3.2.29 \(\int \frac {(a+b \tanh ^{-1}(c x^3))^3}{x^7} \, dx\) [129]

Optimal. Leaf size=136 \[ \frac {1}{2} b c^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2-\frac {b c \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{2 x^3}+\frac {1}{6} c^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^3}{6 x^6}+b^2 c^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \log \left (2-\frac {2}{1+c x^3}\right )-\frac {1}{2} b^3 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right ) \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6039, 6037, 6129, 6135, 6079, 2497, 6095} \begin {gather*} b^2 c^2 \log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {1}{2} b c^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2+\frac {1}{6} c^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^3-\frac {b c \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{2 x^3}-\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^3}{6 x^6}-\frac {1}{2} b^3 c^2 \text {Li}_2\left (\frac {2}{c x^3+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])^3/x^7,x]

[Out]

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

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 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 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 )^3}{x^7} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{8 x^7}+\frac {3 b \left (-2 a+b \log \left (1-c x^3\right )\right )^2 \log \left (1+c x^3\right )}{8 x^7}-\frac {3 b^2 \left (-2 a+b \log \left (1-c x^3\right )\right ) \log ^2\left (1+c x^3\right )}{8 x^7}+\frac {b^3 \log ^3\left (1+c x^3\right )}{8 x^7}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{x^7} \, dx+\frac {1}{8} (3 b) \int \frac {\left (-2 a+b \log \left (1-c x^3\right )\right )^2 \log \left (1+c x^3\right )}{x^7} \, dx-\frac {1}{8} \left (3 b^2\right ) \int \frac {\left (-2 a+b \log \left (1-c x^3\right )\right ) \log ^2\left (1+c x^3\right )}{x^7} \, dx+\frac {1}{8} b^3 \int \frac {\log ^3\left (1+c x^3\right )}{x^7} \, dx\\ &=\frac {1}{24} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^3}{x^3} \, dx,x,x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{24} b^3 \text {Subst}\left (\int \frac {\log ^3(1+c x)}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^2 (1-c x)} \, dx,x,x^3\right )+\frac {1}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^2 (1+c x)} \, dx,x,x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}-\frac {1}{16} b \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x \left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (-\frac {1}{c}+\frac {x}{c}\right )^2} \, dx,x,1+c x^3\right )\\ &=-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}-\frac {1}{16} b \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{\left (\frac {1}{c}-\frac {x}{c}\right )^2} \, dx,x,1-c x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{16} b^3 \text {Subst}\left (\int \frac {\log ^2(x)}{\left (-\frac {1}{c}+\frac {x}{c}\right )^2} \, dx,x,1+c x^3\right )-\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x \left (\frac {1}{c}-\frac {x}{c}\right )} \, dx,x,1-c x^3\right )-\frac {1}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x \left (-\frac {1}{c}+\frac {x}{c}\right )} \, dx,x,1+c x^3\right )\\ &=-\frac {b c \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{16 x^3}-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 c \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{16 x^3}-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{16} (b c) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^3\right )-\frac {1}{8} \left (b^2 c\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}{16} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{-\frac {1}{c}+\frac {x}{c}} \, dx,x,1+c x^3\right )+\frac {1}{8} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{-\frac {1}{c}+\frac {x}{c}} \, dx,x,1+c x^3\right )-\frac {1}{16} \left (b c^2\right ) \text {Subst}\left (\int \frac {(2 a-b \log (x))^2}{x} \, dx,x,1-c x^3\right )+\frac {1}{16} \left (b^3 c^2\right ) \text {Subst}\left (\int \frac {\log ^2(x)}{x} \, dx,x,1+c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} b c^2 \log \left (c x^3\right ) \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 )^3}{48 x^6}-\frac {b^3 c \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{16 x^3}-\frac {1}{16} b^3 c^2 \log \left (-c x^3\right ) \log ^2\left (1+c x^3\right )-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}-\frac {1}{8} b^3 c^2 \text {Li}_2\left (-c x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )+\frac {1}{8} \left (b^3 c\right ) \text {Subst}\left (\int \frac {\log (x)}{\frac {1}{c}-\frac {x}{c}} \, dx,x,1-c x^3\right )+\frac {1}{16} c^2 \text {Subst}\left (\int x^2 \, dx,x,2 a-b \log \left (1-c x^3\right )\right )+\frac {1}{8} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) (2 a-b \log (x))}{x} \, dx,x,1-c x^3\right )+\frac {1}{16} \left (b^3 c^2\right ) \text {Subst}\left (\int x^2 \, dx,x,\log \left (1+c x^3\right )\right )+\frac {1}{8} \left (b^3 c^2\right ) \text {Subst}\left (\int \frac {\log (1-x) \log (x)}{x} \, dx,x,1+c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} b c^2 \log \left (c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2+\frac {1}{48} c^2 \left (2 a-b \log \left (1-c x^3\right )\right )^3-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 c \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{16 x^3}-\frac {1}{16} b^3 c^2 \log \left (-c x^3\right ) \log ^2\left (1+c x^3\right )+\frac {1}{48} b^3 c^2 \log ^3\left (1+c x^3\right )-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}-\frac {1}{8} b^3 c^2 \text {Li}_2\left (-c x^3\right )+\frac {1}{8} b^3 c^2 \text {Li}_2\left (c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a-b \log \left (1-c x^3\right )\right ) \text {Li}_2\left (1-c x^3\right )-\frac {1}{8} b^3 c^2 \log \left (1+c x^3\right ) \text {Li}_2\left (1+c x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} \left (b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x^3\right )+\frac {1}{8} \left (b^3 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x^3\right )\\ &=\frac {3}{4} a b^2 c^2 \log (x)-\frac {b c \left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{16 x^3}+\frac {1}{16} b c^2 \log \left (c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2+\frac {1}{48} c^2 \left (2 a-b \log \left (1-c x^3\right )\right )^3-\frac {\left (2 a-b \log \left (1-c x^3\right )\right )^3}{48 x^6}-\frac {b^3 c \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{16 x^3}-\frac {1}{16} b^3 c^2 \log \left (-c x^3\right ) \log ^2\left (1+c x^3\right )+\frac {1}{48} b^3 c^2 \log ^3\left (1+c x^3\right )-\frac {b^3 \log ^3\left (1+c x^3\right )}{48 x^6}-\frac {1}{8} b^3 c^2 \text {Li}_2\left (-c x^3\right )+\frac {1}{8} b^3 c^2 \text {Li}_2\left (c x^3\right )-\frac {1}{8} b^2 c^2 \left (2 a-b \log \left (1-c x^3\right )\right ) \text {Li}_2\left (1-c x^3\right )-\frac {1}{8} b^3 c^2 \log \left (1+c x^3\right ) \text {Li}_2\left (1+c x^3\right )-\frac {1}{8} b^3 c^2 \text {Li}_3\left (1-c x^3\right )+\frac {1}{8} b^3 c^2 \text {Li}_3\left (1+c x^3\right )+\frac {1}{8} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^3} \, dx,x,x^3\right )-\frac {1}{8} b^2 \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^3} \, dx,x,x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 218, normalized size = 1.60 \begin {gather*} \frac {6 b^2 \left (-1+c x^3\right ) \left (a+a c x^3+b c x^3\right ) \tanh ^{-1}\left (c x^3\right )^2+2 b^3 \left (-1+c^2 x^6\right ) \tanh ^{-1}\left (c x^3\right )^3-6 b \tanh ^{-1}\left (c x^3\right ) \left (a^2+2 a b c x^3-2 b^2 c^2 x^6 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )\right )+a \left (-2 a^2-6 a b c x^3-3 a b c^2 x^6 \log \left (1-c x^3\right )+3 a b c^2 x^6 \log \left (1+c x^3\right )+12 b^2 c^2 x^6 \log \left (\frac {c x^3}{\sqrt {1-c^2 x^6}}\right )\right )-6 b^3 c^2 x^6 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )}{12 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])^3/x^7,x]

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{3}\right )\right )^{3}}{x^{7}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))^3/x^7,x)

[Out]

int((a+b*arctanh(c*x^3))^3/x^7,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))^3/x^7,x, algorithm="maxima")

[Out]

1/4*((c*log(c*x^3 + 1) - c*log(c*x^3 - 1) - 2/x^3)*c - 2*arctanh(c*x^3)/x^6)*a^2*b + 1/8*((2*(log(c*x^3 - 1) -
2)*log(c*x^3 + 1) - log(c*x^3 + 1)^2 - log(c*x^3 - 1)^2 - 4*log(c*x^3 - 1) + 24*log(x))*c^2 + 4*(c*log(c*x^3
+ 1) - c*log(c*x^3 - 1) - 2/x^3)*c*arctanh(c*x^3))*a*b^2 - 1/48*b^3*(((c^2*x^6 - 1)*log(-c*x^3 + 1)^3 + 3*(2*c
*x^3 - (c^2*x^6 - 1)*log(c*x^3 + 1))*log(-c*x^3 + 1)^2)/x^6 + 6*integrate(-((c*x^3 - 1)*log(c*x^3 + 1)^3 + 3*(
2*c^2*x^6 - (c*x^3 - 1)*log(c*x^3 + 1)^2 - (c^3*x^9 - c*x^3)*log(c*x^3 + 1))*log(-c*x^3 + 1))/(c*x^10 - x^7),
x)) - 1/2*a*b^2*arctanh(c*x^3)^2/x^6 - 1/6*a^3/x^6

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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))^3/x^7,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x^3)^3 + 3*a*b^2*arctanh(c*x^3)^2 + 3*a^2*b*arctanh(c*x^3) + a^3)/x^7, 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))**3/x**7,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))^3/x^7,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^3) + a)^3/x^7, 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 )}^3}{x^7} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))^3/x^7,x)

[Out]

int((a + b*atanh(c*x^3))^3/x^7, x)

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