3.129 \(\int \frac{\tanh ^{-1}(a x)^3}{c+a c x} \, dx\)

Optimal. Leaf size=104 \[ \frac{3 \text{PolyLog}\left (4,1-\frac{2}{a x+1}\right )}{4 a c}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{a x+1}\right )}{2 a c}+\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{a x+1}\right )}{2 a c}-\frac{\log \left (\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{a c} \]

[Out]

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

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Rubi [A]  time = 0.164142, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5918, 5948, 6056, 6060, 6610} \[ \frac{3 \text{PolyLog}\left (4,1-\frac{2}{a x+1}\right )}{4 a c}+\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{a x+1}\right )}{2 a c}+\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{a x+1}\right )}{2 a c}-\frac{\log \left (\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{a c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(c + a*c*x),x]

[Out]

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

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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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 6060

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 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]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)^3}{c+a c x} \, dx &=-\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1+a x}\right )}{a c}+\frac{3 \int \frac{\tanh ^{-1}(a x)^2 \log \left (\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=-\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1+a x}\right )}{a c}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+a x}\right )}{2 a c}-\frac{3 \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (1-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=-\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1+a x}\right )}{a c}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+a x}\right )}{2 a c}+\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+a x}\right )}{2 a c}-\frac{3 \int \frac{\text{Li}_3\left (1-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{2 c}\\ &=-\frac{\tanh ^{-1}(a x)^3 \log \left (\frac{2}{1+a x}\right )}{a c}+\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (1-\frac{2}{1+a x}\right )}{2 a c}+\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (1-\frac{2}{1+a x}\right )}{2 a c}+\frac{3 \text{Li}_4\left (1-\frac{2}{1+a x}\right )}{4 a c}\\ \end{align*}

Mathematica [A]  time = 0.0836102, size = 82, normalized size = 0.79 \[ \frac{6 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text{PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{4 a c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^3/(c + a*c*x),x]

[Out]

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

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Maple [C]  time = 0.191, size = 703, normalized size = 6.8 \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*c*x+c),x)

[Out]

1/a/c*arctanh(a*x)^3*ln(a*x+1)-2/a/c*arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/2/a/c*arctanh(a*x)^4+1/2*
I/a/c*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))-1/2*I/a/c*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))^2-1/2*I/a/c*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3-I/
a/c*arctanh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*I/a/c*arctanh(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/a/c*arct
anh(a*x)^3*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2-1/2*I/a/c*arctanh(a*x)^3*Pi*c
sgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3-1/a/c*arctanh(a*x)^3*ln(2)-3/2/a/c*arctanh(a*x)^2*po
lylog(2,-(a*x+1)^2/(-a^2*x^2+1))+3/2/a/c*arctanh(a*x)*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))-3/4/a/c*polylog(4,-(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{\log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3}}{8 \, a c} + \frac{1}{8} \, \int \frac{6 \, a x \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} +{\left (a x - 1\right )} \log \left (a x + 1\right )^{3} - 3 \,{\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )}{a^{2} c x^{2} - c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(a*x)^3/(a*c*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**3/(a*c*x+c),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctanh(a*x)^3/(a*c*x + c), x)