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

Optimal. Leaf size=157 \[ \frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{a}-\frac{2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{2 \tanh ^{-1}(a x)^3}{3 a}-x \tanh ^{-1}(a x)-\frac{2 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a} \]

[Out]

-(x*ArcTanh[a*x]) + ((1 - a^2*x^2)*ArcTanh[a*x]^2)/(2*a) + (2*ArcTanh[a*x]^3)/(3*a) + (2*x*ArcTanh[a*x]^3)/3 +
(x*(1 - a^2*x^2)*ArcTanh[a*x]^3)/3 - (2*ArcTanh[a*x]^2*Log[2/(1 - a*x)])/a - Log[1 - a^2*x^2]/(2*a) - (2*ArcT
anh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/a + PolyLog[3, 1 - 2/(1 - a*x)]/a

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Rubi [A]  time = 0.192105, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.471, Rules used = {5944, 5910, 5984, 5918, 5948, 6058, 6610, 260} \[ \frac{\text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{a}-\frac{2 \tanh ^{-1}(a x) \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{2 \tanh ^{-1}(a x)^3}{3 a}-x \tanh ^{-1}(a x)-\frac{2 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a} \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)*ArcTanh[a*x]^3,x]

[Out]

-(x*ArcTanh[a*x]) + ((1 - a^2*x^2)*ArcTanh[a*x]^2)/(2*a) + (2*ArcTanh[a*x]^3)/(3*a) + (2*x*ArcTanh[a*x]^3)/3 +
(x*(1 - a^2*x^2)*ArcTanh[a*x]^3)/3 - (2*ArcTanh[a*x]^2*Log[2/(1 - a*x)])/a - Log[1 - a^2*x^2]/(2*a) - (2*ArcT
anh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/a + PolyLog[3, 1 - 2/(1 - a*x)]/a

Rule 5944

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

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5984

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

Rule 5918

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

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]

Rule 6058

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rule 260

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

Rubi steps

\begin{align*} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3 \, dx &=\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac{2}{3} \int \tanh ^{-1}(a x)^3 \, dx-\int \tanh ^{-1}(a x) \, dx\\ &=-x \tanh ^{-1}(a x)+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+a \int \frac{x}{1-a^2 x^2} \, dx-(2 a) \int \frac{x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=-x \tanh ^{-1}(a x)+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2 \tanh ^{-1}(a x)^3}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\frac{\log \left (1-a^2 x^2\right )}{2 a}-2 \int \frac{\tanh ^{-1}(a x)^2}{1-a x} \, dx\\ &=-x \tanh ^{-1}(a x)+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2 \tanh ^{-1}(a x)^3}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\frac{2 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}+4 \int \frac{\tanh ^{-1}(a x) \log \left (\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-x \tanh ^{-1}(a x)+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2 \tanh ^{-1}(a x)^3}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\frac{2 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}-\frac{2 \tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a}+2 \int \frac{\text{Li}_2\left (1-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-x \tanh ^{-1}(a x)+\frac{\left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{2 a}+\frac{2 \tanh ^{-1}(a x)^3}{3 a}+\frac{2}{3} x \tanh ^{-1}(a x)^3+\frac{1}{3} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3-\frac{2 \tanh ^{-1}(a x)^2 \log \left (\frac{2}{1-a x}\right )}{a}-\frac{\log \left (1-a^2 x^2\right )}{2 a}-\frac{2 \tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1-a x}\right )}{a}+\frac{\text{Li}_3\left (1-\frac{2}{1-a x}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.293761, size = 134, normalized size = 0.85 \[ -\frac{-12 \tanh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )-6 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \log \left (1-a^2 x^2\right )+2 a^3 x^3 \tanh ^{-1}(a x)^3+3 a^2 x^2 \tanh ^{-1}(a x)^2-6 a x \tanh ^{-1}(a x)^3+4 \tanh ^{-1}(a x)^3-3 \tanh ^{-1}(a x)^2+6 a x \tanh ^{-1}(a x)+12 \tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{6 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(6*a*x*ArcTanh[a*x] - 3*ArcTanh[a*x]^2 + 3*a^2*x^2*ArcTanh[a*x]^2 + 4*ArcTanh[a*x]^3 - 6*a*x*ArcTanh[a*x]^3 +
2*a^3*x^3*ArcTanh[a*x]^3 + 12*ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*x])] + 3*Log[1 - a^2*x^2] - 12*ArcTanh[a
*x]*PolyLog[2, -E^(-2*ArcTanh[a*x])] - 6*PolyLog[3, -E^(-2*ArcTanh[a*x])])/(6*a)

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Maple [C]  time = 0.551, size = 829, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*arctanh(a*x)^3*x^3+x*arctanh(a*x)^3-1/2*a*arctanh(a*x)^2*x^2+1/a*arctanh(a*x)^2*ln(a*x-1)+1/a*arctanh
(a*x)^2*ln(a*x+1)-2/a*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))-1/2*I/a*arctanh(a*x)^2*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-1/2*I/a*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1
)^2/(-a^2*x^2+1)+1))^3*Pi-1/2*I/a*arctanh(a*x)^2*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*cs
gn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*Pi-I/a*arctanh(a*x)^2*Pi+1/2*I/a*arctanh(a*x)^2*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-I/a*arctanh(a*x)^2*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+1/2*I/a*arctanh(a*x)^2*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-1/2*I/a*arctanh(a*x)^2*csgn(I*
(a*x+1)^2/(a^2*x^2-1))^3*Pi+I/a*arctanh(a*x)^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*Pi-I/a*arctanh(a*x)^2*csgn
(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*Pi+2/3*arctanh(a*x)^3/a-2/a*arctanh(a*x)^2*ln(2)-x*arctanh(a*x)+1/2*arctanh(a
*x)^2/a-arctanh(a*x)/a+1/a*ln((a*x+1)^2/(-a^2*x^2+1)+1)-2/a*arctanh(a*x)*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))+1/
a*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - 12 \, a x - 6 \,{\left (a^{3} x^{3} - 3 \, a x - 2\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{2}}{48 \, a} - \frac{{\left (\log \left (-a x + 1\right )^{3} - 3 \, \log \left (-a x + 1\right )^{2} + 6 \, \log \left (-a x + 1\right ) - 6\right )}{\left (a x - 1\right )}}{8 \, a} + \frac{4 \,{\left (9 \, \log \left (-a x + 1\right )^{3} - 9 \, \log \left (-a x + 1\right )^{2} + 6 \, \log \left (-a x + 1\right ) - 2\right )}{\left (a x - 1\right )}^{3} + 27 \,{\left (4 \, \log \left (-a x + 1\right )^{3} - 6 \, \log \left (-a x + 1\right )^{2} + 6 \, \log \left (-a x + 1\right ) - 3\right )}{\left (a x - 1\right )}^{2} + 108 \,{\left (\log \left (-a x + 1\right )^{3} - 3 \, \log \left (-a x + 1\right )^{2} + 6 \, \log \left (-a x + 1\right ) - 6\right )}{\left (a x - 1\right )}}{864 \, a} + \frac{1}{8} \, \int -\frac{3 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right )^{3} +{\left (2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - 9 \,{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right )^{2} - 12 \, a x - 6 \,{\left (a^{3} x^{3} - 3 \, a x - 2\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )}{3 \,{\left (a x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*a^3*x^3 - 3*a^2*x^2 - 12*a*x - 6*(a^3*x^3 - 3*a*x - 2)*log(a*x + 1))*log(-a*x + 1)^2/a - 1/8*(log(-a*x
+ 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1)/a + 1/864*(4*(9*log(-a*x + 1)^3 - 9*log(-a*x + 1)
^2 + 6*log(-a*x + 1) - 2)*(a*x - 1)^3 + 27*(4*log(-a*x + 1)^3 - 6*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 3)*(a*x
- 1)^2 + 108*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1))/a + 1/8*integrate(-1/3*(3*
(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1)^3 + (2*a^3*x^3 - 3*a^2*x^2 - 9*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*
x + 1)^2 - 12*a*x - 6*(a^3*x^3 - 3*a*x - 2)*log(a*x + 1))*log(-a*x + 1))/(a*x - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a**2*x**2*atanh(a*x)**3, x) - Integral(-atanh(a*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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