3.4.44 \(\int \frac {1}{(1-a^2 x^2)^3 \tanh ^{-1}(a x)^6} \, dx\) [344]

Optimal. Leaf size=257 \[ -\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {32}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {16 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a} \]

[Out]

-1/5/a/(-a^2*x^2+1)^2/arctanh(a*x)^5-1/5*x/(-a^2*x^2+1)^2/arctanh(a*x)^4-4/15/a/(-a^2*x^2+1)^2/arctanh(a*x)^3+
1/5/a/(-a^2*x^2+1)/arctanh(a*x)^3-8/15*x/(-a^2*x^2+1)^2/arctanh(a*x)^2+1/5*x/(-a^2*x^2+1)/arctanh(a*x)^2-32/15
/a/(-a^2*x^2+1)^2/arctanh(a*x)+8/5/a/(-a^2*x^2+1)/arctanh(a*x)+1/5*(a^2*x^2+1)/a/(-a^2*x^2+1)/arctanh(a*x)+2/1
5*Shi(2*arctanh(a*x))/a+16/15*Shi(4*arctanh(a*x))/a

________________________________________________________________________________________

Rubi [A]
time = 0.91, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 49, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6113, 6179, 6175, 6143, 6181, 5556, 12, 3379} \begin {gather*} \frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}+\frac {a^2 x^2+1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {32}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {16 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^6),x]

[Out]

-1/5*1/(a*(1 - a^2*x^2)^2*ArcTanh[a*x]^5) - x/(5*(1 - a^2*x^2)^2*ArcTanh[a*x]^4) - 4/(15*a*(1 - a^2*x^2)^2*Arc
Tanh[a*x]^3) + 1/(5*a*(1 - a^2*x^2)*ArcTanh[a*x]^3) - (8*x)/(15*(1 - a^2*x^2)^2*ArcTanh[a*x]^2) + x/(5*(1 - a^
2*x^2)*ArcTanh[a*x]^2) - 32/(15*a*(1 - a^2*x^2)^2*ArcTanh[a*x]) + 8/(5*a*(1 - a^2*x^2)*ArcTanh[a*x]) + (1 + a^
2*x^2)/(5*a*(1 - a^2*x^2)*ArcTanh[a*x]) + (2*SinhIntegral[2*ArcTanh[a*x]])/(15*a) + (16*SinhIntegral[4*ArcTanh
[a*x]])/(15*a)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 6113

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

Rule 6143

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTa
nh[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2))), x] + (Dist[4/(b^2*(p + 1)*(p + 2)), Int[x*((a + b*ArcTanh[c*x])
^(p + 2)/(d + e*x^2)^2), x], x] + Simp[(1 + c^2*x^2)*((a + b*ArcTanh[c*x])^(p + 2)/(b^2*e*(p + 1)*(p + 2)*(d +
e*x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[p, -1] && NeQ[p, -2]

Rule 6175

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

Rule 6179

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

Rule 6181

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(
m + 1), Subst[Int[(a + b*x)^p*(Sinh[x]^m/Cosh[x]^(m + 2*(q + 1))), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c,
d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^6} \, dx &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}+\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^5} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}+\frac {1}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx+\frac {1}{5} \left (3 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {1}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {3}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^4} \, dx-\frac {3}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4} \, dx+\frac {1}{15} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {2 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{15} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx-\frac {1}{5} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \, dx+\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx+\frac {1}{5} \left (2 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+2 \left (\frac {2}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\right )-\frac {2}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\frac {1}{15} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx-\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{5} \left (6 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {6}{5} \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx-\frac {6}{5} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx+\frac {8 \text {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {4 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {1}{5} (4 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {1}{5} (8 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\right )\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {8 \text {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {4 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {4 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8 \text {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {1}{5} (12 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{5} (24 a) \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}+\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{15 a}-\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}-\frac {4 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8 \text {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {12 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {24 \text {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {4 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{15 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a}-\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\right )-\frac {12 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}+\frac {24 \text {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{5 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{5 a}\right )+\frac {3 \text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{5 a}\\ &=-\frac {1}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^5}-\frac {x}{5 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4}-\frac {4}{15 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {1}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {8 x}{15 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {x}{5 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {4}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {8}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1+a^2 x^2}{5 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {2 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a}+2 \left (-\frac {2}{5 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{5 a}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{5 a}\right )\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 166, normalized size = 0.65 \begin {gather*} -\frac {3+3 a x \tanh ^{-1}(a x)+\tanh ^{-1}(a x)^2+3 a^2 x^2 \tanh ^{-1}(a x)^2+5 a x \tanh ^{-1}(a x)^3+3 a^3 x^3 \tanh ^{-1}(a x)^3+5 \tanh ^{-1}(a x)^4+24 a^2 x^2 \tanh ^{-1}(a x)^4+3 a^4 x^4 \tanh ^{-1}(a x)^4-2 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )-16 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^5 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{15 a \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^6),x]

[Out]

-1/15*(3 + 3*a*x*ArcTanh[a*x] + ArcTanh[a*x]^2 + 3*a^2*x^2*ArcTanh[a*x]^2 + 5*a*x*ArcTanh[a*x]^3 + 3*a^3*x^3*A
rcTanh[a*x]^3 + 5*ArcTanh[a*x]^4 + 24*a^2*x^2*ArcTanh[a*x]^4 + 3*a^4*x^4*ArcTanh[a*x]^4 - 2*(-1 + a^2*x^2)^2*A
rcTanh[a*x]^5*SinhIntegral[2*ArcTanh[a*x]] - 16*(-1 + a^2*x^2)^2*ArcTanh[a*x]^5*SinhIntegral[4*ArcTanh[a*x]])/
(a*(-1 + a^2*x^2)^2*ArcTanh[a*x]^5)

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Maple [A]
time = 3.71, size = 182, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {-\frac {3}{40 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{10 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{20 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{15}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )^{2}}-\frac {4 \cosh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {16 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{15}}{a}\)	\(182\)
default	\(\frac {-\frac {3}{40 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{10 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{20 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{15}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{5}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{40 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )^{2}}-\frac {4 \cosh \left (4 \arctanh \left (a x \right )\right )}{15 \arctanh \left (a x \right )}+\frac {16 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{15}}{a}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^3/arctanh(a*x)^6,x,method=_RETURNVERBOSE)

[Out]

1/a*(-3/40/arctanh(a*x)^5-1/10/arctanh(a*x)^5*cosh(2*arctanh(a*x))-1/20/arctanh(a*x)^4*sinh(2*arctanh(a*x))-1/
30/arctanh(a*x)^3*cosh(2*arctanh(a*x))-1/30*sinh(2*arctanh(a*x))/arctanh(a*x)^2-1/15/arctanh(a*x)*cosh(2*arcta
nh(a*x))+2/15*Shi(2*arctanh(a*x))-1/40/arctanh(a*x)^5*cosh(4*arctanh(a*x))-1/40/arctanh(a*x)^4*sinh(4*arctanh(
a*x))-1/30/arctanh(a*x)^3*cosh(4*arctanh(a*x))-1/15/arctanh(a*x)^2*sinh(4*arctanh(a*x))-4/15/arctanh(a*x)*cosh
(4*arctanh(a*x))+16/15*Shi(4*arctanh(a*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(1/(-a^2*x^2+1)^3/arctanh(a*x)^6,x, algorithm="maxima")

[Out]

-2/15*((3*a^4*x^4 + 24*a^2*x^2 + 5)*log(a*x + 1)^4 + (3*a^4*x^4 + 24*a^2*x^2 + 5)*log(-a*x + 1)^4 + 2*(3*a^3*x
^3 + 5*a*x)*log(a*x + 1)^3 - 2*(3*a^3*x^3 + 5*a*x + 2*(3*a^4*x^4 + 24*a^2*x^2 + 5)*log(a*x + 1))*log(-a*x + 1)
^3 + 24*a*x*log(a*x + 1) + 4*(3*a^2*x^2 + 1)*log(a*x + 1)^2 + 2*(6*a^2*x^2 + 3*(3*a^4*x^4 + 24*a^2*x^2 + 5)*lo
g(a*x + 1)^2 + 3*(3*a^3*x^3 + 5*a*x)*log(a*x + 1) + 2)*log(-a*x + 1)^2 - 2*(2*(3*a^4*x^4 + 24*a^2*x^2 + 5)*log
(a*x + 1)^3 + 3*(3*a^3*x^3 + 5*a*x)*log(a*x + 1)^2 + 12*a*x + 4*(3*a^2*x^2 + 1)*log(a*x + 1))*log(-a*x + 1) +
48)/((a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1)^5 - 5*(a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1)^4*log(-a*x + 1) + 10*
(a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1)^3*log(-a*x + 1)^2 - 10*(a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1)^2*log(-a*
x + 1)^3 + 5*(a^5*x^4 - 2*a^3*x^2 + a)*log(a*x + 1)*log(-a*x + 1)^4 - (a^5*x^4 - 2*a^3*x^2 + a)*log(-a*x + 1)^
5) + integrate(-8/15*(15*a^3*x^3 + 17*a*x)/((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x + 1) - (a^6*x^6 - 3*
a^4*x^4 + 3*a^2*x^2 - 1)*log(-a*x + 1)), x)

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Fricas [A]
time = 0.38, size = 341, normalized size = 1.33 \begin {gather*} \frac {{\left (8 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 8 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} - 2 \, {\left (3 \, a^{4} x^{4} + 24 \, a^{2} x^{2} + 5\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} - 4 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} - 48 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 8 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 96}{15 \, {\left (a^{5} x^{4} - 2 \, a^{3} x^{2} + a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^3/arctanh(a*x)^6,x, algorithm="fricas")

[Out]

1/15*((8*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 8*(a^4*x^4 - 2*
a^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) + (a^4*x^4 - 2*a^2*x^2 + 1)*log_integra
l(-(a*x + 1)/(a*x - 1)) - (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x -
1))^5 - 2*(3*a^4*x^4 + 24*a^2*x^2 + 5)*log(-(a*x + 1)/(a*x - 1))^4 - 4*(3*a^3*x^3 + 5*a*x)*log(-(a*x + 1)/(a*x
- 1))^3 - 48*a*x*log(-(a*x + 1)/(a*x - 1)) - 8*(3*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 96)/((a^5*x^4 -
2*a^3*x^2 + a)*log(-(a*x + 1)/(a*x - 1))^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{a^{6} x^{6} \operatorname {atanh}^{6}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{6}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{6}{\left (a x \right )} - \operatorname {atanh}^{6}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**3/atanh(a*x)**6,x)

[Out]

-Integral(1/(a**6*x**6*atanh(a*x)**6 - 3*a**4*x**4*atanh(a*x)**6 + 3*a**2*x**2*atanh(a*x)**6 - atanh(a*x)**6),
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(1/(-a^2*x^2+1)^3/arctanh(a*x)^6,x, algorithm="giac")

[Out]

integrate(-1/((a^2*x^2 - 1)^3*arctanh(a*x)^6), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^6\,{\left (a^2\,x^2-1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(atanh(a*x)^6*(a^2*x^2 - 1)^3),x)

[Out]

-int(1/(atanh(a*x)^6*(a^2*x^2 - 1)^3), x)

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