3.244 \(\int \frac {\tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {\tanh ^{-1}(a x)^4}{4 a} \]

[Out]

1/4*arctanh(a*x)^4/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 = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {5948} \[ \frac {\tanh ^{-1}(a x)^4}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(1 - a^2*x^2),x]

[Out]

ArcTanh[a*x]^4/(4*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)^3}{1-a^2 x^2} \, dx &=\frac {\tanh ^{-1}(a x)^4}{4 a}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ \frac {\tanh ^{-1}(a x)^4}{4 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[a*x]^4/(4*a)

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fricas [A]  time = 0.75, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{4}}{64 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*log(-(a*x + 1)/(a*x - 1))^4/a

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giac [A]  time = 0.14, size = 22, normalized size = 1.69 \[ \frac {\log \left (-\frac {a x + 1}{a x - 1}\right )^{4}}{64 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*log(-(a*x + 1)/(a*x - 1))^4/a

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maple [A]  time = 0.02, size = 12, normalized size = 0.92 \[ \frac {\arctanh \left (a x \right )^{4}}{4 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arctanh(a*x)^4/a

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maxima [B]  time = 0.33, size = 209, normalized size = 16.08 \[ \frac {1}{2} \, {\left (\frac {\log \left (a x + 1\right )}{a} - \frac {\log \left (a x - 1\right )}{a}\right )} \operatorname {artanh}\left (a x\right )^{3} + \frac {1}{64} \, a {\left (\frac {8 \, {\left (\log \left (a x + 1\right )^{3} - 3 \, \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) + 3 \, \log \left (a x + 1\right ) \log \left (a x - 1\right )^{2} - \log \left (a x - 1\right )^{3}\right )} \operatorname {artanh}\left (a x\right )}{a^{2}} - \frac {\log \left (a x + 1\right )^{4} - 4 \, \log \left (a x + 1\right )^{3} \log \left (a x - 1\right ) + 6 \, \log \left (a x + 1\right )^{2} \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x + 1\right ) \log \left (a x - 1\right )^{3} + \log \left (a x - 1\right )^{4}}{a^{2}}\right )} - \frac {3 \, {\left (\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2}\right )} \operatorname {artanh}\left (a x\right )^{2}}{8 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(a*x + 1)/a - log(a*x - 1)/a)*arctanh(a*x)^3 + 1/64*a*(8*(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)*arctanh(a*x)/a^2 - (log(a*x + 1)^4 - 4*log(a*x + 1)^3*log
(a*x - 1) + 6*log(a*x + 1)^2*log(a*x - 1)^2 - 4*log(a*x + 1)*log(a*x - 1)^3 + log(a*x - 1)^4)/a^2) - 3/8*(log(
a*x + 1)^2 - 2*log(a*x + 1)*log(a*x - 1) + log(a*x - 1)^2)*arctanh(a*x)^2/a

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mupad [B]  time = 0.90, size = 90, normalized size = 6.92 \[ \frac {{\ln \left (a\,x+1\right )}^4}{64\,a}+\frac {{\ln \left (1-a\,x\right )}^4}{64\,a}-\frac {\ln \left (a\,x+1\right )\,{\ln \left (1-a\,x\right )}^3}{16\,a}-\frac {{\ln \left (a\,x+1\right )}^3\,\ln \left (1-a\,x\right )}{16\,a}+\frac {3\,{\ln \left (a\,x+1\right )}^2\,{\ln \left (1-a\,x\right )}^2}{32\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^3/(a^2*x^2 - 1),x)

[Out]

log(a*x + 1)^4/(64*a) + log(1 - a*x)^4/(64*a) - (log(a*x + 1)*log(1 - a*x)^3)/(16*a) - (log(a*x + 1)^3*log(1 -
a*x))/(16*a) + (3*log(a*x + 1)^2*log(1 - a*x)^2)/(32*a)

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sympy [A]  time = 0.96, size = 10, normalized size = 0.77 \[ \begin {cases} \frac {\operatorname {atanh}^{4}{\left (a x \right )}}{4 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**3/(-a**2*x**2+1),x)

[Out]

Piecewise((atanh(a*x)**4/(4*a), Ne(a, 0)), (0, True))

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