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

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

[Out]

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

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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 = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {5948} \[ \frac {\tanh ^{-1}(a x)^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2),x]

[Out]

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

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \frac {\tanh ^{-1}(a x)^2}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*log(-(a*x + 1)/(a*x - 1))^2/a

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giac [A]  time = 0.21, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*log(-(a*x + 1)/(a*x - 1))^2/a

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maple [A]  time = 0.02, size = 12, normalized size = 0.92 \[ \frac {\arctanh \left (a x \right )^{2}}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 65, normalized size = 5.00 \[ \frac {1}{2} \, {\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right ) - \frac {\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.87, size = 23, normalized size = 1.77 \[ \frac {{\left (\ln \left (a\,x+1\right )-\ln \left (1-a\,x\right )\right )}^2}{8\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)/(a^2*x^2 - 1),x)

[Out]

(log(a*x + 1) - log(1 - a*x))^2/(8*a)

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sympy [A]  time = 0.90, size = 10, normalized size = 0.77 \[ \begin {cases} \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{2 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((atanh(a*x)**2/(2*a), Ne(a, 0)), (0, True))

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