3.3.17 \(\int (1-a^2 x^2)^2 \tanh ^{-1}(a x)^3 \, dx\) [217]

Optimal. Leaf size=248 \[ -\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8 \tanh ^{-1}(a x)^3}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {8 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{5 a}-\frac {\log \left (1-a^2 x^2\right )}{2 a}-\frac {8 \tanh ^{-1}(a x) \text {PolyLog}\left (2,1-\frac {2}{1-a x}\right )}{5 a}+\frac {4 \text {PolyLog}\left (3,1-\frac {2}{1-a x}\right )}{5 a} \]

[Out]

1/20*(a^2*x^2-1)/a-x*arctanh(a*x)-1/10*x*(-a^2*x^2+1)*arctanh(a*x)+2/5*(-a^2*x^2+1)*arctanh(a*x)^2/a+3/20*(-a^
2*x^2+1)^2*arctanh(a*x)^2/a+8/15*arctanh(a*x)^3/a+8/15*x*arctanh(a*x)^3+4/15*x*(-a^2*x^2+1)*arctanh(a*x)^3+1/5
*x*(-a^2*x^2+1)^2*arctanh(a*x)^3-8/5*arctanh(a*x)^2*ln(2/(-a*x+1))/a-1/2*ln(-a^2*x^2+1)/a-8/5*arctanh(a*x)*pol
ylog(2,1-2/(-a*x+1))/a+4/5*polylog(3,1-2/(-a*x+1))/a

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Rubi [A]
time = 0.19, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6091, 6021, 6131, 6055, 6095, 6205, 6745, 266, 6089} \begin {gather*} -\frac {1-a^2 x^2}{20 a}-\frac {\log \left (1-a^2 x^2\right )}{2 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {4 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{5 a}-\frac {8 \text {Li}_2\left (1-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)}{5 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {8 \tanh ^{-1}(a x)^3}{15 a}-x \tanh ^{-1}(a x)-\frac {8 \log \left (\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{5 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^2*ArcTanh[a*x]^3,x]

[Out]

-1/20*(1 - a^2*x^2)/a - x*ArcTanh[a*x] - (x*(1 - a^2*x^2)*ArcTanh[a*x])/10 + (2*(1 - a^2*x^2)*ArcTanh[a*x]^2)/
(5*a) + (3*(1 - a^2*x^2)^2*ArcTanh[a*x]^2)/(20*a) + (8*ArcTanh[a*x]^3)/(15*a) + (8*x*ArcTanh[a*x]^3)/15 + (4*x
*(1 - a^2*x^2)*ArcTanh[a*x]^3)/15 + (x*(1 - a^2*x^2)^2*ArcTanh[a*x]^3)/5 - (8*ArcTanh[a*x]^2*Log[2/(1 - a*x)])
/(5*a) - Log[1 - a^2*x^2]/(2*a) - (8*ArcTanh[a*x]*PolyLog[2, 1 - 2/(1 - a*x)])/(5*a) + (4*PolyLog[3, 1 - 2/(1
- a*x)])/(5*a)

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 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 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 6089

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

Rule 6091

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

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 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 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3 \, dx &=\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {3}{10} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx+\frac {4}{5} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3 \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {1}{5} \int \tanh ^{-1}(a x) \, dx+\frac {8}{15} \int \tanh ^{-1}(a x)^3 \, dx-\frac {4}{5} \int \tanh ^{-1}(a x) \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3+\frac {1}{5} a \int \frac {x}{1-a^2 x^2} \, dx+\frac {1}{5} (4 a) \int \frac {x}{1-a^2 x^2} \, dx-\frac {1}{5} (8 a) \int \frac {x \tanh ^{-1}(a x)^2}{1-a^2 x^2} \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8 \tanh ^{-1}(a x)^3}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {\log \left (1-a^2 x^2\right )}{2 a}-\frac {8}{5} \int \frac {\tanh ^{-1}(a x)^2}{1-a x} \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8 \tanh ^{-1}(a x)^3}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {8 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{5 a}-\frac {\log \left (1-a^2 x^2\right )}{2 a}+\frac {16}{5} \int \frac {\tanh ^{-1}(a x) \log \left (\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8 \tanh ^{-1}(a x)^3}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {8 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{5 a}-\frac {\log \left (1-a^2 x^2\right )}{2 a}-\frac {8 \tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{5 a}+\frac {8}{5} \int \frac {\text {Li}_2\left (1-\frac {2}{1-a x}\right )}{1-a^2 x^2} \, dx\\ &=-\frac {1-a^2 x^2}{20 a}-x \tanh ^{-1}(a x)-\frac {1}{10} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}{5 a}+\frac {3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}{20 a}+\frac {8 \tanh ^{-1}(a x)^3}{15 a}+\frac {8}{15} x \tanh ^{-1}(a x)^3+\frac {4}{15} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3+\frac {1}{5} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3-\frac {8 \tanh ^{-1}(a x)^2 \log \left (\frac {2}{1-a x}\right )}{5 a}-\frac {\log \left (1-a^2 x^2\right )}{2 a}-\frac {8 \tanh ^{-1}(a x) \text {Li}_2\left (1-\frac {2}{1-a x}\right )}{5 a}+\frac {4 \text {Li}_3\left (1-\frac {2}{1-a x}\right )}{5 a}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 183, normalized size = 0.74 \begin {gather*} \frac {-3+3 a^2 x^2-66 a x \tanh ^{-1}(a x)+6 a^3 x^3 \tanh ^{-1}(a x)+33 \tanh ^{-1}(a x)^2-42 a^2 x^2 \tanh ^{-1}(a x)^2+9 a^4 x^4 \tanh ^{-1}(a x)^2-32 \tanh ^{-1}(a x)^3+60 a x \tanh ^{-1}(a x)^3-40 a^3 x^3 \tanh ^{-1}(a x)^3+12 a^5 x^5 \tanh ^{-1}(a x)^3-96 \tanh ^{-1}(a x)^2 \log \left (1+e^{-2 \tanh ^{-1}(a x)}\right )-30 \log \left (1-a^2 x^2\right )+96 \tanh ^{-1}(a x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(a x)}\right )+48 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(a x)}\right )}{60 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - a^2*x^2)^2*ArcTanh[a*x]^3,x]

[Out]

(-3 + 3*a^2*x^2 - 66*a*x*ArcTanh[a*x] + 6*a^3*x^3*ArcTanh[a*x] + 33*ArcTanh[a*x]^2 - 42*a^2*x^2*ArcTanh[a*x]^2
+ 9*a^4*x^4*ArcTanh[a*x]^2 - 32*ArcTanh[a*x]^3 + 60*a*x*ArcTanh[a*x]^3 - 40*a^3*x^3*ArcTanh[a*x]^3 + 12*a^5*x
^5*ArcTanh[a*x]^3 - 96*ArcTanh[a*x]^2*Log[1 + E^(-2*ArcTanh[a*x])] - 30*Log[1 - a^2*x^2] + 96*ArcTanh[a*x]*Pol
yLog[2, -E^(-2*ArcTanh[a*x])] + 48*PolyLog[3, -E^(-2*ArcTanh[a*x])])/(60*a)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 30.23, size = 828, normalized size = 3.34

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(828\)
default	\(\text {Expression too large to display}\)	\(828\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*x^2+1)^2*arctanh(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/a*(-2/3*arctanh(a*x)^3*a^3*x^3-4/5*I*Pi*arctanh(a*x)^2+arctanh(a*x)^3*a*x+2/5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2
-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^
2-1/20+1/20*a^2*x^2-11/10*a*x*arctanh(a*x)+1/10*a^3*x^3*arctanh(a*x)+3/20*a^4*x^4*arctanh(a*x)^2-7/10*a^2*x^2*
arctanh(a*x)^2+8/15*arctanh(a*x)^3+11/20*arctanh(a*x)^2-arctanh(a*x)+4/5*arctanh(a*x)^2*ln(a*x+1)+4/5*arctanh(
a*x)^2*ln(a*x-1)-4/5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)^2-2/
5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a
*x)^2+2/5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctan
h(a*x)^2-2/5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))*csgn(I*(a*x+1)/(-a^2*x^2+1)^(1/2))^2*arctanh(a*x)^2+ln((a*x+1)
^2/(-a^2*x^2+1)+1)+4/5*polylog(3,-(a*x+1)^2/(-a^2*x^2+1))-2/5*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a
^2*x^2+1)+1))^3*arctanh(a*x)^2+4/5*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^2-2/5*I*Pi*csgn(I*(a
*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^2-4/5*I*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^2+1/5*arctanh
(a*x)^3*a^5*x^5-8/5*arctanh(a*x)^2*ln(2)-8/5*arctanh(a*x)^2*ln((a*x+1)/(-a^2*x^2+1)^(1/2))-8/5*arctanh(a*x)*po
lylog(2,-(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((-a^2*x^2+1)^2*arctanh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2400*(36*a^5*x^5 - 45*a^4*x^4 - 140*a^3*x^3 + 210*a^2*x^2 + 480*a*x - 60*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x +
8)*log(a*x + 1))*log(-a*x + 1)^2/a - 1/8*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1
)/a - 1/1440000*(288*(125*log(-a*x + 1)^3 - 75*log(-a*x + 1)^2 + 30*log(-a*x + 1) - 6)*(a*x - 1)^5 + 5625*(32*
log(-a*x + 1)^3 - 24*log(-a*x + 1)^2 + 12*log(-a*x + 1) - 3)*(a*x - 1)^4 + 40000*(9*log(-a*x + 1)^3 - 9*log(-a
*x + 1)^2 + 6*log(-a*x + 1) - 2)*(a*x - 1)^3 + 90000*(4*log(-a*x + 1)^3 - 6*log(-a*x + 1)^2 + 6*log(-a*x + 1)
- 3)*(a*x - 1)^2 + 180000*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(a*x - 1))/a + 1/432*(4*
(9*log(-a*x + 1)^3 - 9*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 2)*(a*x - 1)^3 + 27*(4*log(-a*x + 1)^3 - 6*log(-a*x
+ 1)^2 + 6*log(-a*x + 1) - 3)*(a*x - 1)^2 + 108*(log(-a*x + 1)^3 - 3*log(-a*x + 1)^2 + 6*log(-a*x + 1) - 6)*(
a*x - 1))/a - 1/8*integrate(-1/150*(150*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*log(a*x + 1)^3 +
(36*a^5*x^5 - 45*a^4*x^4 - 140*a^3*x^3 + 210*a^2*x^2 - 450*(a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x -
1)*log(a*x + 1)^2 + 480*a*x - 60*(3*a^5*x^5 - 10*a^3*x^3 + 15*a*x + 8)*log(a*x + 1))*log(-a*x + 1))/(a*x - 1)
, 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((-a^2*x^2+1)^2*arctanh(a*x)^3,x, algorithm="fricas")

[Out]

integral((a^4*x^4 - 2*a^2*x^2 + 1)*arctanh(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*x**2+1)**2*atanh(a*x)**3,x)

[Out]

Integral((a*x - 1)**2*(a*x + 1)**2*atanh(a*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((-a^2*x^2+1)^2*arctanh(a*x)^3,x, algorithm="giac")

[Out]

integrate((a^2*x^2 - 1)^2*arctanh(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^3*(a^2*x^2 - 1)^2,x)

[Out]

int(atanh(a*x)^3*(a^2*x^2 - 1)^2, x)

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