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

Optimal. Leaf size=239 \[ -\frac {3 a \text {Li}_4\left (\frac {2}{a x+1}-1\right )}{c}-\frac {3 a \text {Li}_5\left (\frac {2}{1-a x}-1\right )}{2 c}+\frac {2 a \text {Li}_2\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)^3}{c}-\frac {6 a \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2}{c}-\frac {3 a \text {Li}_3\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)^2}{c}-\frac {6 a \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {3 a \text {Li}_4\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {a \tanh ^{-1}(a x)^4}{c}-\frac {\tanh ^{-1}(a x)^4}{c x}+\frac {a \log \left (2-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{c}+\frac {4 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]

[Out]

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

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Rubi [A]  time = 0.55, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {5934, 5916, 5988, 5932, 5948, 6056, 6060, 6610, 6058, 6062} \[ -\frac {3 a \text {PolyLog}\left (4,\frac {2}{a x+1}-1\right )}{c}-\frac {3 a \text {PolyLog}\left (5,\frac {2}{1-a x}-1\right )}{2 c}+\frac {2 a \tanh ^{-1}(a x)^3 \text {PolyLog}\left (2,\frac {2}{1-a x}-1\right )}{c}-\frac {6 a \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )}{c}-\frac {3 a \tanh ^{-1}(a x)^2 \text {PolyLog}\left (3,\frac {2}{1-a x}-1\right )}{c}-\frac {6 a \tanh ^{-1}(a x) \text {PolyLog}\left (3,\frac {2}{a x+1}-1\right )}{c}+\frac {3 a \tanh ^{-1}(a x) \text {PolyLog}\left (4,\frac {2}{1-a x}-1\right )}{c}+\frac {a \tanh ^{-1}(a x)^4}{c}-\frac {\tanh ^{-1}(a x)^4}{c x}+\frac {a \log \left (2-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{c}+\frac {4 a \log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5934

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

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 5988

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

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

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Mathematica [C]  time = 0.49, size = 172, normalized size = 0.72 \[ -\frac {a \left (-2 \left (\tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x)^2 \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )+3 \left (\tanh ^{-1}(a x)+2\right ) \tanh ^{-1}(a x) \text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )-3 \tanh ^{-1}(a x) \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )-3 \text {Li}_4\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {3}{2} \text {Li}_5\left (e^{2 \tanh ^{-1}(a x)}\right )+\frac {\tanh ^{-1}(a x)^4}{a x}+\tanh ^{-1}(a x)^4+\tanh ^{-1}(a x)^4 \left (-\log \left (1-e^{2 \tanh ^{-1}(a x)}\right )\right )-4 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\frac {i \pi ^5}{160}-\frac {\pi ^4}{16}\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a*(-1/16*Pi^4 + (I/160)*Pi^5 + ArcTanh[a*x]^4 + ArcTanh[a*x]^4/(a*x) - 4*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTan
h[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*Ar
cTanh[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))/c)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{4}}{a c x^{3} - c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arctanh(a*x)^4/(a*c*x^3 - c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.03, size = 583, normalized size = 2.44 \[ -\frac {a \arctanh \left (a x \right )^{4}}{c}-\frac {\arctanh \left (a x \right )^{4}}{c x}+\frac {a \arctanh \left (a x \right )^{4} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \arctanh \left (a x \right ) \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \polylog \left (5, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {a \arctanh \left (a x \right )^{4} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \arctanh \left (a x \right ) \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \polylog \left (5, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \arctanh \left (a x \right ) \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \polylog \left (4, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {4 a \arctanh \left (a x \right )^{3} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {12 a \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {24 a \arctanh \left (a x \right ) \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {24 a \polylog \left (4, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*arctanh(a*x)^4/c-arctanh(a*x)^4/c/x+a/c*arctanh(a*x)^4*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+4*a/c*arctanh(a*x)^
3*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-12*a/c*arctanh(a*x)^2*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+24*a/c*arc
tanh(a*x)*polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))-24*a/c*polylog(5,(a*x+1)/(-a^2*x^2+1)^(1/2))+a/c*arctanh(a*x)^
4*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+4*a/c*arctanh(a*x)^3*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-12*a/c*arctanh(
a*x)^2*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+24*a/c*arctanh(a*x)*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))-24*a/
c*polylog(5,-(a*x+1)/(-a^2*x^2+1)^(1/2))+4*a/c*arctanh(a*x)^3*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+12*a/c*arctanh(
a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-24*a/c*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+24*a/c*
polylog(4,(a*x+1)/(-a^2*x^2+1)^(1/2))+4*a/c*arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+12*a/c*arctanh(a*x
)^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-24*a/c*arctanh(a*x)*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+24*a/c*p
olylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^4}{x^2\,\left (c-a\,c\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^4/(x^2*(c - a*c*x)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\operatorname {atanh}^{4}{\left (a x \right )}}{a x^{3} - x^{2}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(atanh(a*x)**4/(a*x**3 - x**2), x)/c

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