3.25 \(\int x^4 (a+b \tanh ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=262 \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{10 c^5}+\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac{9 a b^2 x}{10 c^4}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}-\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{b^3 x^2}{20 c^3}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c^5}+\frac{9 b^3 x \tanh ^{-1}(c x)}{10 c^4} \]

[Out]

(9*a*b^2*x)/(10*c^4) + (b^3*x^2)/(20*c^3) + (9*b^3*x*ArcTanh[c*x])/(10*c^4) + (b^2*x^3*(a + b*ArcTanh[c*x]))/(
10*c^2) - (9*b*(a + b*ArcTanh[c*x])^2)/(20*c^5) + (3*b*x^2*(a + b*ArcTanh[c*x])^2)/(10*c^3) + (3*b*x^4*(a + b*
ArcTanh[c*x])^2)/(20*c) + (a + b*ArcTanh[c*x])^3/(5*c^5) + (x^5*(a + b*ArcTanh[c*x])^3)/5 - (3*b*(a + b*ArcTan
h[c*x])^2*Log[2/(1 - c*x)])/(5*c^5) + (b^3*Log[1 - c^2*x^2])/(2*c^5) - (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[2,
1 - 2/(1 - c*x)])/(5*c^5) + (3*b^3*PolyLog[3, 1 - 2/(1 - c*x)])/(10*c^5)

________________________________________________________________________________________

Rubi [A]  time = 0.773954, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {5916, 5980, 266, 43, 5910, 260, 5948, 5984, 5918, 6058, 6610} \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{10 c^5}+\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac{9 a b^2 x}{10 c^4}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}-\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{b^3 x^2}{20 c^3}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c^5}+\frac{9 b^3 x \tanh ^{-1}(c x)}{10 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x])^3,x]

[Out]

(9*a*b^2*x)/(10*c^4) + (b^3*x^2)/(20*c^3) + (9*b^3*x*ArcTanh[c*x])/(10*c^4) + (b^2*x^3*(a + b*ArcTanh[c*x]))/(
10*c^2) - (9*b*(a + b*ArcTanh[c*x])^2)/(20*c^5) + (3*b*x^2*(a + b*ArcTanh[c*x])^2)/(10*c^3) + (3*b*x^4*(a + b*
ArcTanh[c*x])^2)/(20*c) + (a + b*ArcTanh[c*x])^3/(5*c^5) + (x^5*(a + b*ArcTanh[c*x])^3)/5 - (3*b*(a + b*ArcTan
h[c*x])^2*Log[2/(1 - c*x)])/(5*c^5) + (b^3*Log[1 - c^2*x^2])/(2*c^5) - (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[2,
1 - 2/(1 - c*x)])/(5*c^5) + (3*b^3*PolyLog[3, 1 - 2/(1 - c*x)])/(10*c^5)

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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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 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 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 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 x^4 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{1}{5} (3 b c) \int \frac{x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{(3 b) \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c}-\frac{(3 b) \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{1}{10} \left (3 b^2\right ) \int \frac{x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac{(3 b) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c^3}-\frac{(3 b) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c^3}\\ &=\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{(3 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx}{5 c^4}+\frac{\left (3 b^2\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^2}-\frac{\left (3 b^2\right ) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{10 c^2}-\frac{\left (3 b^2\right ) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}\\ &=\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{5 c^5}+\frac{\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^4}-\frac{\left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{10 c^4}+\frac{\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^4}-\frac{\left (3 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{5 c^4}+\frac{\left (6 b^2\right ) \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac{b^3 \int \frac{x^3}{1-c^2 x^2} \, dx}{10 c}\\ &=\frac{9 a b^2 x}{10 c^4}+\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{5 c^5}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^5}+\frac{\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{10 c^4}+\frac{\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{5 c^4}+\frac{\left (3 b^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac{b^3 \operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )}{20 c}\\ &=\frac{9 a b^2 x}{10 c^4}+\frac{9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{5 c^5}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^5}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{10 c^5}-\frac{\left (3 b^3\right ) \int \frac{x}{1-c^2 x^2} \, dx}{10 c^3}-\frac{\left (3 b^3\right ) \int \frac{x}{1-c^2 x^2} \, dx}{5 c^3}-\frac{b^3 \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{20 c}\\ &=\frac{9 a b^2 x}{10 c^4}+\frac{b^3 x^2}{20 c^3}+\frac{9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac{b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac{9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac{3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{5 c^5}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c^5}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{5 c^5}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{10 c^5}\\ \end{align*}

Mathematica [A]  time = 0.735381, size = 383, normalized size = 1.46 \[ \frac{12 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)\right )+6 b^3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+3 a^2 b c^4 x^4+6 a^2 b c^2 x^2+6 a^2 b \log \left (1-c^2 x^2\right )+12 a^2 b c^5 x^5 \tanh ^{-1}(c x)+4 a^3 c^5 x^5+2 a b^2 c^3 x^3+12 a b^2 c^5 x^5 \tanh ^{-1}(c x)^2+6 a b^2 c^4 x^4 \tanh ^{-1}(c x)+12 a b^2 c^2 x^2 \tanh ^{-1}(c x)+18 a b^2 c x-12 a b^2 \tanh ^{-1}(c x)^2-18 a b^2 \tanh ^{-1}(c x)-24 a b^2 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+b^3 c^2 x^2+10 b^3 \log \left (1-c^2 x^2\right )+4 b^3 c^5 x^5 \tanh ^{-1}(c x)^3+3 b^3 c^4 x^4 \tanh ^{-1}(c x)^2+2 b^3 c^3 x^3 \tanh ^{-1}(c x)+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^2+18 b^3 c x \tanh ^{-1}(c x)-4 b^3 \tanh ^{-1}(c x)^3-9 b^3 \tanh ^{-1}(c x)^2-12 b^3 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-b^3}{20 c^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x])^3,x]

[Out]

(-b^3 + 18*a*b^2*c*x + 6*a^2*b*c^2*x^2 + b^3*c^2*x^2 + 2*a*b^2*c^3*x^3 + 3*a^2*b*c^4*x^4 + 4*a^3*c^5*x^5 - 18*
a*b^2*ArcTanh[c*x] + 18*b^3*c*x*ArcTanh[c*x] + 12*a*b^2*c^2*x^2*ArcTanh[c*x] + 2*b^3*c^3*x^3*ArcTanh[c*x] + 6*
a*b^2*c^4*x^4*ArcTanh[c*x] + 12*a^2*b*c^5*x^5*ArcTanh[c*x] - 12*a*b^2*ArcTanh[c*x]^2 - 9*b^3*ArcTanh[c*x]^2 +
6*b^3*c^2*x^2*ArcTanh[c*x]^2 + 3*b^3*c^4*x^4*ArcTanh[c*x]^2 + 12*a*b^2*c^5*x^5*ArcTanh[c*x]^2 - 4*b^3*ArcTanh[
c*x]^3 + 4*b^3*c^5*x^5*ArcTanh[c*x]^3 - 24*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*b^3*ArcTanh[c*
x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 6*a^2*b*Log[1 - c^2*x^2] + 10*b^3*Log[1 - c^2*x^2] + 12*b^2*(a + b*ArcTanh
[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*b^3*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(20*c^5)

________________________________________________________________________________________

Maple [C]  time = 0.823, size = 1275, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x))^3,x)

[Out]

3/20*I/c^5*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I*(c*x+1)^2/(c^2
*x^2-1))*Pi-3/20*I/c^5*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*P
i-1/20/c^5*b^3+1/5*x^5*a^3+1/20*b^3*x^2/c^3+3/10/c^5*b^3*arctanh(c*x)^2*ln(c*x+1)-3/5/c^5*b^3*arctanh(c*x)^2*l
n((c*x+1)/(-c^2*x^2+1)^(1/2))-3/5/c^5*b^3*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3/5*x^5*a^2*b*arctan
h(c*x)+3/5*x^5*a*b^2*arctanh(c*x)^2+1/10*a*b^2*x^3/c^2+9/10*a*b^2*x/c^4+9/10*b^3*x*arctanh(c*x)/c^4+3/20*I/c^5
*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1)
)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*Pi-3/10*I/c^5*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*
x+1)^2/(c^2*x^2-1))^2*Pi-3/20*I/c^5*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^
2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*Pi-3/10*I/c^5*b^3*arctanh(c*x)^2*Pi+3/5/c^5*a*b^2*arctanh(c*x)*ln(c*x-1
)+3/5/c^5*a*b^2*arctanh(c*x)*ln(c*x+1)-3/10/c^5*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+3/10/c^5*a*b^2*ln(-1/2*c*x+1/2
)*ln(c*x+1)-3/10/c^5*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+3/10/c*a*b^2*x^4*arctanh(c*x)+3/5/c^3*a*b^2*arctan
h(c*x)*x^2+3/20/c*x^4*a^2*b+3/10/c^3*a^2*b*x^2+3/20/c*b^3*arctanh(c*x)^2*x^4+3/10/c^3*b^3*arctanh(c*x)^2*x^2+1
/10/c^2*b^3*arctanh(c*x)*x^3+9/20/c^5*a*b^2*ln(c*x-1)-9/20/c^5*a*b^2*ln(c*x+1)+3/10/c^5*a^2*b*ln(c*x-1)+3/10/c
^5*a^2*b*ln(c*x+1)+3/20/c^5*a*b^2*ln(c*x-1)^2-3/20/c^5*a*b^2*ln(c*x+1)^2-3/5/c^5*b^3*arctanh(c*x)^2*ln(2)-3/5/
c^5*a*b^2*dilog(1/2+1/2*c*x)+3/10/c^5*b^3*arctanh(c*x)^2*ln(c*x-1)+1/5*x^5*b^3*arctanh(c*x)^3+1/5/c^5*b^3*arct
anh(c*x)^3-9/20/c^5*b^3*arctanh(c*x)^2+3/10/c^5*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/c^5*b^3*ln((c*x+1)^2/
(-c^2*x^2+1)+1)+1/c^5*b^3*arctanh(c*x)-3/10*I/c^5*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*Pi-3
/20*I/c^5*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*Pi-3/20*I/c^5*b^3*arct
anh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*Pi+3/10*I/c^5*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))
^2*Pi

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{5} \, a^{3} x^{5} + \frac{3}{20} \,{\left (4 \, x^{5} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{c^{2} x^{4} + 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )}\right )} a^{2} b - \frac{2 \,{\left (b^{3} c^{5} x^{5} - b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \,{\left (4 \, a b^{2} c^{5} x^{5} + b^{3} c^{4} x^{4} + 2 \, b^{3} c^{2} x^{2} + 2 \,{\left (b^{3} c^{5} x^{5} + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{80 \, c^{5}} - \int -\frac{5 \,{\left (b^{3} c^{5} x^{5} - b^{3} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{3} + 30 \,{\left (a b^{2} c^{5} x^{5} - a b^{2} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{2} - 3 \,{\left (4 \, a b^{2} c^{5} x^{5} + b^{3} c^{4} x^{4} + 2 \, b^{3} c^{2} x^{2} + 5 \,{\left (b^{3} c^{5} x^{5} - b^{3} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{2} - 2 \,{\left (10 \, a b^{2} c^{4} x^{4} -{\left (10 \, a b^{2} c^{5} + b^{3} c^{5}\right )} x^{5} - b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{40 \,{\left (c^{5} x - c^{4}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/5*a^3*x^5 + 3/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*a^2*b - 1/80*(2*(
b^3*c^5*x^5 - b^3)*log(-c*x + 1)^3 - 3*(4*a*b^2*c^5*x^5 + b^3*c^4*x^4 + 2*b^3*c^2*x^2 + 2*(b^3*c^5*x^5 + b^3)*
log(c*x + 1))*log(-c*x + 1)^2)/c^5 - integrate(-1/40*(5*(b^3*c^5*x^5 - b^3*c^4*x^4)*log(c*x + 1)^3 + 30*(a*b^2
*c^5*x^5 - a*b^2*c^4*x^4)*log(c*x + 1)^2 - 3*(4*a*b^2*c^5*x^5 + b^3*c^4*x^4 + 2*b^3*c^2*x^2 + 5*(b^3*c^5*x^5 -
b^3*c^4*x^4)*log(c*x + 1)^2 - 2*(10*a*b^2*c^4*x^4 - (10*a*b^2*c^5 + b^3*c^5)*x^5 - b^3)*log(c*x + 1))*log(-c*
x + 1))/(c^5*x - c^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x^{4} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} x^{4} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b x^{4} \operatorname{artanh}\left (c x\right ) + a^{3} x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^4*arctanh(c*x)^3 + 3*a*b^2*x^4*arctanh(c*x)^2 + 3*a^2*b*x^4*arctanh(c*x) + a^3*x^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x))**3,x)

[Out]

Integral(x**4*(a + b*atanh(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3*x^4, x)