3.121 \(\int (1+c x)^2 (a+b \tanh ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=240 \[ -\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}+\frac{2 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{c}-\frac{6 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+a b^2 x+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{4 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c}+b^3 x \tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.442961, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.722, Rules used = {5928, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 1586, 6058, 6610} \[ -\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}-\frac{3 b^3 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c}+\frac{2 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{c}-\frac{6 b^2 \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+a b^2 x+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+\frac{(c x+1)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{4 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c}+b^3 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + c*x)^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

Rule 5928

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

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 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 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 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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 (1+c x)^2 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-b \int \left (-3 \left (a+b \tanh ^{-1}(c x)\right )^2-c x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{4 (1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2}\right ) \, dx\\ &=\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}+(3 b) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx-(4 b) \int \frac{(1+c x) \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx+(b c) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-(4 b) \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx-\left (6 b^2 c\right ) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\left (b^2 c^2\right ) \int \frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2}{c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}+b^2 \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx-b^2 \int \frac{a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (6 b^2\right ) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (8 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\\ &=a b^2 x+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+b^3 \int \tanh ^{-1}(c x) \, dx+\left (4 b^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (6 b^3\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b^2 x+b^3 x \tanh ^{-1}(c x)+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}-\frac{4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\frac{2 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{c}-\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c}-\left (b^3 c\right ) \int \frac{x}{1-c^2 x^2} \, dx\\ &=a b^2 x+b^3 x \tanh ^{-1}(c x)+\frac{5 b \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c}+3 b x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{1}{2} b c x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac{(1+c x)^3 \left (a+b \tanh ^{-1}(c x)\right )^3}{3 c}-\frac{6 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c}-\frac{4 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-c x}\right )}{c}+\frac{b^3 \log \left (1-c^2 x^2\right )}{2 c}-\frac{3 b^3 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}-\frac{4 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c}+\frac{2 b^3 \text{Li}_3\left (1-\frac{2}{1-c x}\right )}{c}\\ \end{align*}

Mathematica [B]  time = 0.988259, size = 488, normalized size = 2.03 \[ \frac{6 b^2 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right ) \left (4 a+4 b \tanh ^{-1}(c x)+3 b\right )+12 b^3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+3 a^2 b c^2 x^2+6 a^2 b c^3 x^3 \tanh ^{-1}(c x)+18 a^2 b c^2 x^2 \tanh ^{-1}(c x)+18 a^2 b c x+21 a^2 b \log (1-c x)+3 a^2 b \log (c x+1)+18 a^2 b c x \tanh ^{-1}(c x)+2 a^3 c^3 x^3+6 a^3 c^2 x^2+6 a^3 c x+18 a b^2 \log \left (1-c^2 x^2\right )+6 a b^2 c^3 x^3 \tanh ^{-1}(c x)^2+18 a b^2 c^2 x^2 \tanh ^{-1}(c x)^2+6 a b^2 c^2 x^2 \tanh ^{-1}(c x)+6 a b^2 c x-42 a b^2 \tanh ^{-1}(c x)^2+18 a b^2 c x \tanh ^{-1}(c x)^2-6 a b^2 \tanh ^{-1}(c x)+36 a b^2 c x \tanh ^{-1}(c x)-48 a b^2 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+3 b^3 \log \left (1-c^2 x^2\right )+2 b^3 c^3 x^3 \tanh ^{-1}(c x)^3+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^3+3 b^3 c^2 x^2 \tanh ^{-1}(c x)^2-14 b^3 \tanh ^{-1}(c x)^3+6 b^3 c x \tanh ^{-1}(c x)^3-21 b^3 \tanh ^{-1}(c x)^2+18 b^3 c x \tanh ^{-1}(c x)^2+6 b^3 c x \tanh ^{-1}(c x)-24 b^3 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-36 b^3 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )}{6 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(1 + c*x)^2*(a + b*ArcTanh[c*x])^3,x]

[Out]

(6*a^3*c*x + 18*a^2*b*c*x + 6*a*b^2*c*x + 6*a^3*c^2*x^2 + 3*a^2*b*c^2*x^2 + 2*a^3*c^3*x^3 - 6*a*b^2*ArcTanh[c*
x] + 18*a^2*b*c*x*ArcTanh[c*x] + 36*a*b^2*c*x*ArcTanh[c*x] + 6*b^3*c*x*ArcTanh[c*x] + 18*a^2*b*c^2*x^2*ArcTanh
[c*x] + 6*a*b^2*c^2*x^2*ArcTanh[c*x] + 6*a^2*b*c^3*x^3*ArcTanh[c*x] - 42*a*b^2*ArcTanh[c*x]^2 - 21*b^3*ArcTanh
[c*x]^2 + 18*a*b^2*c*x*ArcTanh[c*x]^2 + 18*b^3*c*x*ArcTanh[c*x]^2 + 18*a*b^2*c^2*x^2*ArcTanh[c*x]^2 + 3*b^3*c^
2*x^2*ArcTanh[c*x]^2 + 6*a*b^2*c^3*x^3*ArcTanh[c*x]^2 - 14*b^3*ArcTanh[c*x]^3 + 6*b^3*c*x*ArcTanh[c*x]^3 + 6*b
^3*c^2*x^2*ArcTanh[c*x]^3 + 2*b^3*c^3*x^3*ArcTanh[c*x]^3 - 48*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])]
- 36*b^3*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 24*b^3*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 21*a
^2*b*Log[1 - c*x] + 3*a^2*b*Log[1 + c*x] + 18*a*b^2*Log[1 - c^2*x^2] + 3*b^3*Log[1 - c^2*x^2] + 6*b^2*(4*a + 3
*b + 4*b*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 12*b^3*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(6*c)

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Maple [C]  time = 0.278, size = 811, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x+1)^2*(a+b*arctanh(c*x))^3,x)

[Out]

3*c*a^2*b*arctanh(c*x)*x^2+8/c*a*b^2*arctanh(c*x)*ln(c*x-1)-4/c*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+c^2*a*b^2*arct
anh(c*x)^2*x^3+3*c*a*b^2*arctanh(c*x)^2*x^2+b^3*arctanh(c*x)/c+b^3*x*arctanh(c*x)+1/2*c*a^2*b*x^2+b^3*x*arctan
h(c*x)^3+1/3/c*b^3*arctanh(c*x)^3+2/c*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/c*a*b^2+3*x*a^2*b+1/3*c^2*x^3*a
^3+c*x^2*a^3+5/2/c*b^3*arctanh(c*x)^2-6/c*b^3*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-6/c*b^3*dilog(1-I*(c*x+1)/
(-c^2*x^2+1)^(1/2))-1/c*b^3*ln((c*x+1)^2/(-c^2*x^2+1)+1)+3*b^3*arctanh(c*x)^2*x+a*b^2*x-4/c*b^3*arctanh(c*x)*p
olylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3*x*a*b^2*arctanh(c*x)^2+1/c*a*b^2*arctanh(c*x)^2+3*x*a^2*b*arctanh(c*x)+2/c
*a*b^2*ln(c*x-1)^2-4/c*a*b^2*dilog(1/2+1/2*c*x)-6/c*b^3*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+4/c*b^
3*arctanh(c*x)^2*ln(c*x-1)-6/c*b^3*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*c*b^3*arctanh(c*x)^2*x^
2+c*b^3*arctanh(c*x)^3*x^2+1/3*c^2*b^3*arctanh(c*x)^3*x^3+6*a*b^2*arctanh(c*x)*x-4/c*b^3*arctanh(c*x)^2*ln(2)+
1/c*a^2*b*arctanh(c*x)+4/c*a^2*b*ln(c*x-1)+7/2/c*a*b^2*ln(c*x-1)+5/2/c*a*b^2*ln(c*x+1)-4*I/c*b^3*Pi*csgn(I/((c
*x+1)^2/(-c^2*x^2+1)+1))^3*arctanh(c*x)^2+4*I/c*b^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+x*a
^3+c*a*b^2*arctanh(c*x)*x^2+c^2*a^2*b*arctanh(c*x)*x^3-4*I/c*b^3*Pi*arctanh(c*x)^2+1/3/c*a^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a^{3} c^{2} x^{3} + \frac{1}{2} \,{\left (2 \, x^{3} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{x^{2}}{c^{2}} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a^{2} b c^{2} + a^{3} c x^{2} + \frac{3}{2} \,{\left (2 \, x^{2} \operatorname{artanh}\left (c x\right ) + c{\left (\frac{2 \, x}{c^{2}} - \frac{\log \left (c x + 1\right )}{c^{3}} + \frac{\log \left (c x - 1\right )}{c^{3}}\right )}\right )} a^{2} b c + a^{3} x + \frac{3 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a^{2} b}{2 \, c} - \frac{{\left (b^{3} c^{3} x^{3} + 3 \, b^{3} c^{2} x^{2} + 3 \, b^{3} c x - 7 \, b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \,{\left (2 \, a b^{2} c^{3} x^{3} +{\left (6 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} + 6 \,{\left (a b^{2} c + b^{3} c\right )} x +{\left (b^{3} c^{3} x^{3} + 3 \, b^{3} c^{2} x^{2} + 3 \, b^{3} c x + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{24 \, c} - \int -\frac{{\left (b^{3} c^{3} x^{3} + b^{3} c^{2} x^{2} - b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c^{3} x^{3} + a b^{2} c^{2} x^{2} - a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} -{\left (4 \, a b^{2} c^{3} x^{3} + 2 \,{\left (6 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} + 3 \,{\left (b^{3} c^{3} x^{3} + b^{3} c^{2} x^{2} - b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} + 12 \,{\left (a b^{2} c + b^{3} c\right )} x + 2 \,{\left ({\left (6 \, a b^{2} c^{3} + b^{3} c^{3}\right )} x^{3} - 6 \, a b^{2} + b^{3} + 3 \,{\left (2 \, a b^{2} c^{2} + b^{3} c^{2}\right )} x^{2} - 3 \,{\left (2 \, a b^{2} c - b^{3} c\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/3*a^3*c^2*x^3 + 1/2*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a^2*b*c^2 + a^3*c*x^2 + 3/2*(2
*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*a^2*b*c + a^3*x + 3/2*(2*c*x*arctanh(c*
x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/24*((b^3*c^3*x^3 + 3*b^3*c^2*x^2 + 3*b^3*c*x - 7*b^3)*log(-c*x + 1)^3 - 3*
(2*a*b^2*c^3*x^3 + (6*a*b^2*c^2 + b^3*c^2)*x^2 + 6*(a*b^2*c + b^3*c)*x + (b^3*c^3*x^3 + 3*b^3*c^2*x^2 + 3*b^3*
c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c - integrate(-1/8*((b^3*c^3*x^3 + b^3*c^2*x^2 - b^3*c*x - b^3)*log(
c*x + 1)^3 + 6*(a*b^2*c^3*x^3 + a*b^2*c^2*x^2 - a*b^2*c*x - a*b^2)*log(c*x + 1)^2 - (4*a*b^2*c^3*x^3 + 2*(6*a*
b^2*c^2 + b^3*c^2)*x^2 + 3*(b^3*c^3*x^3 + b^3*c^2*x^2 - b^3*c*x - b^3)*log(c*x + 1)^2 + 12*(a*b^2*c + b^3*c)*x
+ 2*((6*a*b^2*c^3 + b^3*c^3)*x^3 - 6*a*b^2 + b^3 + 3*(2*a*b^2*c^2 + b^3*c^2)*x^2 - 3*(2*a*b^2*c - b^3*c)*x)*l
og(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3} \left (c x + 1\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)**2*(a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3*(c*x + 1)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x + 1\right )}^{2}{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x+1)^2*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((c*x + 1)^2*(b*arctanh(c*x) + a)^3, x)