^2 - log(a*x - 1)^3)/a

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mupad [B]  time = 0.95, size = 68, normalized size = 5.23 \[ \frac {{\ln \left (a\,x+1\right )}^3}{24\,a}-\frac {{\ln \left (1-a\,x\right )}^3}{24\,a}+\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^2}{8\,a}-\frac {{\ln \left (a\,x+1\right )}^2\,\ln \left (1-a\,x\right )}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^2/(a^2*x^2 - 1),x)

[Out]

log(a*x + 1)^3/(24*a) - log(1 - a*x)^3/(24*a) + (log(a*x + 1)*log(1 - a*x)^2)/(8*a) - (log(a*x + 1)^2*log(1 -
a*x))/(8*a)

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sympy [A]  time = 0.93, size = 10, normalized size = 0.77 \[ \begin {cases} \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((atanh(a*x)**3/(3*a), Ne(a, 0)), (0, True))

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