3.2.27 \(\int \frac {x^2 \tanh ^{-1}(a x)^3}{c+a c x} \, dx\) [127]

Optimal. Leaf size=309 \[ \frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^3 c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}-\frac {3 \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,1-\frac {2}{1+a x}\right )}{2 a^3 c}-\frac {3 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (3,1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \text {PolyLog}\left (4,1-\frac {2}{1+a x}\right )}{4 a^3 c} \]

[Out]

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

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Rubi [A]
time = 0.46, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {6077, 6037, 6127, 6021, 6131, 6055, 2449, 2352, 6095, 6205, 6745, 6203, 6207} \begin {gather*} -\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{a x+1}\right )}{4 a^3 c}+\frac {3 \text {Li}_2\left (1-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3 c}+\frac {3 \text {Li}_3\left (1-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)}{2 a^3 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}-\frac {\log \left (\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{a^3 c}+\frac {3 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{a^3 c}-\frac {3 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTanh[a*x]^3)/(c + a*c*x),x]

[Out]

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

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 6077

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d*(f/e), Int[(f*x)^(m - 1)*((a + b*ArcTanh[c*x])^p/(d
+ e*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 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 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 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 6207

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-(a
+ b*ArcTanh[c*x])^p)*(PolyLog[k + 1, u]/(2*c*d)), x] + Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog
[k + 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[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 \frac {x^2 \tanh ^{-1}(a x)^3}{c+a c x} \, dx &=-\frac {\int \frac {x \tanh ^{-1}(a x)^3}{c+a c x} \, dx}{a}+\frac {\int x \tanh ^{-1}(a x)^3 \, dx}{a c}\\ &=\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}+\frac {\int \frac {\tanh ^{-1}(a x)^3}{c+a c x} \, dx}{a^2}-\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 c}-\frac {\int \tanh ^{-1}(a x)^3 \, dx}{a^2 c}\\ &=-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}+\frac {3 \int \tanh ^{-1}(a x)^2 \, dx}{2 a^2 c}-\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{2 a^2 c}+\frac {3 \int \frac {\tanh ^{-1}(a x)^2 \log \left (\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{a^2 c}+\frac {3 \int \frac {x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{a c}\\ &=\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \int \frac {\tanh ^{-1}(a x)^2}{1-a x} \, dx}{a^2 c}-\frac {3 \int \frac {\tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{a^2 c}-\frac {3 \int \frac {x \tanh ^{-1}(a x)}{1-a^2 x^2} \, dx}{a c}\\ &=\frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}+\frac {3 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^3 c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}-\frac {3 \int \frac {\text {Li}_3\left (1-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{2 a^2 c}-\frac {3 \int \frac {\tanh ^{-1}(a x)}{1-a x} \, dx}{a^2 c}-\frac {6 \int \frac {\tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^2 c}\\ &=\frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^3 c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1+a x}\right )}{4 a^3 c}+\frac {3 \int \frac {\log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^2 c}-\frac {3 \int \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx}{a^2 c}\\ &=\frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^3 c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1+a x}\right )}{4 a^3 c}-\frac {3 \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-a x}\right )}{a^3 c}\\ &=\frac {3 \tanh ^{-1}(a x)^2}{2 a^3 c}+\frac {3 x \tanh ^{-1}(a x)^2}{2 a^2 c}-\frac {3 \tanh ^{-1}(a x)^3}{2 a^3 c}-\frac {x \tanh ^{-1}(a x)^3}{a^2 c}+\frac {x^2 \tanh ^{-1}(a x)^3}{2 a c}-\frac {3 \tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{a^3 c}-\frac {\tanh ^{-1}(a x)^3 \log \left (\frac {2}{1+a x}\right )}{a^3 c}-\frac {3 \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{a^3 c}+\frac {3 \tanh ^{-1}(a x)^2 \text {Li}_2\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}-\frac {3 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{2 a^3 c}+\frac {3 \tanh ^{-1}(a x) \text {Li}_3\left (1-\frac {2}{1+a x}\right )}{2 a^3 c}+\frac {3 \text {Li}_4\left (1-\frac {2}{1+a x}\right )}{4 a^3 c}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 172, normalized size = 0.56 \begin {gather*} \frac {-6 \tanh ^{-1}(a x)^2+6 a x \tanh ^{-1}(a x)^2+2 \tanh ^{-1}(a x)^3-4 a x \tanh ^{-1}(a x)^3+2 a^2 x^2 \tanh ^{-1}(a x)^3-12 \tanh ^{-1}(a x) \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )+12 \tanh ^{-1}(a x)^2 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-4 \tanh ^{-1}(a x)^3 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )+6 \left (-1+\tanh ^{-1}(a x)\right )^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+6 \left (-1+\tanh ^{-1}(a x)\right ) \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )+3 \text {PolyLog}\left (4,-e^{-2 \tanh ^{-1}(a x)}\right )}{4 a^3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*ArcTanh[a*x]^2 + 6*a*x*ArcTanh[a*x]^2 + 2*ArcTanh[a*x]^3 - 4*a*x*ArcTanh[a*x]^3 + 2*a^2*x^2*ArcTanh[a*x]^3
- 12*ArcTanh[a*x]*Log[1 + E^(-2*ArcTanh[a*x])] + 12*ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*x])] - 4*ArcTanh[a
*x]^3*Log[1 + E^(-2*ArcTanh[a*x])] + 6*(-1 + ArcTanh[a*x])^2*PolyLog[2, -E^(-2*ArcTanh[a*x])] + 6*(-1 + ArcTan
h[a*x])*PolyLog[3, -E^(-2*ArcTanh[a*x])] + 3*PolyLog[4, -E^(-2*ArcTanh[a*x])])/(4*a^3*c)

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Maple [A]
time = 23.01, size = 352, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {\frac {\arctanh \left (a x \right )^{2} \left (a x \arctanh \left (a x \right )-\arctanh \left (a x \right )+3\right ) \left (a x -1\right )}{2 c}+\frac {\arctanh \left (a x \right )^{4}}{2 c}-\frac {\arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}-\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}-\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4 c}+\frac {3 \arctanh \left (a x \right )^{2}}{c}-\frac {3 \arctanh \left (a x \right ) \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}-\frac {3 \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}-\frac {2 \arctanh \left (a x \right )^{3}}{c}+\frac {3 \arctanh \left (a x \right )^{2} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{c}-\frac {3 \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}}{a^{3}}\)	\(352\)
default	\(\frac {\frac {\arctanh \left (a x \right )^{2} \left (a x \arctanh \left (a x \right )-\arctanh \left (a x \right )+3\right ) \left (a x -1\right )}{2 c}+\frac {\arctanh \left (a x \right )^{4}}{2 c}-\frac {\arctanh \left (a x \right )^{3} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}-\frac {3 \arctanh \left (a x \right )^{2} \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}-\frac {3 \polylog \left (4, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{4 c}+\frac {3 \arctanh \left (a x \right )^{2}}{c}-\frac {3 \arctanh \left (a x \right ) \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}-\frac {3 \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}-\frac {2 \arctanh \left (a x \right )^{3}}{c}+\frac {3 \arctanh \left (a x \right )^{2} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}+1\right )}{c}+\frac {3 \arctanh \left (a x \right ) \polylog \left (2, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{c}-\frac {3 \polylog \left (3, -\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}\right )}{2 c}}{a^{3}}\)	\(352\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(a*x)^3/(a*c*x+c),x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

-1/16*(a^2*x^2 - 2*a*x + 2*log(a*x + 1))*log(-a*x + 1)^3/(a^3*c) + 1/8*integrate(1/2*(2*(a^3*x^3 - a^2*x^2)*lo
g(a*x + 1)^3 - 6*(a^3*x^3 - a^2*x^2)*log(a*x + 1)^2*log(-a*x + 1) + 3*(a^3*x^3 - a^2*x^2 - 2*a*x + 2*(a^3*x^3
- a^2*x^2 + a*x + 1)*log(a*x + 1))*log(-a*x + 1)^2)/(a^4*c*x^2 - a^2*c), 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^2*arctanh(a*x)^3/(a*c*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {x^{2} \operatorname {atanh}^{3}{\left (a x \right )}}{a x + 1}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*atanh(a*x)**3/(a*x + 1), x)/c

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atanh(a*x)^3)/(c + a*c*x),x)

[Out]

int((x^2*atanh(a*x)^3)/(c + a*c*x), x)

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