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

Optimal. Leaf size=91 \[ -\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 c^2}+\frac{a b x^2}{2 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac{b^2 \log \left (1-c^2 x^4\right )}{4 c^2}+\frac{b^2 x^2 \tanh ^{-1}\left (c x^2\right )}{2 c} \]

[Out]

(a*b*x^2)/(2*c) + (b^2*x^2*ArcTanh[c*x^2])/(2*c) - (a + b*ArcTanh[c*x^2])^2/(4*c^2) + (x^4*(a + b*ArcTanh[c*x^
2])^2)/4 + (b^2*Log[1 - c^2*x^4])/(4*c^2)

________________________________________________________________________________________

Rubi [C]  time = 0.96549, antiderivative size = 524, normalized size of antiderivative = 5.76, number of steps used = 44, number of rules used = 16, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6099, 2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304, 2395, 43, 2439, 2416, 2394, 2393, 2391} \[ \frac{b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-c x^2\right )\right )}{8 c^2}+\frac{b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (c x^2+1\right )\right )}{8 c^2}+\frac{\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac{b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{16 c^2}-\frac{b \log \left (\frac{1}{2} \left (c x^2+1\right )\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 c^2}+\frac{3 a b x^2}{4 c}-\frac{1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac{1}{8} b x^4 \log \left (c x^2+1\right ) \left (2 a-b \log \left (1-c x^2\right )\right )+\frac{b^2 \left (1-c x^2\right )^2}{32 c^2}+\frac{b^2 \left (c x^2+1\right )^2}{32 c^2}+\frac{b^2 \left (c x^2+1\right )^2 \log ^2\left (c x^2+1\right )}{16 c^2}-\frac{b^2 \left (c x^2+1\right ) \log ^2\left (c x^2+1\right )}{8 c^2}+\frac{3 b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c^2}-\frac{b^2 \log \left (1-c x^2\right )}{16 c^2}-\frac{b^2 \left (c x^2+1\right )^2 \log \left (c x^2+1\right )}{16 c^2}+\frac{3 b^2 \left (c x^2+1\right ) \log \left (c x^2+1\right )}{8 c^2}+\frac{b^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (c x^2+1\right )}{8 c^2}-\frac{b^2 \log \left (c x^2+1\right )}{16 c^2}+\frac{1}{16} b^2 x^4 \log \left (c x^2+1\right )-\frac{1}{16} b^2 x^4 \]

Warning: Unable to verify antiderivative.

[In]

Int[x^3*(a + b*ArcTanh[c*x^2])^2,x]

[Out]

(3*a*b*x^2)/(4*c) - (b^2*x^4)/16 + (b^2*(1 - c*x^2)^2)/(32*c^2) + (b^2*(1 + c*x^2)^2)/(32*c^2) - (b^2*Log[1 -
c*x^2])/(16*c^2) + (3*b^2*(1 - c*x^2)*Log[1 - c*x^2])/(8*c^2) - (b*x^4*(2*a - b*Log[1 - c*x^2]))/16 + (b*(1 -
c*x^2)^2*(2*a - b*Log[1 - c*x^2]))/(16*c^2) - ((1 - c*x^2)*(2*a - b*Log[1 - c*x^2])^2)/(8*c^2) + ((1 - c*x^2)^
2*(2*a - b*Log[1 - c*x^2])^2)/(16*c^2) - (b*(2*a - b*Log[1 - c*x^2])*Log[(1 + c*x^2)/2])/(8*c^2) - (b^2*Log[1
+ c*x^2])/(16*c^2) + (b^2*x^4*Log[1 + c*x^2])/16 + (3*b^2*(1 + c*x^2)*Log[1 + c*x^2])/(8*c^2) - (b^2*(1 + c*x^
2)^2*Log[1 + c*x^2])/(16*c^2) + (b^2*Log[(1 - c*x^2)/2]*Log[1 + c*x^2])/(8*c^2) + (b*x^4*(2*a - b*Log[1 - c*x^
2])*Log[1 + c*x^2])/8 - (b^2*(1 + c*x^2)*Log[1 + c*x^2]^2)/(8*c^2) + (b^2*(1 + c*x^2)^2*Log[1 + c*x^2]^2)/(16*
c^2) + (b^2*PolyLog[2, (1 - c*x^2)/2])/(8*c^2) + (b^2*PolyLog[2, (1 + c*x^2)/2])/(8*c^2)

Rule 6099

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^
m*(a + (b*Log[1 + c*x^n])/2 - (b*Log[1 - c*x^n])/2)^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0] &&
IntegerQ[m] && IntegerQ[n]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 0]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2296

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

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

Rule 2391

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

Rubi steps

\begin{align*} \int x^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac{1}{4} x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac{1}{2} b x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac{1}{4} b^2 x^3 \log ^2\left (1+c x^2\right )\right ) \, dx\\ &=\frac{1}{4} \int x^3 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \, dx-\frac{1}{2} b \int x^3 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right ) \, dx+\frac{1}{4} b^2 \int x^3 \log ^2\left (1+c x^2\right ) \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int x (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )-\frac{1}{4} b \operatorname{Subst}\left (\int x (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^2\right )+\frac{1}{8} b^2 \operatorname{Subst}\left (\int x \log ^2(1+c x) \, dx,x,x^2\right )\\ &=\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \left (\frac{(2 a-b \log (1-c x))^2}{c}-\frac{(1-c x) (2 a-b \log (1-c x))^2}{c}\right ) \, dx,x,x^2\right )+\frac{1}{8} b^2 \operatorname{Subst}\left (\int \left (-\frac{\log ^2(1+c x)}{c}+\frac{(1+c x) \log ^2(1+c x)}{c}\right ) \, dx,x,x^2\right )+\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{x^2 (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2 \log (1+c x)}{1-c x} \, dx,x,x^2\right )\\ &=\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac{\operatorname{Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )}{8 c}-\frac{\operatorname{Subst}\left (\int (1-c x) (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )}{8 c}-\frac{b^2 \operatorname{Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )}{8 c}+\frac{b^2 \operatorname{Subst}\left (\int (1+c x) \log ^2(1+c x) \, dx,x,x^2\right )}{8 c}+\frac{1}{8} (b c) \operatorname{Subst}\left (\int \left (-\frac{-2 a+b \log (1-c x)}{c^2}+\frac{x (-2 a+b \log (1-c x))}{c}+\frac{-2 a+b \log (1-c x)}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (-\frac{\log (1+c x)}{c^2}-\frac{x \log (1+c x)}{c}-\frac{\log (1+c x)}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac{1}{8} b \operatorname{Subst}\left (\int x (-2 a+b \log (1-c x)) \, dx,x,x^2\right )+\frac{1}{8} b^2 \operatorname{Subst}\left (\int x \log (1+c x) \, dx,x,x^2\right )-\frac{\operatorname{Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{8 c^2}+\frac{\operatorname{Subst}\left (\int x (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{8 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{8 c^2}+\frac{b^2 \operatorname{Subst}\left (\int x \log ^2(x) \, dx,x,1+c x^2\right )}{8 c^2}-\frac{b \operatorname{Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^2\right )}{8 c}+\frac{b \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )}{8 c}+\frac{b^2 \operatorname{Subst}\left (\int \log (1+c x) \, dx,x,x^2\right )}{8 c}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log (1+c x)}{-1+c x} \, dx,x,x^2\right )}{8 c}\\ &=\frac{a b x^2}{4 c}-\frac{1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac{\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )}{8 c^2}+\frac{1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac{b^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac{b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}+\frac{b \operatorname{Subst}\left (\int x (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{8 c^2}-\frac{b \operatorname{Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{4 c^2}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{8 c^2}-\frac{b^2 \operatorname{Subst}\left (\int x \log (x) \, dx,x,1+c x^2\right )}{8 c^2}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )}{8 c}-\frac{b^2 \operatorname{Subst}\left (\int \log (1-c x) \, dx,x,x^2\right )}{8 c}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )}{8 c}+\frac{1}{16} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-c x} \, dx,x,x^2\right )-\frac{1}{16} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+c x} \, dx,x,x^2\right )\\ &=\frac{3 a b x^2}{4 c}-\frac{3 b^2 x^2}{8 c}+\frac{b^2 \left (1-c x^2\right )^2}{32 c^2}+\frac{b^2 \left (1+c x^2\right )^2}{32 c^2}-\frac{1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac{b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{16 c^2}-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac{\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )}{8 c^2}+\frac{1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac{3 b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac{b^2 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{16 c^2}+\frac{b^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac{b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{8 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{8 c^2}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{8 c^2}+\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{4 c^2}+\frac{1}{16} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{x}{c}-\frac{1}{c^2 (-1+c x)}\right ) \, dx,x,x^2\right )-\frac{1}{16} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2}+\frac{x}{c}+\frac{1}{c^2 (1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac{3 a b x^2}{4 c}-\frac{b^2 x^4}{16}+\frac{b^2 \left (1-c x^2\right )^2}{32 c^2}+\frac{b^2 \left (1+c x^2\right )^2}{32 c^2}-\frac{b^2 \log \left (1-c x^2\right )}{16 c^2}+\frac{3 b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c^2}-\frac{1}{16} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right )+\frac{b \left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )}{16 c^2}-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c^2}+\frac{\left (1-c x^2\right )^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )}{8 c^2}-\frac{b^2 \log \left (1+c x^2\right )}{16 c^2}+\frac{1}{16} b^2 x^4 \log \left (1+c x^2\right )+\frac{3 b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c^2}-\frac{b^2 \left (1+c x^2\right )^2 \log \left (1+c x^2\right )}{16 c^2}+\frac{b^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 c^2}+\frac{1}{8} b x^4 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c^2}+\frac{b^2 \left (1+c x^2\right )^2 \log ^2\left (1+c x^2\right )}{16 c^2}+\frac{b^2 \text{Li}_2\left (\frac{1}{2} \left (1-c x^2\right )\right )}{8 c^2}+\frac{b^2 \text{Li}_2\left (\frac{1}{2} \left (1+c x^2\right )\right )}{8 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0603613, size = 106, normalized size = 1.16 \[ \frac{a^2 c^2 x^4+2 a b c x^2+b (a+b) \log \left (1-c x^2\right )-a b \log \left (c x^2+1\right )+2 b c x^2 \tanh ^{-1}\left (c x^2\right ) \left (a c x^2+b\right )+b^2 \left (c^2 x^4-1\right ) \tanh ^{-1}\left (c x^2\right )^2+b^2 \log \left (c x^2+1\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*c*x^2 + a^2*c^2*x^4 + 2*b*c*x^2*(b + a*c*x^2)*ArcTanh[c*x^2] + b^2*(-1 + c^2*x^4)*ArcTanh[c*x^2]^2 + b*
(a + b)*Log[1 - c*x^2] - a*b*Log[1 + c*x^2] + b^2*Log[1 + c*x^2])/(4*c^2)

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Maple [B]  time = 0.163, size = 247, normalized size = 2.7 \begin{align*}{\frac{{b}^{2} \left ({c}^{2}{x}^{4}-1 \right ) \left ( \ln \left ( c{x}^{2}+1 \right ) \right ) ^{2}}{16\,{c}^{2}}}+{\frac{b \left ( -{x}^{4}b\ln \left ( -c{x}^{2}+1 \right ){c}^{2}+2\,a{c}^{2}{x}^{4}+2\,bc{x}^{2}+b\ln \left ( -c{x}^{2}+1 \right ) \right ) \ln \left ( c{x}^{2}+1 \right ) }{8\,{c}^{2}}}+{\frac{{b}^{2}{x}^{4} \left ( \ln \left ( -c{x}^{2}+1 \right ) \right ) ^{2}}{16}}-{\frac{ab{x}^{4}\ln \left ( -c{x}^{2}+1 \right ) }{4}}+{\frac{{a}^{2}{x}^{4}}{4}}-{\frac{{b}^{2}{x}^{2}\ln \left ( -c{x}^{2}+1 \right ) }{4\,c}}+{\frac{{x}^{2}ab}{2\,c}}-{\frac{{b}^{2} \left ( \ln \left ( -c{x}^{2}+1 \right ) \right ) ^{2}}{16\,{c}^{2}}}+{\frac{b\ln \left ( -c{x}^{2}+1 \right ) a}{4\,{c}^{2}}}+{\frac{{b}^{2}\ln \left ( -c{x}^{2}+1 \right ) }{4\,{c}^{2}}}-{\frac{b\ln \left ( -c{x}^{2}-1 \right ) a}{4\,{c}^{2}}}+{\frac{{b}^{2}\ln \left ( -c{x}^{2}-1 \right ) }{4\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^2*(c^2*x^4-1)/c^2*ln(c*x^2+1)^2+1/8*b*(-x^4*b*ln(-c*x^2+1)*c^2+2*a*c^2*x^4+2*b*c*x^2+b*ln(-c*x^2+1))/c^
2*ln(c*x^2+1)+1/16*b^2*x^4*ln(-c*x^2+1)^2-1/4*a*b*x^4*ln(-c*x^2+1)+1/4*a^2*x^4-1/4/c*b^2*x^2*ln(-c*x^2+1)+1/2*
a*b*x^2/c-1/16/c^2*b^2*ln(-c*x^2+1)^2+1/4/c^2*b*ln(-c*x^2+1)*a+1/4/c^2*b^2*ln(-c*x^2+1)-1/4/c^2*b*ln(-c*x^2-1)
*a+1/4/c^2*b^2*ln(-c*x^2-1)

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Maxima [B]  time = 0.976063, size = 251, normalized size = 2.76 \begin{align*} \frac{1}{4} \, b^{2} x^{4} \operatorname{artanh}\left (c x^{2}\right )^{2} + \frac{1}{4} \, a^{2} x^{4} + \frac{1}{4} \,{\left (2 \, x^{4} \operatorname{artanh}\left (c x^{2}\right ) + c{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{\log \left (c x^{2} + 1\right )}{c^{3}} + \frac{\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} a b + \frac{1}{16} \,{\left (4 \, c{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{\log \left (c x^{2} + 1\right )}{c^{3}} + \frac{\log \left (c x^{2} - 1\right )}{c^{3}}\right )} \operatorname{artanh}\left (c x^{2}\right ) - \frac{2 \,{\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right )}{c^{2}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4*arctanh(c*x^2)^2 + 1/4*a^2*x^4 + 1/4*(2*x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + l
og(c*x^2 - 1)/c^3))*a*b + 1/16*(4*c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3)*arctanh(c*x^2) - (2*
(log(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1))/c^2)*b^2

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Fricas [A]  time = 2.11289, size = 292, normalized size = 3.21 \begin{align*} \frac{4 \, a^{2} c^{2} x^{4} + 8 \, a b c x^{2} +{\left (b^{2} c^{2} x^{4} - b^{2}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \,{\left (a b - b^{2}\right )} \log \left (c x^{2} + 1\right ) + 4 \,{\left (a b + b^{2}\right )} \log \left (c x^{2} - 1\right ) + 4 \,{\left (a b c^{2} x^{4} + b^{2} c x^{2}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*a^2*c^2*x^4 + 8*a*b*c*x^2 + (b^2*c^2*x^4 - b^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 - 4*(a*b - b^2)*log(c*
x^2 + 1) + 4*(a*b + b^2)*log(c*x^2 - 1) + 4*(a*b*c^2*x^4 + b^2*c*x^2)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2

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Sympy [A]  time = 19.1199, size = 163, normalized size = 1.79 \begin{align*} \begin{cases} \frac{a^{2} x^{4}}{4} + \frac{a b x^{4} \operatorname{atanh}{\left (c x^{2} \right )}}{2} + \frac{a b x^{2}}{2 c} - \frac{a b \operatorname{atanh}{\left (c x^{2} \right )}}{2 c^{2}} + \frac{b^{2} x^{4} \operatorname{atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac{b^{2} x^{2} \operatorname{atanh}{\left (c x^{2} \right )}}{2 c} + \frac{b^{2} \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{2 c^{2}} + \frac{b^{2} \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{2 c^{2}} - \frac{b^{2} \operatorname{atanh}^{2}{\left (c x^{2} \right )}}{4 c^{2}} - \frac{b^{2} \operatorname{atanh}{\left (c x^{2} \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\\frac{a^{2} x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x**4/4 + a*b*x**4*atanh(c*x**2)/2 + a*b*x**2/(2*c) - a*b*atanh(c*x**2)/(2*c**2) + b**2*x**4*at
anh(c*x**2)**2/4 + b**2*x**2*atanh(c*x**2)/(2*c) + b**2*log(x - I*sqrt(1/c))/(2*c**2) + b**2*log(x + I*sqrt(1/
c))/(2*c**2) - b**2*atanh(c*x**2)**2/(4*c**2) - b**2*atanh(c*x**2)/(2*c**2), Ne(c, 0)), (a**2*x**4/4, True))

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Giac [A]  time = 1.38177, size = 186, normalized size = 2.04 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{a b x^{2}}{2 \, c} + \frac{1}{16} \,{\left (b^{2} x^{4} - \frac{b^{2}}{c^{2}}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )^{2} + \frac{1}{4} \,{\left (a b x^{4} + \frac{b^{2} x^{2}}{c}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) - \frac{{\left (a b - b^{2}\right )} \log \left (c x^{2} + 1\right )}{4 \, c^{2}} + \frac{{\left (a b + b^{2}\right )} \log \left (c x^{2} - 1\right )}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 1/2*a*b*x^2/c + 1/16*(b^2*x^4 - b^2/c^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 + 1/4*(a*b*x^4 + b^2*x^
2/c)*log(-(c*x^2 + 1)/(c*x^2 - 1)) - 1/4*(a*b - b^2)*log(c*x^2 + 1)/c^2 + 1/4*(a*b + b^2)*log(c*x^2 - 1)/c^2