3.1.25 \(\int x^4 (a+b \tanh ^{-1}(c x))^3 \, dx\) [25]

Optimal. Leaf size=262 \[ \frac {9 a b^2 x}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{10 c^5} \]

[Out]

9/10*a*b^2*x/c^4+1/20*b^3*x^2/c^3+9/10*b^3*x*arctanh(c*x)/c^4+1/10*b^2*x^3*(a+b*arctanh(c*x))/c^2-9/20*b*(a+b*
arctanh(c*x))^2/c^5+3/10*b*x^2*(a+b*arctanh(c*x))^2/c^3+3/20*b*x^4*(a+b*arctanh(c*x))^2/c+1/5*(a+b*arctanh(c*x
))^3/c^5+1/5*x^5*(a+b*arctanh(c*x))^3-3/5*b*(a+b*arctanh(c*x))^2*ln(2/(-c*x+1))/c^5+1/2*b^3*ln(-c^2*x^2+1)/c^5
-3/5*b^2*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))/c^5+3/10*b^3*polylog(3,1-2/(-c*x+1))/c^5

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Rubi [A]
time = 0.55, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6037, 6127, 272, 45, 6021, 266, 6095, 6131, 6055, 6205, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{5 c^5}+\frac {9 a b^2 x}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}-\frac {3 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{5 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTanh[c*x])^3,x]

[Out]

(9*a*b^2*x)/(10*c^4) + (b^3*x^2)/(20*c^3) + (9*b^3*x*ArcTanh[c*x])/(10*c^4) + (b^2*x^3*(a + b*ArcTanh[c*x]))/(
10*c^2) - (9*b*(a + b*ArcTanh[c*x])^2)/(20*c^5) + (3*b*x^2*(a + b*ArcTanh[c*x])^2)/(10*c^3) + (3*b*x^4*(a + b*
ArcTanh[c*x])^2)/(20*c) + (a + b*ArcTanh[c*x])^3/(5*c^5) + (x^5*(a + b*ArcTanh[c*x])^3)/5 - (3*b*(a + b*ArcTan
h[c*x])^2*Log[2/(1 - c*x)])/(5*c^5) + (b^3*Log[1 - c^2*x^2])/(2*c^5) - (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[2,
1 - 2/(1 - c*x)])/(5*c^5) + (3*b^3*PolyLog[3, 1 - 2/(1 - c*x)])/(10*c^5)

Rule 45

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 266

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

Rule 272

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

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 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 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 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]

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^4 \left (a+b \tanh ^{-1}(c x)\right )^3 \, dx &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{5} (3 b c) \int \frac {x^5 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {(3 b) \int x^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c}-\frac {(3 b) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c}\\ &=\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {1}{10} \left (3 b^2\right ) \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx+\frac {(3 b) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{5 c^3}-\frac {(3 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{1-c^2 x^2} \, dx}{5 c^3}\\ &=\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {(3 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1-c x} \, dx}{5 c^4}+\frac {\left (3 b^2\right ) \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^2}-\frac {\left (3 b^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{10 c^2}-\frac {\left (3 b^2\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{5 c^2}\\ &=\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{10 c^4}-\frac {\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{10 c^4}+\frac {\left (3 b^2\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{5 c^4}-\frac {\left (3 b^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{5 c^4}+\frac {\left (6 b^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac {b^3 \int \frac {x^3}{1-c^2 x^2} \, dx}{10 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{10 c^4}+\frac {\left (3 b^3\right ) \int \tanh ^{-1}(c x) \, dx}{5 c^4}+\frac {\left (3 b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{5 c^4}-\frac {b^3 \text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )}{20 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}-\frac {\left (3 b^3\right ) \int \frac {x}{1-c^2 x^2} \, dx}{10 c^3}-\frac {\left (3 b^3\right ) \int \frac {x}{1-c^2 x^2} \, dx}{5 c^3}-\frac {b^3 \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{20 c}\\ &=\frac {9 a b^2 x}{10 c^4}+\frac {b^3 x^2}{20 c^3}+\frac {9 b^3 x \tanh ^{-1}(c x)}{10 c^4}+\frac {b^2 x^3 \left (a+b \tanh ^{-1}(c x)\right )}{10 c^2}-\frac {9 b \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c^5}+\frac {3 b x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{10 c^3}+\frac {3 b x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{20 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{5 c^5}+\frac {1}{5} x^5 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{5 c^5}+\frac {b^3 \log \left (1-c^2 x^2\right )}{2 c^5}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{5 c^5}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{10 c^5}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 383, normalized size = 1.46 \begin {gather*} \frac {-b^3+18 a b^2 c x+6 a^2 b c^2 x^2+b^3 c^2 x^2+2 a b^2 c^3 x^3+3 a^2 b c^4 x^4+4 a^3 c^5 x^5-18 a b^2 \tanh ^{-1}(c x)+18 b^3 c x \tanh ^{-1}(c x)+12 a b^2 c^2 x^2 \tanh ^{-1}(c x)+2 b^3 c^3 x^3 \tanh ^{-1}(c x)+6 a b^2 c^4 x^4 \tanh ^{-1}(c x)+12 a^2 b c^5 x^5 \tanh ^{-1}(c x)-12 a b^2 \tanh ^{-1}(c x)^2-9 b^3 \tanh ^{-1}(c x)^2+6 b^3 c^2 x^2 \tanh ^{-1}(c x)^2+3 b^3 c^4 x^4 \tanh ^{-1}(c x)^2+12 a b^2 c^5 x^5 \tanh ^{-1}(c x)^2-4 b^3 \tanh ^{-1}(c x)^3+4 b^3 c^5 x^5 \tanh ^{-1}(c x)^3-24 a b^2 \tanh ^{-1}(c x) \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-12 b^3 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+6 a^2 b \log \left (1-c^2 x^2\right )+10 b^3 \log \left (1-c^2 x^2\right )+12 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+6 b^3 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )}{20 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcTanh[c*x])^3,x]

[Out]

(-b^3 + 18*a*b^2*c*x + 6*a^2*b*c^2*x^2 + b^3*c^2*x^2 + 2*a*b^2*c^3*x^3 + 3*a^2*b*c^4*x^4 + 4*a^3*c^5*x^5 - 18*
a*b^2*ArcTanh[c*x] + 18*b^3*c*x*ArcTanh[c*x] + 12*a*b^2*c^2*x^2*ArcTanh[c*x] + 2*b^3*c^3*x^3*ArcTanh[c*x] + 6*
a*b^2*c^4*x^4*ArcTanh[c*x] + 12*a^2*b*c^5*x^5*ArcTanh[c*x] - 12*a*b^2*ArcTanh[c*x]^2 - 9*b^3*ArcTanh[c*x]^2 +
6*b^3*c^2*x^2*ArcTanh[c*x]^2 + 3*b^3*c^4*x^4*ArcTanh[c*x]^2 + 12*a*b^2*c^5*x^5*ArcTanh[c*x]^2 - 4*b^3*ArcTanh[
c*x]^3 + 4*b^3*c^5*x^5*ArcTanh[c*x]^3 - 24*a*b^2*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] - 12*b^3*ArcTanh[c*
x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 6*a^2*b*Log[1 - c^2*x^2] + 10*b^3*Log[1 - c^2*x^2] + 12*b^2*(a + b*ArcTanh
[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 6*b^3*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(20*c^5)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.39, size = 1188, normalized size = 4.53

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1188\)
default	\(\text {Expression too large to display}\)	\(1188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x))^3,x,method=_RETURNVERBOSE)

[Out]

1/c^5*(-3/20*I*b^3*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))-1/20*b
^3+1/5*c^5*x^5*b^3*arctanh(c*x)^3+3/20*a^2*b*c^4*x^4+3/10*a^2*b*c^2*x^2+3/20*b^3*c^4*x^4*arctanh(c*x)^2+3/10*b
^3*arctanh(c*x)^2*c^2*x^2+1/10*b^3*arctanh(c*x)*c^3*x^3+9/10*b^3*arctanh(c*x)*c*x-3/10*I*b^3*Pi*arctanh(c*x)^2
+3/20*I*b^3*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^
2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))-3/5*a*b^2*dilog(1/2*c*x+1/2)-3/5*b^3*arctanh(c*x)*polylog(2,-(c*x+1)
^2/(-c^2*x^2+1))-3/5*b^3*ln(2)*arctanh(c*x)^2+3/5*c^5*x^5*a*b^2*arctanh(c*x)^2+3/5*c^5*x^5*a^2*b*arctanh(c*x)+
3/10*a*b^2*c^4*x^4*arctanh(c*x)+3/5*a*b^2*c^2*x^2*arctanh(c*x)-3/20*I*b^3*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(
c^2*x^2-1))^3-3/10*I*b^3*Pi*arctanh(c*x)^2*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-3/20*I*b^3*Pi*arctanh(c*x)^2*c
sgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+3/10*I*b^3*Pi*arctanh(c*x)^2*csgn(I/(1+(c*x+1)^2/(-c
^2*x^2+1)))^2-b^3*ln(1+(c*x+1)^2/(-c^2*x^2+1))+3/10*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+1/20*c^2*b^3*x^2+3/
10*a^2*b*ln(c*x-1)+3/10*a^2*b*ln(c*x+1)-3/5*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+3/10*b^3*arctanh
(c*x)^2*ln(c*x-1)+3/10*b^3*arctanh(c*x)^2*ln(c*x+1)+3/20*a*b^2*ln(c*x-1)^2-3/20*a*b^2*ln(c*x+1)^2+9/20*a*b^2*l
n(c*x-1)-9/20*a*b^2*ln(c*x+1)+3/5*a*b^2*arctanh(c*x)*ln(c*x+1)-3/10*a*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)+3/10*a*b^2
*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/10*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+3/5*a*b^2*arctanh(c*x)*ln(c*x-1)+1/5*a
^3*c^5*x^5+b^3*arctanh(c*x)-9/20*b^3*arctanh(c*x)^2+1/5*b^3*arctanh(c*x)^3-3/10*I*b^3*Pi*arctanh(c*x)^2*csgn(I
*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2+3/20*I*b^3*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^
2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-3/20*I*b^3*Pi*arctanh(c*x)^2*csgn(I/(1+(c
*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+1/10*a*b^2*c^3*x^3+9/10*b^2*
c*x*a)

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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^4*(a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

1/5*a^3*x^5 + 3/20*(4*x^5*arctanh(c*x) + c*((c^2*x^4 + 2*x^2)/c^4 + 2*log(c^2*x^2 - 1)/c^6))*a^2*b - 1/80*(2*(
b^3*c^5*x^5 - b^3)*log(-c*x + 1)^3 - 3*(4*a*b^2*c^5*x^5 + b^3*c^4*x^4 + 2*b^3*c^2*x^2 + 2*(b^3*c^5*x^5 + b^3)*
log(c*x + 1))*log(-c*x + 1)^2)/c^5 - integrate(-1/40*(5*(b^3*c^5*x^5 - b^3*c^4*x^4)*log(c*x + 1)^3 + 30*(a*b^2
*c^5*x^5 - a*b^2*c^4*x^4)*log(c*x + 1)^2 - 3*(4*a*b^2*c^5*x^5 + b^3*c^4*x^4 + 2*b^3*c^2*x^2 + 5*(b^3*c^5*x^5 -
b^3*c^4*x^4)*log(c*x + 1)^2 - 2*(10*a*b^2*c^4*x^4 - (10*a*b^2*c^5 + b^3*c^5)*x^5 - b^3)*log(c*x + 1))*log(-c*
x + 1))/(c^5*x - c^4), 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^4*(a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^4*arctanh(c*x)^3 + 3*a*b^2*x^4*arctanh(c*x)^2 + 3*a^2*b*x^4*arctanh(c*x) + a^3*x^4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atanh(c*x))**3,x)

[Out]

Integral(x**4*(a + b*atanh(c*x))**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^4*(a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3*x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*atanh(c*x))^3,x)

[Out]

int(x^4*(a + b*atanh(c*x))^3, x)

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