3.2.17 \(\int x^8 (a+b \tanh ^{-1}(c x^3))^2 \, dx\) [117]

Optimal. Leaf size=146 \[ \frac {b^2 x^3}{9 c^2}-\frac {b^2 \tanh ^{-1}\left (c x^3\right )}{9 c^3}+\frac {b x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{9 c}+\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{9 c^3}+\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2-\frac {2 b \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \log \left (\frac {2}{1-c x^3}\right )}{9 c^3}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{9 c^3} \]

[Out]

1/9*b^2*x^3/c^2-1/9*b^2*arctanh(c*x^3)/c^3+1/9*b*x^6*(a+b*arctanh(c*x^3))/c+1/9*(a+b*arctanh(c*x^3))^2/c^3+1/9
*x^9*(a+b*arctanh(c*x^3))^2-2/9*b*(a+b*arctanh(c*x^3))*ln(2/(-c*x^3+1))/c^3-1/9*b^2*polylog(2,1-2/(-c*x^3+1))/
c^3

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Rubi [A]
time = 0.18, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6039, 6037, 6127, 327, 212, 6131, 6055, 2449, 2352} \begin {gather*} \frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{9 c^3}-\frac {2 b \log \left (\frac {2}{1-c x^3}\right ) \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{9 c^3}+\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2+\frac {b x^6 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{9 c}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x^3}\right )}{9 c^3}-\frac {b^2 \tanh ^{-1}\left (c x^3\right )}{9 c^3}+\frac {b^2 x^3}{9 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*ArcTanh[c*x^3])^2,x]

[Out]

(b^2*x^3)/(9*c^2) - (b^2*ArcTanh[c*x^3])/(9*c^3) + (b*x^6*(a + b*ArcTanh[c*x^3]))/(9*c) + (a + b*ArcTanh[c*x^3
])^2/(9*c^3) + (x^9*(a + b*ArcTanh[c*x^3])^2)/9 - (2*b*(a + b*ArcTanh[c*x^3])*Log[2/(1 - c*x^3)])/(9*c^3) - (b
^2*PolyLog[2, 1 - 2/(1 - c*x^3)])/(9*c^3)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

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

Rule 2352

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

Rule 2449

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 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -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 6127

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 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]

Rubi steps

\begin {align*} \int x^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^8 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{2} b x^8 \left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{4} b^2 x^8 \log ^2\left (1+c x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int x^8 \left (2 a-b \log \left (1-c x^3\right )\right )^2 \, dx-\frac {1}{2} b \int x^8 \left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right ) \, dx+\frac {1}{4} b^2 \int x^8 \log ^2\left (1+c x^3\right ) \, dx\\ &=\frac {1}{12} \text {Subst}\left (\int x^2 (2 a-b \log (1-c x))^2 \, dx,x,x^3\right )-\frac {1}{6} b \text {Subst}\left (\int x^2 (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^3\right )+\frac {1}{12} b^2 \text {Subst}\left (\int x^2 \log ^2(1+c x) \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {x^3 (2 a-b \log (1-c x))}{1-c x} \, dx,x,x^3\right )+\frac {1}{18} (b c) \text {Subst}\left (\int \frac {x^3 (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^3 \log (1+c x)}{1-c x} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^3 \log (1+c x)}{1+c x} \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )+\frac {1}{18} b \text {Subst}\left (\int \frac {\left (\frac {1}{c}-\frac {x}{c}\right )^3 (2 a-b \log (x))}{x} \, dx,x,1-c x^3\right )+\frac {1}{18} (b c) \text {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{c^3}-\frac {x (-2 a+b \log (1-c x))}{c^2}+\frac {x^2 (-2 a+b \log (1-c x))}{c}-\frac {-2 a+b \log (1-c x)}{c^3 (1+c x)}\right ) \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c^3}-\frac {x \log (1+c x)}{c^2}-\frac {x^2 \log (1+c x)}{c}-\frac {\log (1+c x)}{c^3 (-1+c x)}\right ) \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \left (\frac {\log (1+c x)}{c^3}-\frac {x \log (1+c x)}{c^2}+\frac {x^2 \log (1+c x)}{c}-\frac {\log (1+c x)}{c^3 (1+c x)}\right ) \, dx,x,x^3\right )\\ &=\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{108} b \left (2 a-b \log \left (1-c x^3\right )\right ) \left (\frac {18 \left (1-c x^3\right )}{c^3}-\frac {9 \left (1-c x^3\right )^2}{c^3}+\frac {2 \left (1-c x^3\right )^3}{c^3}-\frac {6 \log \left (1-c x^3\right )}{c^3}\right )+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )+\frac {1}{18} b \text {Subst}\left (\int x^2 (-2 a+b \log (1-c x)) \, dx,x,x^3\right )+\frac {1}{18} b^2 \text {Subst}\left (\int \frac {x \left (-18+9 x-2 x^2\right )+6 \log (x)}{6 c^3 x} \, dx,x,1-c x^3\right )+\frac {b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^3\right )}{18 c^2}-\frac {b \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^3\right )}{18 c^2}+\frac {b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^3\right )}{18 c^2}+\frac {b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{1+c x} \, dx,x,x^3\right )}{18 c^2}-\frac {b \text {Subst}\left (\int x (-2 a+b \log (1-c x)) \, dx,x,x^3\right )}{18 c}+2 \frac {b^2 \text {Subst}\left (\int x \log (1+c x) \, dx,x,x^3\right )}{18 c}\\ &=-\frac {a b x^3}{9 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^3\right )\right )}{36 c}-\frac {1}{54} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{108} b \left (2 a-b \log \left (1-c x^3\right )\right ) \left (\frac {18 \left (1-c x^3\right )}{c^3}-\frac {9 \left (1-c x^3\right )^2}{c^3}+\frac {2 \left (1-c x^3\right )^3}{c^3}-\frac {6 \log \left (1-c x^3\right )}{c^3}\right )+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 c^3}+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )-\frac {1}{36} b^2 \text {Subst}\left (\int \frac {x^2}{1-c x} \, dx,x,x^3\right )+2 \left (\frac {b^2 x^6 \log \left (1+c x^3\right )}{36 c}-\frac {1}{36} b^2 \text {Subst}\left (\int \frac {x^2}{1+c x} \, dx,x,x^3\right )\right )+\frac {b^2 \text {Subst}\left (\int \frac {x \left (-18+9 x-2 x^2\right )+6 \log (x)}{x} \, dx,x,1-c x^3\right )}{108 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c x^3\right )}{18 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^3\right )}{18 c^2}+\frac {b^2 \text {Subst}\left (\int \log (1-c x) \, dx,x,x^3\right )}{18 c^2}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^3\right )}{18 c^2}+\frac {1}{54} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x^3}{1-c x} \, dx,x,x^3\right )\\ &=-\frac {a b x^3}{9 c^2}+\frac {b x^6 \left (2 a-b \log \left (1-c x^3\right )\right )}{36 c}-\frac {1}{54} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{108} b \left (2 a-b \log \left (1-c x^3\right )\right ) \left (\frac {18 \left (1-c x^3\right )}{c^3}-\frac {9 \left (1-c x^3\right )^2}{c^3}+\frac {2 \left (1-c x^3\right )^3}{c^3}-\frac {6 \log \left (1-c x^3\right )}{c^3}\right )+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 c^3}+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \log ^2\left (1+c x^3\right )}{36 c^3}+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )-\frac {1}{36} b^2 \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {x}{c}-\frac {1}{c^2 (-1+c x)}\right ) \, dx,x,x^3\right )+2 \left (\frac {b^2 x^6 \log \left (1+c x^3\right )}{36 c}-\frac {1}{36} b^2 \text {Subst}\left (\int \left (-\frac {1}{c^2}+\frac {x}{c}+\frac {1}{c^2 (1+c x)}\right ) \, dx,x,x^3\right )\right )+\frac {b^2 \text {Subst}\left (\int \left (-18+9 x-2 x^2+\frac {6 \log (x)}{x}\right ) \, dx,x,1-c x^3\right )}{108 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^3\right )}{18 c^3}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^3\right )}{18 c^3}-\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^3\right )}{18 c^3}+\frac {1}{54} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {1}{c^3}-\frac {x}{c^2}-\frac {x^2}{c}-\frac {1}{c^3 (-1+c x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {a b x^3}{9 c^2}+\frac {13 b^2 x^3}{108 c^2}+\frac {b^2 x^6}{216 c}-\frac {b^2 x^9}{162}+\frac {b^2 \left (1-c x^3\right )^2}{24 c^3}-\frac {b^2 \left (1-c x^3\right )^3}{162 c^3}+\frac {b^2 \log \left (1-c x^3\right )}{108 c^3}-\frac {b^2 \left (1-c x^3\right ) \log \left (1-c x^3\right )}{18 c^3}+\frac {b x^6 \left (2 a-b \log \left (1-c x^3\right )\right )}{36 c}-\frac {1}{54} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{108} b \left (2 a-b \log \left (1-c x^3\right )\right ) \left (\frac {18 \left (1-c x^3\right )}{c^3}-\frac {9 \left (1-c x^3\right )^2}{c^3}+\frac {2 \left (1-c x^3\right )^3}{c^3}-\frac {6 \log \left (1-c x^3\right )}{c^3}\right )+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 c^3}+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \log ^2\left (1+c x^3\right )}{36 c^3}+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )+2 \left (\frac {b^2 x^3}{36 c^2}-\frac {b^2 x^6}{72 c}-\frac {b^2 \log \left (1+c x^3\right )}{36 c^3}+\frac {b^2 x^6 \log \left (1+c x^3\right )}{36 c}\right )-\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^3\right )\right )}{18 c^3}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}+\frac {b^2 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1-c x^3\right )}{18 c^3}\\ &=-\frac {a b x^3}{9 c^2}+\frac {13 b^2 x^3}{108 c^2}+\frac {b^2 x^6}{216 c}-\frac {b^2 x^9}{162}+\frac {b^2 \left (1-c x^3\right )^2}{24 c^3}-\frac {b^2 \left (1-c x^3\right )^3}{162 c^3}+\frac {b^2 \log \left (1-c x^3\right )}{108 c^3}-\frac {b^2 \left (1-c x^3\right ) \log \left (1-c x^3\right )}{18 c^3}+\frac {b^2 \log ^2\left (1-c x^3\right )}{36 c^3}+\frac {b x^6 \left (2 a-b \log \left (1-c x^3\right )\right )}{36 c}-\frac {1}{54} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right )+\frac {1}{36} x^9 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{108} b \left (2 a-b \log \left (1-c x^3\right )\right ) \left (\frac {18 \left (1-c x^3\right )}{c^3}-\frac {9 \left (1-c x^3\right )^2}{c^3}+\frac {2 \left (1-c x^3\right )^3}{c^3}-\frac {6 \log \left (1-c x^3\right )}{c^3}\right )+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 c^3}+\frac {1}{18} b x^9 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \log ^2\left (1+c x^3\right )}{36 c^3}+\frac {1}{36} b^2 x^9 \log ^2\left (1+c x^3\right )+2 \left (\frac {b^2 x^3}{36 c^2}-\frac {b^2 x^6}{72 c}-\frac {b^2 \log \left (1+c x^3\right )}{36 c^3}+\frac {b^2 x^6 \log \left (1+c x^3\right )}{36 c}\right )-\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^3\right )\right )}{18 c^3}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^3\right )\right )}{18 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 132, normalized size = 0.90 \begin {gather*} \frac {b^2 c x^3+a b c^2 x^6+a^2 c^3 x^9+b^2 \left (-1+c^3 x^9\right ) \tanh ^{-1}\left (c x^3\right )^2+b \tanh ^{-1}\left (c x^3\right ) \left (-b+b c^2 x^6+2 a c^3 x^9-2 b \log \left (1+e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )\right )+a b \log \left (-1+c^2 x^6\right )+b^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )}{9 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*ArcTanh[c*x^3])^2,x]

[Out]

(b^2*c*x^3 + a*b*c^2*x^6 + a^2*c^3*x^9 + b^2*(-1 + c^3*x^9)*ArcTanh[c*x^3]^2 + b*ArcTanh[c*x^3]*(-b + b*c^2*x^
6 + 2*a*c^3*x^9 - 2*b*Log[1 + E^(-2*ArcTanh[c*x^3])]) + a*b*Log[-1 + c^2*x^6] + b^2*PolyLog[2, -E^(-2*ArcTanh[
c*x^3])])/(9*c^3)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int x^{8} \left (a +b \arctanh \left (c \,x^{3}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a+b*arctanh(c*x^3))^2,x)

[Out]

int(x^8*(a+b*arctanh(c*x^3))^2,x)

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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^8*(a+b*arctanh(c*x^3))^2,x, algorithm="maxima")

[Out]

1/9*a^2*x^9 + 1/9*(2*x^9*arctanh(c*x^3) + (x^6/c^2 + log(c^2*x^6 - 1)/c^4)*c)*a*b + 1/648*(18*x^9*log(-c*x^3 +
1)^2 - 2*c^4*(2*(c^2*x^9 + 3*x^3)/c^6 - 3*log(c*x^3 + 1)/c^7 + 3*log(c*x^3 - 1)/c^7) + 3*(x^6/c^4 + log(c^2*x
^6 - 1)/c^6)*c^3 + 1944*c^3*integrate(1/9*x^11*log(c*x^3 + 1)/(c^4*x^6 - c^2), x) - 9*c^2*(2*x^3/c^4 - log(c*x
^3 + 1)/c^5 + log(c*x^3 - 1)/c^5) - 6*c*((2*c^2*x^9 + 3*c*x^6 + 6*x^3)/c^3 + 6*log(c*x^3 - 1)/c^4)*log(-c*x^3
+ 1) + 972*c*integrate(1/9*x^5*log(c*x^3 + 1)/(c^4*x^6 - c^2), x) + 6*(3*c^3*x^9*log(c*x^3 + 1)^2 + (2*c^3*x^9
- 3*c^2*x^6 + 6*c*x^3 - 6*(c^3*x^9 + 1)*log(c*x^3 + 1))*log(-c*x^3 + 1))/c^3 + (4*c^3*x^9 + 15*c^2*x^6 + 66*c
*x^3 + 18*log(c*x^3 - 1)^2 + 66*log(c*x^3 - 1))/c^3 - 18*log(9*c^4*x^6 - 9*c^2)/c^3 + 972*integrate(1/9*x^2*lo
g(c*x^3 + 1)/(c^4*x^6 - c^2), x))*b^2

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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^8*(a+b*arctanh(c*x^3))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^8*arctanh(c*x^3)^2 + 2*a*b*x^8*arctanh(c*x^3) + a^2*x^8, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(a+b*atanh(c*x**3))**2,x)

[Out]

Timed out

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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^8*(a+b*arctanh(c*x^3))^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^3) + a)^2*x^8, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^8\,{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*atanh(c*x^3))^2,x)

[Out]

int(x^8*(a + b*atanh(c*x^3))^2, x)

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