3.2.35 \(\int \frac {x \tanh ^{-1}(a x)^4}{c-a c x} \, dx\) [135]

Optimal. Leaf size=261 \[ -\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {4 \tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {6 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \text {PolyLog}\left (5,1-\frac {2}{1-a x}\right )}{2 a^2 c} \]

[Out]

-arctanh(a*x)^4/a^2/c-x*arctanh(a*x)^4/a/c+4*arctanh(a*x)^3*ln(2/(-a*x+1))/a^2/c+arctanh(a*x)^4*ln(2/(-a*x+1))
/a^2/c+6*arctanh(a*x)^2*polylog(2,1-2/(-a*x+1))/a^2/c+2*arctanh(a*x)^3*polylog(2,1-2/(-a*x+1))/a^2/c-6*arctanh
(a*x)*polylog(3,1-2/(-a*x+1))/a^2/c-3*arctanh(a*x)^2*polylog(3,1-2/(-a*x+1))/a^2/c+3*polylog(4,1-2/(-a*x+1))/a
^2/c+3*arctanh(a*x)*polylog(4,1-2/(-a*x+1))/a^2/c-3/2*polylog(5,1-2/(-a*x+1))/a^2/c

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Rubi [A]
time = 0.36, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {6077, 6021, 6131, 6055, 6095, 6205, 6209, 6745} \begin {gather*} \frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a^2 c}+\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^2 c}+\frac {6 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^2 c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^2 c}-\frac {6 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^2 c}-\frac {\tanh ^{-1}(a x)^4}{a^2 c}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a^2 c}+\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*ArcTanh[a*x]^4)/(c - a*c*x),x]

[Out]

-(ArcTanh[a*x]^4/(a^2*c)) - (x*ArcTanh[a*x]^4)/(a*c) + (4*ArcTanh[a*x]^3*Log[2/(1 - a*x)])/(a^2*c) + (ArcTanh[
a*x]^4*Log[2/(1 - a*x)])/(a^2*c) + (6*ArcTanh[a*x]^2*PolyLog[2, 1 - 2/(1 - a*x)])/(a^2*c) + (2*ArcTanh[a*x]^3*
PolyLog[2, 1 - 2/(1 - a*x)])/(a^2*c) - (6*ArcTanh[a*x]*PolyLog[3, 1 - 2/(1 - a*x)])/(a^2*c) - (3*ArcTanh[a*x]^
2*PolyLog[3, 1 - 2/(1 - a*x)])/(a^2*c) + (3*PolyLog[4, 1 - 2/(1 - a*x)])/(a^2*c) + (3*ArcTanh[a*x]*PolyLog[4,
1 - 2/(1 - a*x)])/(a^2*c) - (3*PolyLog[5, 1 - 2/(1 - a*x)])/(2*a^2*c)

Rule 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 1])

Rule 6055

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 6077

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d*(f/e), Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d
+ e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 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 6131

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6205

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 6209

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 {x \tanh ^{-1}(a x)^4}{c-a c x} \, dx &=\frac {\int \frac {\tanh ^{-1}(a x)^4}{c-a c x} \, dx}{a}-\frac {\int \tanh ^{-1}(a x)^4 \, dx}{a c}\\ &=-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {4 \int \frac {x \tanh ^{-1}(a x)^3}{1-a^2 x^2} \, dx}{c}-\frac {4 \int \frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a c}\\ &=-\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {4 \int \frac {\tanh ^{-1}(a x)^3}{1-a x} \, dx}{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}{a c}\\ &=-\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {4 \tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 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}{a c}-\frac {12 \int \frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a c}\\ &=-\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {4 \tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {6 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \int \frac {\text {Li}_4\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a c}-\frac {12 \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a c}\\ &=-\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {4 \tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {6 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {6 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a^2 c}+\frac {6 \int \frac {\text {Li}_3\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a c}\\ &=-\frac {\tanh ^{-1}(a x)^4}{a^2 c}-\frac {x \tanh ^{-1}(a x)^4}{a c}+\frac {4 \tanh ^{-1}(a x)^3 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {\tanh ^{-1}(a x)^4 \log \left (\frac {2}{1-a x}\right )}{a^2 c}+\frac {6 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {6 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^2 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^2 c}-\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a^2 c}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 172, normalized size = 0.66 \begin {gather*} -\frac {-\tanh ^{-1}(a x)^4+a x \tanh ^{-1}(a x)^4-\frac {2}{5} \tanh ^{-1}(a x)^5-4 \tanh ^{-1}(a x)^3 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-\tanh ^{-1}(a x)^4 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )+2 \tanh ^{-1}(a x)^2 \left (3+\tanh ^{-1}(a x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x) \left (2+\tanh ^{-1}(a x)\right ) \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x) \text {PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {PolyLog}\left (5,-e^{-2 \tanh ^{-1}(a x)}\right )}{a^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 7.06, size = 381, normalized size = 1.46

method	result	size
derivativedivides	\(\frac {-\frac {\arctanh \left (a x \right )^{4} a x}{c}-\frac {\arctanh \left (a x \right )^{4} \ln \left (a x -1\right )}{c}+\frac {2 \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-3 \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+3 \arctanh \left (a x \right ) \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-\frac {3 \polylog \left (5, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}+i \pi \arctanh \left (a x \right )^{4}-i \pi \mathrm {csgn}\left (\frac {i}{\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1}\right )^{2} \arctanh \left (a x \right )^{4}-\arctanh \left (a x \right )^{4}+4 \arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+6 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-6 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+\ln \left (2\right ) \arctanh \left (a x \right )^{4}+i \pi \mathrm {csgn}\left (\frac {i}{\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1}\right )^{3} \arctanh \left (a x \right )^{4}}{c}}{a^{2}}\)	\(381\)
default	\(\frac {-\frac {\arctanh \left (a x \right )^{4} a x}{c}-\frac {\arctanh \left (a x \right )^{4} \ln \left (a x -1\right )}{c}+\frac {2 \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-3 \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+3 \arctanh \left (a x \right ) \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-\frac {3 \polylog \left (5, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2}+i \pi \arctanh \left (a x \right )^{4}-i \pi \mathrm {csgn}\left (\frac {i}{\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1}\right )^{2} \arctanh \left (a x \right )^{4}-\arctanh \left (a x \right )^{4}+4 \arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )+6 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )-6 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )+\ln \left (2\right ) \arctanh \left (a x \right )^{4}+i \pi \mathrm {csgn}\left (\frac {i}{\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1}\right )^{3} \arctanh \left (a x \right )^{4}}{c}}{a^{2}}\)	\(381\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-1/c*arctanh(a*x)^4*a*x-1/c*arctanh(a*x)^4*ln(a*x-1)+4/c*(1/2*arctanh(a*x)^3*polylog(2,-(a*x+1)^2/(-a^2
*x^2+1))-3/4*arctanh(a*x)^2*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))+3/4*arctanh(a*x)*polylog(4,-(a*x+1)^2/(-a^2*x^2
+1))-3/8*polylog(5,-(a*x+1)^2/(-a^2*x^2+1))+1/4*I*Pi*arctanh(a*x)^4-1/4*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1)
)^2*arctanh(a*x)^4-1/4*arctanh(a*x)^4+arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)+1)+3/2*arctanh(a*x)^2*polylog(2
,-(a*x+1)^2/(-a^2*x^2+1))-3/2*arctanh(a*x)*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))+3/4*polylog(4,-(a*x+1)^2/(-a^2*x
^2+1))+1/4*ln(2)*arctanh(a*x)^4+1/4*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^4))

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

[Out]

-1/80*(log(-a*x + 1)^5 + 5*(log(-a*x + 1)^4 - 4*log(-a*x + 1)^3 + 12*log(-a*x + 1)^2 - 24*log(-a*x + 1) + 24)*
(a*x - 1))/(a^2*c) + 1/16*integrate(-(x*log(a*x + 1)^4 - 4*x*log(a*x + 1)^3*log(-a*x + 1) + 6*x*log(a*x + 1)^2
*log(-a*x + 1)^2 - 4*x*log(a*x + 1)*log(-a*x + 1)^3)/(a*c*x - c), 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(x*arctanh(a*x)^4/(-a*c*x+c),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*atanh(a*x)^4)/(c - a*c*x),x)

[Out]

int((x*atanh(a*x)^4)/(c - a*c*x), x)

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