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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.498128, antiderivative size = 261, normalized size of antiderivative = 1., 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 = {5930, 5910, 5984, 5918, 5948, 6058, 6062, 6610} \[ \frac{3 \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{a^2 c}-\frac{3 \text{PolyLog}\left (5,1-\frac{2}{1-a x}\right )}{2 a^2 c}+\frac{2 \tanh ^{-1}(a x)^3 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^2 c}+\frac{6 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,1-\frac{2}{1-a x}\right )}{a^2 c}-\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{a^2 c}-\frac{6 \tanh ^{-1}(a x) \text{PolyLog}\left (3,1-\frac{2}{1-a x}\right )}{a^2 c}+\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (4,1-\frac{2}{1-a x}\right )}{a^2 c}-\frac{\tanh ^{-1}(a x)^4}{a^2 c}+\frac{\log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^4}{a^2 c}+\frac{4 \log \left (\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^3}{a^2 c}-\frac{x \tanh ^{-1}(a x)^4}{a c} \]

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 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 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 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 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{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*}

Mathematica [A]  time = 0.269772, size = 172, normalized size = 0.66 \[ -\frac{2 \left (\tanh ^{-1}(a x)+3\right ) \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \left (\tanh ^{-1}(a x)+2\right ) \tanh ^{-1}(a x) \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \tanh ^{-1}(a x) \text{PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text{PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )+\frac{3}{2} \text{PolyLog}\left (5,-e^{-2 \tanh ^{-1}(a x)}\right )-\frac{2}{5} \tanh ^{-1}(a x)^5+a x \tanh ^{-1}(a x)^4-\tanh ^{-1}(a x)^4-\tanh ^{-1}(a x)^4 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )-4 \tanh ^{-1}(a x)^3 \log \left (e^{-2 \tanh ^{-1}(a x)}+1\right )}{a^2 c} \]

Warning: Unable to verify antiderivative.

[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))

________________________________________________________________________________________

Maple [C]  time = 0.223, size = 454, normalized size = 1.7 \begin{align*} -{\frac{x \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{ac}}-{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}\ln \left ( ax-1 \right ) }{{a}^{2}c}}+2\,{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}{{a}^{2}c}{\it polylog} \left ( 2,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }-3\,{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{{a}^{2}c}{\it polylog} \left ( 3,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }+3\,{\frac{{\it Artanh} \left ( ax \right ) }{{a}^{2}c}{\it polylog} \left ( 4,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }-{\frac{3}{2\,{a}^{2}c}{\it polylog} \left ( 5,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }-{\frac{i\pi \, \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{{a}^{2}c} \left ({\it csgn} \left ({i \left ({\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}}+1 \right ) ^{-1}} \right ) \right ) ^{2}}+{\frac{i\pi \, \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{{a}^{2}c}}-{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{{a}^{2}c}}+4\,{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{3}}{{a}^{2}c}\ln \left ({\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}}+1 \right ) }+6\,{\frac{ \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}}{{a}^{2}c}{\it polylog} \left ( 2,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }-6\,{\frac{{\it Artanh} \left ( ax \right ) }{{a}^{2}c}{\it polylog} \left ( 3,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }+3\,{\frac{1}{{a}^{2}c}{\it polylog} \left ( 4,-{\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}} \right ) }+{\frac{\ln \left ( 2 \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{{a}^{2}c}}+{\frac{i\pi \, \left ({\it Artanh} \left ( ax \right ) \right ) ^{4}}{{a}^{2}c} \left ({\it csgn} \left ({i \left ({\frac{ \left ( ax+1 \right ) ^{2}}{-{a}^{2}{x}^{2}+1}}+1 \right ) ^{-1}} \right ) \right ) ^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{x \operatorname{artanh}\left (a x\right )^{4}}{a c x - c}, x\right ) \end{align*}

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)

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x \operatorname{artanh}\left (a x\right )^{4}}{a c x - c}\,{d x} \end{align*}

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)