3.3.45 \(\int \frac {\tanh ^{-1}(a x)^3}{x (1-a^2 x^2)} \, dx\) [245]

Optimal. Leaf size=91 \[ \frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {PolyLog}\left (3,-1+\frac {2}{1+a x}\right )-\frac {3}{4} \text {PolyLog}\left (4,-1+\frac {2}{1+a x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6135, 6079, 6095, 6203, 6207, 6745} \begin {gather*} -\frac {3}{4} \text {Li}_4\left (\frac {2}{a x+1}-1\right )-\frac {3}{2} \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2-\frac {3}{2} \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)+\frac {1}{4} \tanh ^{-1}(a x)^4+\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 6207

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-(a
+ b*ArcTanh[c*x])^p)*(PolyLog[k + 1, u]/(2*c*d)), x] + Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog
[k + 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[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 \left (1-a^2 x^2\right )} \, dx &=\frac {1}{4} \tanh ^{-1}(a x)^4+\int \frac {\tanh ^{-1}(a x)^3}{x (1+a x)} \, dx\\ &=\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-(3 a) \int \frac {\tanh ^{-1}(a x)^2 \log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )+(3 a) \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )+\frac {1}{2} (3 a) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1+a x}\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1+a x}\right )-\frac {3}{4} \text {Li}_4\left (-1+\frac {2}{1+a x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 83, normalized size = 0.91 \begin {gather*} -\frac {1}{4} \tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-\frac {3}{2} \tanh ^{-1}(a x) \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{4} \text {PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 81.98, size = 1245, normalized size = 13.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1245\)
default	\(\text {Expression too large to display}\)	\(1245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(-a^2*x^2+1),x, algorithm="maxima")

[Out]

1/16*log(a*x + 1)*log(-a*x + 1)^3 + 1/64*log(-a*x + 1)^4 - 1/8*integrate(1/2*(3*(a^2*x^2 + a*x + 2)*log(a*x +
1)*log(-a*x + 1)^2 + 2*log(a*x + 1)^3 - 6*log(a*x + 1)^2*log(-a*x + 1))/(a^2*x^3 - x), 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/(-a^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-arctanh(a*x)^3/(a^2*x^3 - x), 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^{3} - x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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