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

Optimal. Leaf size=115 \[ -\frac{3}{8 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac{3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}+\frac{3 \tanh ^{-1}(a x)^2}{8 a} \]

[Out]

-3/(8*a*(1 - a^2*x^2)) + (3*x*ArcTanh[a*x])/(4*(1 - a^2*x^2)) + (3*ArcTanh[a*x]^2)/(8*a) - (3*ArcTanh[a*x]^2)/
(4*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^3)/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^4/(8*a)

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Rubi [A]  time = 0.0962471, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {5956, 5994, 261} \[ -\frac{3}{8 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}-\frac{3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac{3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}+\frac{3 \tanh ^{-1}(a x)^2}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(1 - a^2*x^2)^2,x]

[Out]

-3/(8*a*(1 - a^2*x^2)) + (3*x*ArcTanh[a*x])/(4*(1 - a^2*x^2)) + (3*ArcTanh[a*x]^2)/(8*a) - (3*ArcTanh[a*x]^2)/
(4*a*(1 - a^2*x^2)) + (x*ArcTanh[a*x]^3)/(2*(1 - a^2*x^2)) + ArcTanh[a*x]^4/(8*a)

Rule 5956

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

Rule 5994

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)^3}{\left (1-a^2 x^2\right )^2} \, dx &=\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}-\frac{1}{2} (3 a) \int \frac{x \tanh ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}+\frac{3}{2} \int \frac{\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^2} \, dx\\ &=\frac{3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac{3 \tanh ^{-1}(a x)^2}{8 a}-\frac{3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}-\frac{1}{4} (3 a) \int \frac{x}{\left (1-a^2 x^2\right )^2} \, dx\\ &=-\frac{3}{8 a \left (1-a^2 x^2\right )}+\frac{3 x \tanh ^{-1}(a x)}{4 \left (1-a^2 x^2\right )}+\frac{3 \tanh ^{-1}(a x)^2}{8 a}-\frac{3 \tanh ^{-1}(a x)^2}{4 a \left (1-a^2 x^2\right )}+\frac{x \tanh ^{-1}(a x)^3}{2 \left (1-a^2 x^2\right )}+\frac{\tanh ^{-1}(a x)^4}{8 a}\\ \end{align*}

Mathematica [A]  time = 0.0577629, size = 71, normalized size = 0.62 \[ \frac{\left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^4+3 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2-4 a x \tanh ^{-1}(a x)^3-6 a x \tanh ^{-1}(a x)+3}{8 a \left (a^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 - 6*a*x*ArcTanh[a*x] + 3*(1 + a^2*x^2)*ArcTanh[a*x]^2 - 4*a*x*ArcTanh[a*x]^3 + (-1 + a^2*x^2)*ArcTanh[a*x]^
4)/(8*a*(-1 + a^2*x^2))

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Maple [C]  time = 0.408, size = 1742, normalized size = 15.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16/(a*x-1)/(a*x+1)/a+1/8*a/(a*x-1)/(a*x+1)*arctanh(a*x)^4*x^2+3/8*a/(a*x-1)/(a*x+1)*arctanh(a*x)^2*x^2-1/4/a
*arctanh(a*x)^3/(a*x-1)-1/4/a*arctanh(a*x)^3*ln(a*x-1)-1/4/a*arctanh(a*x)^3/(a*x+1)+1/4/a*arctanh(a*x)^3*ln(a*
x+1)-1/2/a*arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/((a*x+1)^2/(-a^2*x^2+1
)+1))^2*arctanh(a*x)^3*Pi-3/4/(a*x-1)/(a*x+1)*arctanh(a*x)*x-1/4*I/a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi+1/8*I*a
/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^
2+1)+1))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*Pi*x^2+1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I*(a*x+1)^2/(a^
2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi*x^2-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*
x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*Pi*x^2-1/4*
I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*Pi*x^2-1
/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*Pi*x^
2-1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1
))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^3*Pi+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(
(a*x+1)^2/(-a^2*x^2+1)+1))^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^3*Pi-1/8*I*a/(a*x-1)/(a*x+1)*arct
anh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*Pi*x^2+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I
*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*arctanh(a*x)^3*Pi-1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)
^2/(-a^2*x^2+1)+1))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*arctanh(a*x)^3*Pi-1/8*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*c
sgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi*x^2+1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I
/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi*x^2-1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1)
)^2*Pi*x^2+1/4*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(
a*x)^3*Pi+3/16/(a*x-1)/(a*x+1)*a*x^2-1/8/a/(a*x-1)/(a*x+1)*arctanh(a*x)^4+3/8/a/(a*x-1)/(a*x+1)*arctanh(a*x)^2
+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^3*Pi+1/8*I/a/(a*x-1)/(a*x+1)*csgn(I*(a*x
+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3*Pi+1/4*I*a/(a*x-1)/(a*x+1)*arctanh(a*x)^3*Pi*x^
2-1/4*I/a/(a*x-1)/(a*x+1)*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3*Pi

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Maxima [B]  time = 1.03263, size = 620, normalized size = 5.39 \begin{align*} -\frac{1}{4} \,{\left (\frac{2 \, x}{a^{2} x^{2} - 1} - \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{3 \,{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} - 2 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) \log \left (a x - 1\right ) +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4\right )} a \operatorname{artanh}\left (a x\right )^{2}}{16 \,{\left (a^{4} x^{2} - a^{2}\right )}} - \frac{1}{128} \,{\left (\frac{{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{4} - 4 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} \log \left (a x - 1\right ) +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{4} + 6 \,{\left (2 \, a^{2} x^{2} +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right )^{2} + 12 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 4 \,{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} + 6 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} \log \left (a x + 1\right ) - 48\right )} a^{2}}{a^{6} x^{2} - a^{4}} - \frac{8 \,{\left ({\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{3} - 3 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x + 1\right )^{2} \log \left (a x - 1\right ) -{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{3} - 12 \, a x + 3 \,{\left (2 \, a^{2} x^{2} +{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )^{2} - 2\right )} \log \left (a x + 1\right ) - 6 \,{\left (a^{2} x^{2} - 1\right )} \log \left (a x - 1\right )\right )} a \operatorname{artanh}\left (a x\right )}{a^{5} x^{2} - a^{3}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^2*x^2 - 1) - log(a*x + 1)/a + log(a*x - 1)/a)*arctanh(a*x)^3 - 3/16*((a^2*x^2 - 1)*log(a*x + 1)^2
- 2*(a^2*x^2 - 1)*log(a*x + 1)*log(a*x - 1) + (a^2*x^2 - 1)*log(a*x - 1)^2 - 4)*a*arctanh(a*x)^2/(a^4*x^2 - a
^2) - 1/128*(((a^2*x^2 - 1)*log(a*x + 1)^4 - 4*(a^2*x^2 - 1)*log(a*x + 1)^3*log(a*x - 1) + (a^2*x^2 - 1)*log(a
*x - 1)^4 + 6*(2*a^2*x^2 + (a^2*x^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1)^2 + 12*(a^2*x^2 - 1)*log(a*x - 1)^2
- 4*((a^2*x^2 - 1)*log(a*x - 1)^3 + 6*(a^2*x^2 - 1)*log(a*x - 1))*log(a*x + 1) - 48)*a^2/(a^6*x^2 - a^4) - 8*(
(a^2*x^2 - 1)*log(a*x + 1)^3 - 3*(a^2*x^2 - 1)*log(a*x + 1)^2*log(a*x - 1) - (a^2*x^2 - 1)*log(a*x - 1)^3 - 12
*a*x + 3*(2*a^2*x^2 + (a^2*x^2 - 1)*log(a*x - 1)^2 - 2)*log(a*x + 1) - 6*(a^2*x^2 - 1)*log(a*x - 1))*a*arctanh
(a*x)/(a^5*x^2 - a^3))*a

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Fricas [A]  time = 1.9237, size = 255, normalized size = 2.22 \begin{align*} -\frac{8 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right )^{3} -{\left (a^{2} x^{2} - 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{4} + 48 \, a x \log \left (-\frac{a x + 1}{a x - 1}\right ) - 12 \,{\left (a^{2} x^{2} + 1\right )} \log \left (-\frac{a x + 1}{a x - 1}\right )^{2} - 48}{128 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(8*a*x*log(-(a*x + 1)/(a*x - 1))^3 - (a^2*x^2 - 1)*log(-(a*x + 1)/(a*x - 1))^4 + 48*a*x*log(-(a*x + 1)/
(a*x - 1)) - 12*(a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 48)/(a^3*x^2 - a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{atanh}^{3}{\left (a x \right )}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )^{3}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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