3.1.26 \(\int x^3 (a+b \tanh ^{-1}(c x))^3 \, dx\) [26]

Optimal. Leaf size=185 \[ \frac {b^3 x}{4 c^3}-\frac {b^3 \tanh ^{-1}(c x)}{4 c^4}+\frac {b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {2 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^4}-\frac {b^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.42, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {6037, 6127, 327, 212, 6131, 6055, 2449, 2352, 6021, 6095} \begin {gather*} -\frac {2 b^2 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4}+\frac {b^2 x^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 c^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{4 c^4}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4}+\frac {3 b x \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c^3}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 c}-\frac {b^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4}-\frac {b^3 \tanh ^{-1}(c x)}{4 c^4}+\frac {b^3 x}{4 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x])^3,x]

[Out]

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

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.35, size = 245, normalized size = 1.32 \begin {gather*} \frac {-2 a b^2+6 a^2 b c x+2 b^3 c x+2 a b^2 c^2 x^2+2 a^2 b c^3 x^3+2 a^3 c^4 x^4+2 b^2 \left (b \left (-4+3 c x+c^3 x^3\right )+3 a \left (-1+c^4 x^4\right )\right ) \tanh ^{-1}(c x)^2+2 b^3 \left (-1+c^4 x^4\right ) \tanh ^{-1}(c x)^3+2 b \tanh ^{-1}(c x) \left (3 a^2 c^4 x^4+b^2 \left (-1+c^2 x^2\right )+2 a b c x \left (3+c^2 x^2\right )-8 b^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+3 a^2 b \log (1-c x)-3 a^2 b \log (1+c x)+8 a b^2 \log \left (1-c^2 x^2\right )+8 b^3 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )}{8 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b^2 + 6*a^2*b*c*x + 2*b^3*c*x + 2*a*b^2*c^2*x^2 + 2*a^2*b*c^3*x^3 + 2*a^3*c^4*x^4 + 2*b^2*(b*(-4 + 3*c*x
+ c^3*x^3) + 3*a*(-1 + c^4*x^4))*ArcTanh[c*x]^2 + 2*b^3*(-1 + c^4*x^4)*ArcTanh[c*x]^3 + 2*b*ArcTanh[c*x]*(3*a
^2*c^4*x^4 + b^2*(-1 + c^2*x^2) + 2*a*b*c*x*(3 + c^2*x^2) - 8*b^2*Log[1 + E^(-2*ArcTanh[c*x])]) + 3*a^2*b*Log[
1 - c*x] - 3*a^2*b*Log[1 + c*x] + 8*a*b^2*Log[1 - c^2*x^2] + 8*b^3*PolyLog[2, -E^(-2*ArcTanh[c*x])])/(8*c^4)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.26, size = 1157, normalized size = 6.25

method	result	size
risch	\(\frac {b^{3} x}{4 c^{3}}+\frac {3 \ln \left (-c x +1\right )^{2} a \,b^{2} x^{4}}{16}-\frac {3 \ln \left (-c x +1\right ) a^{2} b \,x^{4}}{8}-\frac {b^{3} \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{c^{4}}+\frac {b^{2} \ln \left (-c x -1\right ) a}{c^{4}}-\frac {3 b^{3} \left (-c x +1\right )^{4} \ln \left (-c x +1\right )^{2}}{128 c^{4}}+\frac {3 b^{3} \left (-c x +1\right )^{4} \ln \left (-c x +1\right )}{256 c^{4}}+\frac {b^{3} \left (-c x +1\right )^{3} \ln \left (-c x +1\right )^{2}}{16 c^{4}}-\frac {b^{3} \left (-c x +1\right )^{3} \ln \left (-c x +1\right )}{12 c^{4}}-\frac {3 b^{3} \left (-c x +1\right )^{2} \ln \left (-c x +1\right )^{2}}{32 c^{4}}+\frac {5 b^{3} \left (-c x +1\right )^{2} \ln \left (-c x +1\right )}{32 c^{4}}-\frac {3 b \ln \left (-c x -1\right ) a^{2}}{8 c^{4}}+\frac {b^{3} \left (c^{4} x^{4}-1\right ) \ln \left (c x +1\right )^{3}}{32 c^{4}}+\frac {b^{2} \left (-3 x^{4} b \ln \left (-c x +1\right ) c^{4}+6 c^{4} x^{4} a +2 b \,c^{3} x^{3}+6 b c x +3 b \ln \left (-c x +1\right )-6 a +8 b \right ) \ln \left (c x +1\right )^{2}}{32 c^{4}}-\frac {b^{3} \left (-c x +1\right ) \ln \left (-c x +1\right )}{2 c^{4}}+\frac {b^{3} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{c^{4}}-\frac {3 x^{4} b^{3} \ln \left (-c x +1\right )}{256}+\left (\frac {3 b^{3} \left (c^{4} x^{4}-1\right ) \ln \left (-c x +1\right )^{2}}{32 c^{4}}-\frac {b^{2} x \left (3 c^{3} x^{3} a +b \,c^{2} x^{2}+3 b \right ) \ln \left (-c x +1\right )}{8 c^{3}}-\frac {b \left (-3 c^{4} x^{4} a^{2}-2 a b \,c^{3} x^{3}-b^{2} c^{2} x^{2}-6 a b c x -3 b \ln \left (-c x +1\right ) a -4 b^{2} \ln \left (-c x +1\right )\right )}{8 c^{4}}\right ) \ln \left (c x +1\right )+\frac {3 b^{2} \left (-c x +1\right )^{2} \ln \left (-c x +1\right ) a}{8 c^{4}}+\frac {3 b^{2} \left (-c x +1\right )^{4} \ln \left (-c x +1\right ) a}{32 c^{4}}-\frac {b^{2} \left (-c x +1\right )^{3} \ln \left (-c x +1\right ) a}{4 c^{4}}-\frac {a^{3}}{4 c^{4}}+\frac {a^{3} x^{4}}{4}+\frac {415 b^{3} \ln \left (-c x +1\right )}{768 c^{4}}+\frac {b^{3} \dilog \left (-\frac {c x}{2}+\frac {1}{2}\right )}{c^{4}}-\frac {a^{2} b}{c^{4}}-\frac {a \,b^{2}}{4 c^{4}}+\frac {b^{3} \ln \left (-c x +1\right )^{3}}{32 c^{4}}-\frac {25 b^{3} \ln \left (-c x +1\right )^{2}}{128 c^{4}}+\frac {3 b^{3} \ln \left (-c x +1\right )^{2} x^{4}}{128}-\frac {3 a \,b^{2} \ln \left (-c x +1\right ) x^{2}}{16 c^{2}}-\frac {3 a \,b^{2} \ln \left (-c x +1\right ) x}{8 c^{3}}-\frac {a \,b^{2} \ln \left (-c x +1\right ) x^{3}}{8 c}-\frac {b^{3} \ln \left (-c x -1\right )}{8 c^{4}}-\frac {b^{3}}{4 c^{4}}-\frac {\ln \left (-c x +1\right )^{3} b^{3} x^{4}}{32}+\frac {a \,b^{2} x^{2}}{4 c^{2}}+\frac {a^{2} b \,x^{3}}{4 c}+\frac {3 a^{2} b x}{4 c^{3}}-\frac {3 a \,b^{2} \ln \left (-c x +1\right ) x^{4}}{32}+\frac {3 a^{2} b \ln \left (-c x +1\right )}{8 c^{4}}-\frac {3 a \,b^{2} \ln \left (-c x +1\right )^{2}}{16 c^{4}}+\frac {25 a \,b^{2} \ln \left (-c x +1\right )}{32 c^{4}}+\frac {b^{3} \ln \left (-c x +1\right )^{2} x^{3}}{32 c}+\frac {3 b^{3} \ln \left (-c x +1\right )^{2} x^{2}}{64 c^{2}}+\frac {3 b^{3} \ln \left (-c x +1\right )^{2} x}{32 c^{3}}-\frac {7 b^{3} \ln \left (-c x +1\right ) x^{3}}{192 c}-\frac {13 b^{3} \ln \left (-c x +1\right ) x^{2}}{128 c^{2}}-\frac {25 b^{3} \ln \left (-c x +1\right ) x}{64 c^{3}}\)	\(1026\)
derivativedivides	\(\text {Expression too large to display}\)	\(1157\)
default	\(\text {Expression too large to display}\)	\(1157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(1/4*c^4*x^4*a^3+3/16*I*b^3*arctanh(c*x)^2*Pi*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)))^2-1/4*b^3+3/4*a*b^2*c^4*x^4*arctanh(c*x)^2+3/4*c^4*x^4*a^2*b*arctanh(c*x)+3/
16*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+3/16*I*b^3*arctanh(c*x)^
2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+3/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-3/8*I*b^
3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^
2*x^2+1)))*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)))+3/8*a^2*b*ln
(c*x-1)-3/8*a^2*b*ln(c*x+1)+3/4*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+3/8*b^3*arctanh(c*x)^2*ln(c*
x-1)-3/8*b^3*arctanh(c*x)^2*ln(c*x+1)-2*b^3*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^3*arctanh(c*x)
*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3/16*a*b^2*ln(c*x-1)^2+3/16*a*b^2*ln(c*x+1)^2+a*b^2*ln(c*x-1)+a*b^2*ln(c*x
+1)+1/4*b^3*c*x-3/4*a*b^2*arctanh(c*x)*ln(c*x+1)-3/8*a*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)+1/4*a^2*b*c^3*x^3+3/4*a^2
*b*c*x+1/4*b^3*c^3*x^3*arctanh(c*x)^2+3/4*b^3*c*x*arctanh(c*x)^2+1/4*b^3*arctanh(c*x)*c^2*x^2+1/4*a*b^2*c^2*x^
2-3/8*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+3/8*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+3/4*a*b^2*arctanh(c*x)*ln(c*
x-1)-1/4*b^3*arctanh(c*x)-2*b^3*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^3*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/
2))+b^3*arctanh(c*x)^2-1/4*b^3*arctanh(c*x)^3-3/8*I*b^3*arctanh(c*x)^2*Pi+1/4*c^4*x^4*b^3*arctanh(c*x)^3-3/16*
I*b^3*arctanh(c*x)^2*Pi*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+3/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2+3/16*I*b^3*a
rctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))+1/2*a*b^2*c^3*x^3*arctanh(
c*x)+3/2*a*b^2*c*x*arctanh(c*x))

________________________________________________________________________________________

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

[Out]

3/4*a*b^2*x^4*arctanh(c*x)^2 + 1/4*a^3*x^4 + 1/8*(6*x^4*arctanh(c*x) + c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x +
1)/c^5 + 3*log(c*x - 1)/c^5))*a^2*b + 1/16*(4*c*(2*(c^2*x^3 + 3*x)/c^4 - 3*log(c*x + 1)/c^5 + 3*log(c*x - 1)/c
^5)*arctanh(c*x) + (4*c^2*x^2 - 2*(3*log(c*x - 1) - 8)*log(c*x + 1) + 3*log(c*x + 1)^2 + 3*log(c*x - 1)^2 + 16
*log(c*x - 1))/c^4)*a*b^2 - 1/9216*(27*c^5*((c^2*x^4 + 2*x^2)/c^7 + 2*log(c^2*x^2 - 1)/c^9) + 74*c^4*(2*(c^2*x
^3 + 3*x)/c^7 - 3*log(c*x + 1)/c^8 + 3*log(c*x - 1)/c^8) + 60*c^3*(x^2/c^5 + log(c^2*x^2 - 1)/c^7) - 221184*c^
3*integrate(1/96*x^3*log(c*x + 1)/(c^5*x^2 - c^3), x) + 1692*c^2*(2*x/c^5 - log(c*x + 1)/c^6 + log(c*x - 1)/c^
6) - 1105920*c*integrate(1/96*x*log(c*x + 1)/(c^5*x^2 - c^3), x) + (9*(32*log(-c*x + 1)^3 - 24*log(-c*x + 1)^2
+ 12*log(-c*x + 1) - 3)*(c*x - 1)^4 + 128*(9*log(-c*x + 1)^3 - 9*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 2)*(c*x
- 1)^3 + 432*(4*log(-c*x + 1)^3 - 6*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 3)*(c*x - 1)^2 + 1152*(log(-c*x + 1)^3
- 3*log(-c*x + 1)^2 + 6*log(-c*x + 1) - 6)*(c*x - 1))/c^4 - 12*(24*(c^4*x^4 - 1)*log(c*x + 1)^3 + 48*(c^3*x^3
+ 3*c*x)*log(c*x + 1)^2 - 6*(3*c^4*x^4 - 4*c^3*x^3 + 6*c^2*x^2 - 12*c*x - 12*(c^4*x^4 - 1)*log(c*x + 1) + 7)*
log(-c*x + 1)^2 + (9*c^4*x^4 + 28*c^3*x^3 - 18*c^2*x^2 - 72*(c^4*x^4 - 1)*log(c*x + 1)^2 + 300*c*x - 96*(c^3*x
^3 + 3*c*x + 4)*log(c*x + 1))*log(-c*x + 1))/c^4 + 1800*log(96*c^5*x^2 - 96*c^3)/c^4 - 442368*integrate(1/96*l
og(c*x + 1)/(c^5*x^2 - c^3), x))*b^3

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

int(x^3*(a + b*atanh(c*x))^3, x)

________________________________________________________________________________________