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

Optimal. Leaf size=104 \[ \frac {3 \text {Li}_4\left (1-\frac {2}{a x+1}\right )}{4 a c}+\frac {3 \text {Li}_2\left (1-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{2 a c}+\frac {3 \text {Li}_3\left (1-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)}{2 a c}-\frac {\log \left (\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{a c} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 104, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.09, size = 82, normalized size = 0.79 \[ \frac {6 \tanh ^{-1}(a x)^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+6 \tanh ^{-1}(a x) \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {Li}_4\left (-e^{-2 \tanh ^{-1}(a x)}\right )-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{4 a c} \]

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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )^{3}}{a c x + c}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (a x\right )^{3}}{a c x + c}\,{d x} \]

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)

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maple [C]  time = 0.28, size = 703, normalized size = 6.76 \[ \frac {\arctanh \left (a x \right )^{3} \ln \left (a x +1\right )}{a c}-\frac {2 \arctanh \left (a x \right )^{3} \ln \left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{a c}+\frac {\arctanh \left (a x \right )^{4}}{2 a c}-\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a c}-\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4 a c}+\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right ) \pi }{2 a c}-\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{2} \pi }{2 a c}-\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right )^{3} \pi }{2 a c}-\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right )^{2} \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \pi }{a c}+\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{2} \pi }{2 a c}-\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{a^{2} x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )^{2} \pi }{2 a c}-\frac {i \arctanh \left (a x \right )^{3} \mathrm {csgn}\left (\frac {i \left (a x +1\right )^{2}}{\left (a^{2} x^{2}-1\right ) \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}\right )^{3} \pi }{2 a c}-\frac {\arctanh \left (a x \right )^{3} \ln \relax (2)}{a c} \]

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)^4-3/2/
a/c*arctanh(a*x)^2*polylog(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))+1/2*I/a/c*arctanh(a*x)^3*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*
(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))*Pi-1/2*I/a/c*arctanh(a*x)^3*cs
gn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi-1/2*I/a/c*arcta
nh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*Pi-I/a/c*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+1/2*I/a/c*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)/
(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi-1/2*I/a/c*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-1/2*I/a/c*arctanh(a*x)^3*csgn(I*(a*x+1)^2/(a^2*x^2-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*Pi-1/a/c
*arctanh(a*x)^3*ln(2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3}}{8 \, a c} + \frac {1}{8} \, \int \frac {6 \, a x \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2} + {\left (a x - 1\right )} \log \left (a x + 1\right )^{3} - 3 \, {\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )}{a^{2} c x^{2} - c}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{c+a\,c\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^3/(c + a*c*x),x)

[Out]

int(atanh(a*x)^3/(c + a*c*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a x + 1}\, dx}{c} \]

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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