3.3.46 \(\int \frac {\tanh ^{-1}(a x)^3}{x^2 (1-a^2 x^2)} \, dx\) [246]

Optimal. Leaf size=90 \[ a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-3 a \tanh ^{-1}(a x) \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right )-\frac {3}{2} a \text {PolyLog}\left (3,-1+\frac {2}{1+a x}\right ) \]

[Out]

a*arctanh(a*x)^3-arctanh(a*x)^3/x+1/4*a*arctanh(a*x)^4+3*a*arctanh(a*x)^2*ln(2-2/(a*x+1))-3*a*arctanh(a*x)*pol
ylog(2,-1+2/(a*x+1))-3/2*a*polylog(3,-1+2/(a*x+1))

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Rubi [A]
time = 0.20, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6129, 6037, 6135, 6079, 6095, 6203, 6745} \begin {gather*} -\frac {3}{2} a \text {Li}_3\left (\frac {2}{a x+1}-1\right )-3 a \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{4} a \tanh ^{-1}(a x)^4+a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+3 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*ArcTanh[a*x]^3 - ArcTanh[a*x]^3/x + (a*ArcTanh[a*x]^4)/4 + 3*a*ArcTanh[a*x]^2*Log[2 - 2/(1 + a*x)] - 3*a*Arc
Tanh[a*x]*PolyLog[2, -1 + 2/(1 + a*x)] - (3*a*PolyLog[3, -1 + 2/(1 + a*x)])/2

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6095

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 6129

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

Rule 6135

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

Rule 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6745

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]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)^3}{x^2 \left (1-a^2 x^2\right )} \, dx &=a^2 \int \frac {\tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx+\int \frac {\tanh ^{-1}(a x)^3}{x^2} \, dx\\ &=-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+(3 a) \int \frac {\tanh ^{-1}(a x)^2}{x \left (1-a^2 x^2\right )} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+(3 a) \int \frac {\tanh ^{-1}(a x)^2}{x (1+a x)} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-\left (6 a^2\right ) \int \frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-3 a \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+\left (3 a^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=a \tanh ^{-1}(a x)^3-\frac {\tanh ^{-1}(a x)^3}{x}+\frac {1}{4} a \tanh ^{-1}(a x)^4+3 a \tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )-3 a \tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} a \text {Li}_3\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.15, size = 93, normalized size = 1.03 \begin {gather*} -a \left (-\frac {i \pi ^3}{8}+\tanh ^{-1}(a x)^3+\frac {\tanh ^{-1}(a x)^3}{a x}-\frac {1}{4} \tanh ^{-1}(a x)^4-3 \tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*((-1/8*I)*Pi^3 + ArcTanh[a*x]^3 + ArcTanh[a*x]^3/(a*x) - ArcTanh[a*x]^4/4 - 3*ArcTanh[a*x]^2*Log[1 - E^(2*
ArcTanh[a*x])] - 3*ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])] + (3*PolyLog[3, E^(2*ArcTanh[a*x])])/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 67.87, size = 810, normalized size = 9.00 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(1/2*arctanh(a*x)^3*ln(a*x+1)-arctanh(a*x)^3/x/a-1/2*arctanh(a*x)^3*ln(a*x-1)+1/2*I*arctanh(a*x)^3*Pi+1/4*I*
Pi*csgn(I/((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)/((a*x+1)^2/(-
a^2*x^2+1)+1))*arctanh(a*x)^3-1/4*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1
)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3-1/4*I*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-
1))*arctanh(a*x)^3-6*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))-6*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/4*arctan
h(a*x)^4-arctanh(a*x)^3-1/2*I*Pi*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*arctanh(a*
x)^3+1/4*I*Pi*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*arctanh
(a*x)^3+1/2*I*arctanh(a*x)^3*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3-1/4*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*
arctanh(a*x)^3-1/4*I*Pi*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)^3-1/2*I*arctan
h(a*x)^3*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2-arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+6*arctanh(a*x)*
polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+3*arctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3*arctanh(a*x)^2*ln(1+
(a*x+1)/(-a^2*x^2+1)^(1/2))+6*arctanh(a*x)*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arctanh(a*x)^3/(a^2*x^4 - x^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a^{2} x^{4} - x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a*x)**3/(a**2*x**4 - x**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{x^2\,\left (a^2\,x^2-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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