3.136 \(\int \frac {\tanh ^{-1}(a x)^4}{c-a c x} \, dx\)

Optimal. Leaf size=131 \[ -\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a c}+\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a c}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a c} \]

[Out]

arctanh(a*x)^4*ln(2/(-a*x+1))/a/c+2*arctanh(a*x)^3*polylog(2,1-2/(-a*x+1))/a/c-3*arctanh(a*x)^2*polylog(3,1-2/
(-a*x+1))/a/c+3*arctanh(a*x)*polylog(4,1-2/(-a*x+1))/a/c-3/2*polylog(5,1-2/(-a*x+1))/a/c

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Rubi [A]  time = 0.21, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5918, 5948, 6058, 6062, 6610} \[ -\frac {3 \text {PolyLog}\left (5,1-\frac {2}{1-a x}\right )}{2 a c}+\frac {2 \tanh ^{-1}(a x)^3 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a c}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{a c}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^4/(c - a*c*x),x]

[Out]

(ArcTanh[a*x]^4*Log[2/(1 - a*x)])/(a*c) + (2*ArcTanh[a*x]^3*PolyLog[2, 1 - 2/(1 - a*x)])/(a*c) - (3*ArcTanh[a*
x]^2*PolyLog[3, 1 - 2/(1 - a*x)])/(a*c) + (3*ArcTanh[a*x]*PolyLog[4, 1 - 2/(1 - a*x)])/(a*c) - (3*PolyLog[5, 1
- 2/(1 - a*x)])/(2*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 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 6062

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)^4}{c-a c x} \, dx &=\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a c}-\frac {4 \int \frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {6 \int \frac {\tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a c}+\frac {6 \int \frac {\tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \int \frac {\text {Li}_4\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a c}-\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a c}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 112, normalized size = 0.85 \[ -\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x) \text {Li}_4\left (-e^{-2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {Li}_5\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\frac {2}{5} \tanh ^{-1}(a x)^5-\tanh ^{-1}(a x)^4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{a c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^4/(c - a*c*x),x]

[Out]

-(((-2*ArcTanh[a*x]^5)/5 - ArcTanh[a*x]^4*Log[1 + E^(-2*ArcTanh[a*x])] + 2*ArcTanh[a*x]^3*PolyLog[2, -E^(-2*Ar
cTanh[a*x])] + 3*ArcTanh[a*x]^2*PolyLog[3, -E^(-2*ArcTanh[a*x])] + 3*ArcTanh[a*x]*PolyLog[4, -E^(-2*ArcTanh[a*
x])] + (3*PolyLog[5, -E^(-2*ArcTanh[a*x])])/2)/(a*c))

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{4}}{a c x - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^4/(-a*c*x+c),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (a x\right )^{4}}{a c x - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^4/(-a*c*x+c),x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)^4/(a*c*x - c), x)

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maple [C]  time = 0.36, size = 285, normalized size = 2.18 \[ -\frac {\arctanh \left (a x \right )^{4} \ln \left (a x -1\right )}{a c}+\frac {i \arctanh \left (a x \right )^{4} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{3}}{a c}-\frac {i \arctanh \left (a x \right )^{4} \pi \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2}}{a c}+\frac {i \arctanh \left (a x \right )^{4} \pi }{a c}+\frac {\arctanh \left (a x \right )^{4} \ln \relax (2)}{a c}+\frac {2 \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a c}-\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a c}-\frac {3 \polylog \left (5, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^4/(-a*c*x+c),x)

[Out]

-1/a/c*arctanh(a*x)^4*ln(a*x-1)+I/a/c*arctanh(a*x)^4*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^3-I/a/c*arctanh(a*x
)^4*Pi*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2+I/a/c*arctanh(a*x)^4*Pi+1/a/c*arctanh(a*x)^4*ln(2)+2/a/c*arctanh(a
*x)^3*polylog(2,-(a*x+1)^2/(-a^2*x^2+1))-3/a/c*arctanh(a*x)^2*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))+3/a/c*arctanh
(a*x)*polylog(4,-(a*x+1)^2/(-a^2*x^2+1))-3/2/a/c*polylog(5,-(a*x+1)^2/(-a^2*x^2+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (-a x + 1\right )^{5}}{80 \, a c} + \frac {1}{16} \, \int -\frac {\log \left (a x + 1\right )^{4} - 4 \, \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right ) + 6 \, \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{2} - 4 \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3}}{a c x - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^4/(-a*c*x+c),x, algorithm="maxima")

[Out]

-1/80*log(-a*x + 1)^5/(a*c) + 1/16*integrate(-(log(a*x + 1)^4 - 4*log(a*x + 1)^3*log(-a*x + 1) + 6*log(a*x + 1
)^2*log(-a*x + 1)^2 - 4*log(a*x + 1)*log(-a*x + 1)^3)/(a*c*x - c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^4}{c-a\,c\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^4/(c - a*c*x),x)

[Out]

int(atanh(a*x)^4/(c - a*c*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\operatorname {atanh}^{4}{\left (a x \right )}}{a x - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**4/(-a*c*x+c),x)

[Out]

-Integral(atanh(a*x)**4/(a*x - 1), x)/c

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