3.1.78 \(\int x (a+b \tanh ^{-1}(c x^2))^3 \, dx\) [78]

Optimal. Leaf size=134 \[ \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 c}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3 b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \log \left (\frac {2}{1-c x^2}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right )}{2 c}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x^2}\right )}{4 c} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6039, 6021, 6131, 6055, 6095, 6205, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{2 c}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3+\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 c}-\frac {3 b \log \left (\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{2 c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x^2])^3,x]

[Out]

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

Rule 6021

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

Rule 6039

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

Rule 6055

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 6095

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 6131

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 6205

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 6745

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 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \, dx &=\int \left (\frac {1}{8} x \left (2 a-b \log \left (1-c x^2\right )\right )^3+\frac {3}{8} b x \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )-\frac {3}{8} b^2 x \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {1}{8} b^3 x \log ^3\left (1+c x^2\right )\right ) \, dx\\ &=\frac {1}{8} \int x \left (2 a-b \log \left (1-c x^2\right )\right )^3 \, dx+\frac {1}{8} (3 b) \int x \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right ) \, dx-\frac {1}{8} \left (3 b^2\right ) \int x \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right ) \, dx+\frac {1}{8} b^3 \int x \log ^3\left (1+c x^2\right ) \, dx\\ &=\frac {1}{16} \text {Subst}\left (\int (2 a-b \log (1-c x))^3 \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int (-2 a+b \log (1-c x))^2 \log (1+c x) \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \text {Subst}\left (\int (-2 a+b \log (1-c x)) \log ^2(1+c x) \, dx,x,x^2\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \log ^3(1+c x) \, dx,x,x^2\right )\\ &=\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )-\frac {\text {Subst}\left (\int (2 a-b \log (x))^3 \, dx,x,1-c x^2\right )}{16 c}+\frac {b^3 \text {Subst}\left (\int \log ^3(x) \, dx,x,1+c x^2\right )}{16 c}-\frac {1}{16} (3 b c) \text {Subst}\left (\int \frac {x (-2 a+b \log (1-c x))^2}{1+c x} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {x (-2 a+b \log (1-c x)) \log (1+c x)}{1-c x} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \frac {x (-2 a+b \log (1-c x)) \log (1+c x)}{1+c x} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \frac {x \log ^2(1+c x)}{1-c x} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {(3 b) \text {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{16 c}-\frac {1}{16} (3 b c) \text {Subst}\left (\int \left (\frac {(-2 a+b \log (1-c x))^2}{c}-\frac {(-2 a+b \log (1-c x))^2}{c (1+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \left (\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c}+\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c (-1+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \text {Subst}\left (\int \left (-\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c}+\frac {(2 a-b \log (1-c x)) \log (1+c x)}{c (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^3 c\right ) \text {Subst}\left (\int \left (-\frac {\log ^2(1+c x)}{c}-\frac {\log ^2(1+c x)}{c (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {3 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )-\frac {3 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {1}{16} (3 b) \text {Subst}\left (\int (-2 a+b \log (1-c x))^2 \, dx,x,x^2\right )+\frac {1}{16} (3 b) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2}{1+c x} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x)) \log (1+c x)}{-1+c x} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x)) \log (1+c x)}{1+c x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3\right ) \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{-1+c x} \, dx,x,x^2\right )-\frac {\left (3 b^2\right ) \text {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{8 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{8 c}\\ &=\frac {3}{4} a b^2 x^2-\frac {3 b^3 x^2}{8}-\frac {3 b \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c}+\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )-\frac {3 b^3 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}+\frac {1}{8} \left (3 b^2\right ) \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right ) \log (1+c x)}{1+c x} \, dx,x,x^2\right )+\frac {(3 b) \text {Subst}\left (\int (-2 a+b \log (x))^2 \, dx,x,1-c x^2\right )}{16 c}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (2-x)) \log (x)}{x} \, dx,x,1+c x^2\right )}{8 c}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log (2-x) (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )}{8 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{8 c}\\ &=\frac {3}{4} a b^2 x^2+\frac {3 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{16 c}+\frac {3 b^3 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{8 c}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {(3 b) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{2-x} \, dx,x,1-c x^2\right )}{16 c}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int (-2 a+b \log (x)) \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2-x}{2}\right ) (-2 a+b \log (x))}{x} \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{2-x} \, dx,x,1+c x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2-x}{2}\right ) \log (x)}{x} \, dx,x,1+c x^2\right )}{8 c}\\ &=\frac {3 b^3 x^2}{8}+\frac {3 b^3 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{8 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{8 c}+\frac {3 b^3 \log \left (1+c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c}+\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) (2 a-b \log (x))}{x} \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) \log (x)}{x} \, dx,x,1+c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{8 c}\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{4 c}+\frac {3 b^3 \log \left (1+c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1-c x^2\right )\right )}{8 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{8 c}\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{8 c}-\frac {3 b \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{16 c}+\frac {3}{16} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 c}+\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{16 c}+\frac {3}{16} b^2 x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )+\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 c}-\frac {3 b^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{4 c}+\frac {3 b^3 \log \left (1+c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1-c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 211, normalized size = 1.57 \begin {gather*} \frac {2 a^3 c x^2+6 a^2 b c x^2 \tanh ^{-1}\left (c x^2\right )-6 a b^2 \tanh ^{-1}\left (c x^2\right )^2+6 a b^2 c x^2 \tanh ^{-1}\left (c x^2\right )^2-2 b^3 \tanh ^{-1}\left (c x^2\right )^3+2 b^3 c x^2 \tanh ^{-1}\left (c x^2\right )^3-12 a b^2 \tanh ^{-1}\left (c x^2\right ) \log \left (1+e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )-6 b^3 \tanh ^{-1}\left (c x^2\right )^2 \log \left (1+e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )+3 a^2 b \log \left (1-c^2 x^4\right )+6 b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )+3 b^3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(124)=248\).
time = 0.32, size = 280, normalized size = 2.09

method	result	size
derivativedivides	\(\frac {a^{3} c \,x^{2}+b^{3} \arctanh \left (c \,x^{2}\right )^{3} c \,x^{2}+b^{3} \arctanh \left (c \,x^{2}\right )^{3}-3 b^{3} \arctanh \left (c \,x^{2}\right )^{2} \ln \left (1+\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )-3 b^{3} \arctanh \left (c \,x^{2}\right ) \polylog \left (2, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )}{2}+3 \arctanh \left (c \,x^{2}\right )^{2} a \,b^{2} c \,x^{2}+3 a \,b^{2} \arctanh \left (c \,x^{2}\right )^{2}-6 \arctanh \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) a \,b^{2}+3 a^{2} b c \,x^{2} \arctanh \left (c \,x^{2}\right )+\frac {3 a^{2} b \ln \left (-c^{2} x^{4}+1\right )}{2}}{2 c}\)	\(280\)
default	\(\frac {a^{3} c \,x^{2}+b^{3} \arctanh \left (c \,x^{2}\right )^{3} c \,x^{2}+b^{3} \arctanh \left (c \,x^{2}\right )^{3}-3 b^{3} \arctanh \left (c \,x^{2}\right )^{2} \ln \left (1+\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )-3 b^{3} \arctanh \left (c \,x^{2}\right ) \polylog \left (2, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right )}{2}+3 \arctanh \left (c \,x^{2}\right )^{2} a \,b^{2} c \,x^{2}+3 a \,b^{2} \arctanh \left (c \,x^{2}\right )^{2}-6 \arctanh \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) a \,b^{2}+3 a^{2} b c \,x^{2} \arctanh \left (c \,x^{2}\right )+\frac {3 a^{2} b \ln \left (-c^{2} x^{4}+1\right )}{2}}{2 c}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*x^2 + 3/4*(2*c*x^2*arctanh(c*x^2) + log(-c^2*x^4 + 1))*a^2*b/c - 1/16*((b^3*c*x^2 - b^3)*log(-c*x^2 +
1)^3 - 3*(2*a*b^2*c*x^2 + (b^3*c*x^2 + b^3)*log(c*x^2 + 1))*log(-c*x^2 + 1)^2)/c - integrate(-1/8*((b^3*c*x^3
- b^3*x)*log(c*x^2 + 1)^3 + 6*(a*b^2*c*x^3 - a*b^2*x)*log(c*x^2 + 1)^2 - 3*(4*a*b^2*c*x^3 + (b^3*c*x^3 - b^3*x
)*log(c*x^2 + 1)^2 + 2*((2*a*b^2*c + b^3*c)*x^3 - (2*a*b^2 - b^3)*x)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^2 -
1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x^2))^3,x)

[Out]

int(x*(a + b*atanh(c*x^2))^3, x)

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