3.134 \(\int \frac {x^2 \tanh ^{-1}(a x)^4}{c-a c x} \, dx\)

Optimal. Leaf size=384 \[ -\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right )}{a^3 c}-\frac {3 \text {Li}_5\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {2 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^3 c}+\frac {6 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3 c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3 c}+\frac {6 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3 c}-\frac {6 \text {Li}_3\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3 c}-\frac {\tanh ^{-1}(a x)^4}{2 a^3 c}-\frac {2 \tanh ^{-1}(a x)^3}{a^3 c}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a^3 c}+\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^3 c}+\frac {6 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3 c}-\frac {x \tanh ^{-1}(a x)^4}{a^2 c}-\frac {2 x \tanh ^{-1}(a x)^3}{a^2 c}-\frac {x^2 \tanh ^{-1}(a x)^4}{2 a c} \]

[Out]

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

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Rubi [A]  time = 0.86, antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {5930, 5916, 5980, 5910, 5984, 5918, 5948, 6058, 6610, 6062} \[ -\frac {3 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{a^3 c}-\frac {3 \text {PolyLog}\left (5,1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {2 \tanh ^{-1}(a x)^3 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {6 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^3 c}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {6 \tanh ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^3 c}-\frac {6 \tanh ^{-1}(a x) \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (4,1-\frac {2}{1-a x}\right )}{a^3 c}-\frac {x \tanh ^{-1}(a x)^4}{a^2 c}-\frac {\tanh ^{-1}(a x)^4}{2 a^3 c}-\frac {2 x \tanh ^{-1}(a x)^3}{a^2 c}-\frac {2 \tanh ^{-1}(a x)^3}{a^3 c}+\frac {\log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a^3 c}+\frac {4 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^3 c}+\frac {6 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3 c}-\frac {x^2 \tanh ^{-1}(a x)^4}{2 a c} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTanh[a*x]^4)/(c - a*c*x),x]

[Out]

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

Rule 5910

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

Rule 5916

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

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 5930

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 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 5980

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

Rule 5984

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

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Mathematica [A]  time = 0.41, size = 233, normalized size = 0.61 \[ -\frac {-\frac {1}{2} \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4+2 \left (\tanh ^{-1}(a x)^2+3 \tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x) \text {Li}_4\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \left (\tanh ^{-1}(a x)+1\right )^2 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {Li}_4\left (-e^{-2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {Li}_5\left (-e^{-2 \tanh ^{-1}(a x)}\right )-\frac {2}{5} \tanh ^{-1}(a x)^5+a x \tanh ^{-1}(a x)^4-\tanh ^{-1}(a x)^4+2 a x \tanh ^{-1}(a x)^3-2 \tanh ^{-1}(a x)^3-\tanh ^{-1}(a x)^4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-6 \tanh ^{-1}(a x)^2 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{a^3 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*ArcTanh[a*x]^4)/(c - a*c*x),x]

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {x^{2} \operatorname {artanh}\left (a x\right )^{4}}{a c x - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(a*x)^4/(-a*c*x+c),x, algorithm="fricas")

[Out]

integral(-x^2*arctanh(a*x)^4/(a*c*x - c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{2} \operatorname {artanh}\left (a x\right )^{4}}{a c x - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(a*x)^4/(-a*c*x+c),x, algorithm="giac")

[Out]

integrate(-x^2*arctanh(a*x)^4/(a*c*x - c), x)

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maple [A]  time = 1.98, size = 496, normalized size = 1.29 \[ -\frac {x^{2} \arctanh \left (a x \right )^{4}}{2 a c}-\frac {x \arctanh \left (a x \right )^{4}}{a^{2} c}-\frac {2 x \arctanh \left (a x \right )^{3}}{a^{2} c}-\frac {\arctanh \left (a x \right )^{4}}{2 a^{3} c}-\frac {2 \arctanh \left (a x \right )^{3}}{a^{3} c}+\frac {\arctanh \left (a x \right )^{4} \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {2 \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}-\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}-\frac {3 \polylog \left (5, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 a^{3} c}+\frac {6 \arctanh \left (a x \right )^{2} \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {6 \arctanh \left (a x \right ) \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}-\frac {3 \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {4 \arctanh \left (a x \right )^{3} \ln \left (1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {6 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}-\frac {6 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c}+\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{a^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(a*x)^4/(-a*c*x+c),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {4 \, \log \left (-a x + 1\right )^{5} + 5 \, {\left (2 \, \log \left (-a x + 1\right )^{4} - 4 \, \log \left (-a x + 1\right )^{3} + 6 \, \log \left (-a x + 1\right )^{2} - 6 \, \log \left (-a x + 1\right ) + 3\right )} {\left (a x - 1\right )}^{2} + 40 \, {\left (\log \left (-a x + 1\right )^{4} - 4 \, \log \left (-a x + 1\right )^{3} + 12 \, \log \left (-a x + 1\right )^{2} - 24 \, \log \left (-a x + 1\right ) + 24\right )} {\left (a x - 1\right )}}{320 \, a^{3} c} + \frac {1}{16} \, \int -\frac {x^{2} \log \left (a x + 1\right )^{4} - 4 \, x^{2} \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right ) + 6 \, x^{2} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{2} - 4 \, x^{2} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3}}{a c x - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(a*x)^4/(-a*c*x+c),x, algorithm="maxima")

[Out]

-1/320*(4*log(-a*x + 1)^5 + 5*(2*log(-a*x + 1)^4 - 4*log(-a*x + 1)^3 + 6*log(-a*x + 1)^2 - 6*log(-a*x + 1) + 3
)*(a*x - 1)^2 + 40*(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^3*c) + 1/16*integrate(-(x^2*log(a*x + 1)^4 - 4*x^2*log(a*x + 1)^3*log(-a*x + 1) + 6*x^2*log(a*x + 1)^2*l
og(-a*x + 1)^2 - 4*x^2*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 = -0.00 \[ \int \frac {x^2\,{\mathrm {atanh}\left (a\,x\right )}^4}{c-a\,c\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atanh(a*x)^4)/(c - a*c*x),x)

[Out]

int((x^2*atanh(a*x)^4)/(c - a*c*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x^{2} \operatorname {atanh}^{4}{\left (a x \right )}}{a x - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(a*x)**4/(-a*c*x+c),x)

[Out]

-Integral(x**2*atanh(a*x)**4/(a*x - 1), x)/c

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