3.2.23 \(\int \frac {(a+b \tanh ^{-1}(c x^3))^2}{x^{10}} \, dx\) [123]

Optimal. Leaf size=144 \[ -\frac {b^2 c^2}{9 x^3}+\frac {1}{9} b^2 c^3 \tanh ^{-1}\left (c x^3\right )-\frac {b c \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{9 x^6}+\frac {1}{9} c^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2-\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{9 x^9}+\frac {2}{9} b c^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{9} b^2 c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6039, 6037, 6129, 331, 212, 6135, 6079, 2497} \begin {gather*} \frac {1}{9} c^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2+\frac {2}{9} b c^3 \log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{9 x^9}-\frac {b c \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{9 x^6}-\frac {1}{9} b^2 c^3 \text {Li}_2\left (\frac {2}{c x^3+1}-1\right )+\frac {1}{9} b^2 c^3 \tanh ^{-1}\left (c x^3\right )-\frac {b^2 c^2}{9 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])^2/x^10,x]

[Out]

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

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 331

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

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6129

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[m, -1]

Rule 6135

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*d*(p + 1)), x] + Dist[1/d, Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{x^{10}} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{4 x^{10}}-\frac {b \left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{2 x^{10}}+\frac {b^2 \log ^2\left (1+c x^3\right )}{4 x^{10}}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{x^{10}} \, dx-\frac {1}{2} b \int \frac {\left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{x^{10}} \, dx+\frac {1}{4} b^2 \int \frac {\log ^2\left (1+c x^3\right )}{x^{10}} \, dx\\ &=\frac {1}{12} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^4} \, dx,x,x^3\right )-\frac {1}{6} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log (1+c x)}{x^4} \, dx,x,x^3\right )+\frac {1}{12} b^2 \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^4} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}+\frac {1}{18} (b c) \text {Subst}\left (\int \frac {2 a-b \log (1-c x)}{x^3 (1-c x)} \, dx,x,x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x^3 (1+c x)} \, dx,x,x^3\right )+\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^3 (1-c x)} \, dx,x,x^3\right )+\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^3 (1+c x)} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}-\frac {1}{18} b \text {Subst}\left (\int \frac {2 a-b \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )^3} \, 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)}{x^3}-\frac {c (-2 a+b \log (1-c x))}{x^2}+\frac {c^2 (-2 a+b \log (1-c x))}{x}-\frac {c^3 (-2 a+b \log (1-c x))}{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)}{x^3}+\frac {c \log (1+c x)}{x^2}+\frac {c^2 \log (1+c x)}{x}-\frac {c^3 \log (1+c x)}{-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)}{x^3}-\frac {c \log (1+c x)}{x^2}+\frac {c^2 \log (1+c x)}{x}-\frac {c^3 \log (1+c x)}{1+c x}\right ) \, dx,x,x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}-\frac {1}{18} b \text {Subst}\left (\int \frac {2 a-b \log (x)}{\left (\frac {1}{c}-\frac {x}{c}\right )^3} \, dx,x,1-c x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {2 a-b \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x^3} \, dx,x,x^3\right )+2 \left (\frac {1}{18} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x^3} \, dx,x,x^3\right )\right )+\frac {1}{18} \left (b c^2\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x^2} \, dx,x,x^3\right )-\frac {1}{18} \left (b c^3\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x} \, dx,x,x^3\right )+2 \left (\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x} \, dx,x,x^3\right )\right )+\frac {1}{18} \left (b c^4\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{1+c x} \, dx,x,x^3\right )\\ &=\frac {1}{3} a b c^3 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^6}+\frac {b c^2 \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {1}{18} b c^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )-\frac {1}{18} b^2 c^3 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}-\frac {1}{9} b^2 c^3 \text {Li}_2\left (-c x^3\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {2 a-b \log (x)}{\left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^3\right )-\frac {1}{36} \left (b^2 c\right ) \text {Subst}\left (\int \frac {1}{x \left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^3\right )-\frac {1}{18} \left (b c^2\right ) \text {Subst}\left (\int \frac {2 a-b \log (x)}{x \left (\frac {1}{c}-\frac {x}{c}\right )} \, dx,x,1-c x^3\right )+\frac {1}{36} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x^2 (1-c x)} \, dx,x,x^3\right )+2 \left (-\frac {b^2 c \log \left (1+c x^3\right )}{36 x^6}+\frac {1}{36} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x^2 (1+c x)} \, dx,x,x^3\right )\right )-\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{x (1-c x)} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+c x^3\right )-\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,x^3\right )+\frac {1}{18} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^3\right )+\frac {1}{18} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^3\right )\\ &=\frac {1}{3} a b c^3 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^6}+\frac {b c^2 \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}-\frac {b c^2 \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {1}{18} b c^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )-\frac {1}{18} b^2 c^3 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {1}{36} b^2 c^3 \log ^2\left (1+c x^3\right )-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}-\frac {1}{9} b^2 c^3 \text {Li}_2\left (-c x^3\right )+\frac {1}{18} b^2 c^3 \text {Li}_2\left (c x^3\right )-\frac {1}{36} \left (b^2 c\right ) \text {Subst}\left (\int \left (\frac {c^2}{(-1+x)^2}-\frac {c^2}{-1+x}+\frac {c^2}{x}\right ) \, dx,x,1-c x^3\right )-\frac {1}{18} \left (b c^2\right ) \text {Subst}\left (\int \frac {2 a-b \log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^3\right )+\frac {1}{36} \left (b^2 c^2\right ) \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {c}{x}-\frac {c^2}{-1+c x}\right ) \, dx,x,x^3\right )+2 \left (-\frac {b^2 c \log \left (1+c x^3\right )}{36 x^6}+\frac {1}{36} \left (b^2 c^2\right ) \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c}{x}+\frac {c^2}{1+c x}\right ) \, dx,x,x^3\right )\right )-\frac {1}{18} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^3\right )-\frac {1}{18} \left (b c^3\right ) \text {Subst}\left (\int \frac {2 a-b \log (x)}{x} \, dx,x,1-c x^3\right )-\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^3\right )-\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^3\right )+\frac {1}{18} \left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^3\right )-\frac {1}{18} \left (b^2 c^4\right ) \text {Subst}\left (\int \frac {1}{1-c x} \, dx,x,x^3\right )\\ &=-\frac {b^2 c^2}{18 x^3}+\frac {2}{3} a b c^3 \log (x)+\frac {1}{6} b^2 c^3 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^6}+\frac {b c^2 \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}-\frac {b c^2 \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}+\frac {1}{36} c^3 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {1}{18} b c^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )-\frac {1}{18} b^2 c^3 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {1}{36} b^2 c^3 \log ^2\left (1+c x^3\right )-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}+2 \left (-\frac {b^2 c^2}{36 x^3}-\frac {1}{12} b^2 c^3 \log (x)+\frac {1}{36} b^2 c^3 \log \left (1+c x^3\right )-\frac {b^2 c \log \left (1+c x^3\right )}{36 x^6}\right )-\frac {1}{9} b^2 c^3 \text {Li}_2\left (-c x^3\right )+\frac {1}{18} b^2 c^3 \text {Li}_2\left (c x^3\right )+\frac {1}{18} b^2 c^3 \text {Li}_2\left (\frac {1}{2} \left (1-c x^3\right )\right )-\frac {1}{18} b^2 c^3 \text {Li}_2\left (\frac {1}{2} \left (1+c x^3\right )\right )+\frac {1}{18} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^3\right )\\ &=-\frac {b^2 c^2}{18 x^3}+\frac {2}{3} a b c^3 \log (x)+\frac {1}{6} b^2 c^3 \log (x)-\frac {b c \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^6}+\frac {b c^2 \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}-\frac {b c^2 \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )}{18 x^3}+\frac {1}{36} c^3 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^2}{36 x^9}-\frac {1}{18} b c^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )-\frac {1}{18} b^2 c^3 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )-\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{18 x^9}-\frac {1}{36} b^2 c^3 \log ^2\left (1+c x^3\right )-\frac {b^2 \log ^2\left (1+c x^3\right )}{36 x^9}+2 \left (-\frac {b^2 c^2}{36 x^3}-\frac {1}{12} b^2 c^3 \log (x)+\frac {1}{36} b^2 c^3 \log \left (1+c x^3\right )-\frac {b^2 c \log \left (1+c x^3\right )}{36 x^6}\right )-\frac {1}{9} b^2 c^3 \text {Li}_2\left (-c x^3\right )+\frac {1}{9} b^2 c^3 \text {Li}_2\left (c x^3\right )+\frac {1}{18} b^2 c^3 \text {Li}_2\left (\frac {1}{2} \left (1-c x^3\right )\right )-\frac {1}{18} b^2 c^3 \text {Li}_2\left (\frac {1}{2} \left (1+c x^3\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.27, size = 159, normalized size = 1.10 \begin {gather*} -\frac {a^2+a b c x^3+b^2 c^2 x^6+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 (2 a+b c x^3-b c^3 x^9-2 b c^3 x^9 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )\right )-2 a b c^3 x^9 \log \left (c x^3\right )+a b c^3 x^9 \log \left (1-c^2 x^6\right )+b^2 c^3 x^9 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])^2/x^10,x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{3}\right )\right )^{2}}{x^{10}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))^2/x^10,x)

[Out]

int((a+b*arctanh(c*x^3))^2/x^10,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((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="maxima")

[Out]

-1/9*((c^2*log(c^2*x^6 - 1) - c^2*log(x^6) + 1/x^6)*c + 2*arctanh(c*x^3)/x^9)*a*b - 1/36*b^2*(log(-c*x^3 + 1)^
2/x^9 + 9*integrate(-1/3*(3*(c*x^3 - 1)*log(c*x^3 + 1)^2 + 2*(c*x^3 - 3*(c*x^3 - 1)*log(c*x^3 + 1))*log(-c*x^3
+ 1))/(c*x^13 - x^10), x)) - 1/9*a^2/x^9

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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((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x^3)^2 + 2*a*b*arctanh(c*x^3) + a^2)/x^10, 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((a+b*atanh(c*x**3))**2/x**10,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((a+b*arctanh(c*x^3))^2/x^10,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^3) + a)^2/x^10, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2}{x^{10}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))^2/x^10,x)

[Out]

int((a + b*atanh(c*x^3))^2/x^10, x)

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