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

Optimal. Leaf size=93 \[ -\frac {3 \text {Li}_4\left (\frac {2}{a x+1}-1\right )}{4 c}-\frac {3 \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2}{2 c}-\frac {3 \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{2 c}+\frac {\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5932, 5948, 6056, 6060, 6610} \[ -\frac {3 \text {PolyLog}\left (4,\frac {2}{a x+1}-1\right )}{4 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )}{2 c}-\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )}{2 c}+\frac {\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(x*(c + a*c*x)),x]

[Out]

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

Rule 5932

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

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Mathematica [A]  time = 0.15, size = 86, normalized size = 0.92 \[ \frac {96 \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )-96 \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )+48 \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )-32 \tanh ^{-1}(a x)^4+64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\pi ^4}{64 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Pi^4 - 32*ArcTanh[a*x]^4 + 64*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] + 96*ArcTanh[a*x]^2*PolyLog[2, E^(2*
ArcTanh[a*x])] - 96*ArcTanh[a*x]*PolyLog[3, E^(2*ArcTanh[a*x])] + 48*PolyLog[4, E^(2*ArcTanh[a*x])])/(64*c)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )^{3}}{a c x^{2} + c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctanh(a*x)^3/(a*c*x^2 + c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.34, size = 1217, normalized size = 13.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^3/(x*(c + a*c*x)),x)

[Out]

int(atanh(a*x)^3/(x*(c + a*c*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a x^{2} + x}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atanh(a*x)**3/(a*x**2 + x), x)/c

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