3.333 \(\int \frac{\sinh ^{-1}(a x)^3}{(c+a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=409 \[ -\frac{9 i \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{5 i \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{5 i \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{9 i \text{PolyLog}\left (4,-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{9 i \text{PolyLog}\left (4,i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{1}{4 a c^3 \sqrt{a^2 x^2+1}}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (a^2 x^2+1\right )}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{a^2 x^2+1}}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (a^2 x^2+1\right )^{3/2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3} \]

[Out]

-1/(4*a*c^3*Sqrt[1 + a^2*x^2]) - (x*ArcSinh[a*x])/(4*c^3*(1 + a^2*x^2)) + ArcSinh[a*x]^2/(4*a*c^3*(1 + a^2*x^2
)^(3/2)) + (9*ArcSinh[a*x]^2)/(8*a*c^3*Sqrt[1 + a^2*x^2]) + (x*ArcSinh[a*x]^3)/(4*c^3*(1 + a^2*x^2)^2) + (3*x*
ArcSinh[a*x]^3)/(8*c^3*(1 + a^2*x^2)) - (5*ArcSinh[a*x]*ArcTan[E^ArcSinh[a*x]])/(a*c^3) + (3*ArcSinh[a*x]^3*Ar
cTan[E^ArcSinh[a*x]])/(4*a*c^3) + (((5*I)/2)*PolyLog[2, (-I)*E^ArcSinh[a*x]])/(a*c^3) - (((9*I)/8)*ArcSinh[a*x
]^2*PolyLog[2, (-I)*E^ArcSinh[a*x]])/(a*c^3) - (((5*I)/2)*PolyLog[2, I*E^ArcSinh[a*x]])/(a*c^3) + (((9*I)/8)*A
rcSinh[a*x]^2*PolyLog[2, I*E^ArcSinh[a*x]])/(a*c^3) + (((9*I)/4)*ArcSinh[a*x]*PolyLog[3, (-I)*E^ArcSinh[a*x]])
/(a*c^3) - (((9*I)/4)*ArcSinh[a*x]*PolyLog[3, I*E^ArcSinh[a*x]])/(a*c^3) - (((9*I)/4)*PolyLog[4, (-I)*E^ArcSin
h[a*x]])/(a*c^3) + (((9*I)/4)*PolyLog[4, I*E^ArcSinh[a*x]])/(a*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.516848, antiderivative size = 409, normalized size of antiderivative = 1., number of steps used = 28, number of rules used = 11, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.579, Rules used = {5690, 5693, 4180, 2531, 6609, 2282, 6589, 5717, 2279, 2391, 261} \[ -\frac{9 i \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x) \text{PolyLog}\left (3,i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{5 i \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{5 i \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{9 i \text{PolyLog}\left (4,-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{9 i \text{PolyLog}\left (4,i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{1}{4 a c^3 \sqrt{a^2 x^2+1}}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (a^2 x^2+1\right )}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{a^2 x^2+1}}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (a^2 x^2+1\right )^{3/2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (a^2 x^2+1\right )}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(4*a*c^3*Sqrt[1 + a^2*x^2]) - (x*ArcSinh[a*x])/(4*c^3*(1 + a^2*x^2)) + ArcSinh[a*x]^2/(4*a*c^3*(1 + a^2*x^2
)^(3/2)) + (9*ArcSinh[a*x]^2)/(8*a*c^3*Sqrt[1 + a^2*x^2]) + (x*ArcSinh[a*x]^3)/(4*c^3*(1 + a^2*x^2)^2) + (3*x*
ArcSinh[a*x]^3)/(8*c^3*(1 + a^2*x^2)) - (5*ArcSinh[a*x]*ArcTan[E^ArcSinh[a*x]])/(a*c^3) + (3*ArcSinh[a*x]^3*Ar
cTan[E^ArcSinh[a*x]])/(4*a*c^3) + (((5*I)/2)*PolyLog[2, (-I)*E^ArcSinh[a*x]])/(a*c^3) - (((9*I)/8)*ArcSinh[a*x
]^2*PolyLog[2, (-I)*E^ArcSinh[a*x]])/(a*c^3) - (((5*I)/2)*PolyLog[2, I*E^ArcSinh[a*x]])/(a*c^3) + (((9*I)/8)*A
rcSinh[a*x]^2*PolyLog[2, I*E^ArcSinh[a*x]])/(a*c^3) + (((9*I)/4)*ArcSinh[a*x]*PolyLog[3, (-I)*E^ArcSinh[a*x]])
/(a*c^3) - (((9*I)/4)*ArcSinh[a*x]*PolyLog[3, I*E^ArcSinh[a*x]])/(a*c^3) - (((9*I)/4)*PolyLog[4, (-I)*E^ArcSin
h[a*x]])/(a*c^3) + (((9*I)/4)*PolyLog[4, I*E^ArcSinh[a*x]])/(a*c^3)

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5693

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sech[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 5717

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 261

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

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac{(3 a) \int \frac{x \sinh ^{-1}(a x)^2}{\left (1+a^2 x^2\right )^{5/2}} \, dx}{4 c^3}+\frac{3 \int \frac{\sinh ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{\int \frac{\sinh ^{-1}(a x)}{\left (1+a^2 x^2\right )^2} \, dx}{2 c^3}-\frac{(9 a) \int \frac{x \sinh ^{-1}(a x)^2}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{8 c^3}+\frac{3 \int \frac{\sinh ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{8 c^2}\\ &=-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{\int \frac{\sinh ^{-1}(a x)}{1+a^2 x^2} \, dx}{4 c^3}-\frac{9 \int \frac{\sinh ^{-1}(a x)}{1+a^2 x^2} \, dx}{4 c^3}+\frac{3 \operatorname{Subst}\left (\int x^3 \text{sech}(x) \, dx,x,\sinh ^{-1}(a x)\right )}{8 a c^3}+\frac{a \int \frac{x}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{4 c^3}\\ &=-\frac{1}{4 a c^3 \sqrt{1+a^2 x^2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{8 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int x^2 \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{8 a c^3}-\frac{\operatorname{Subst}\left (\int x \text{sech}(x) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}-\frac{9 \operatorname{Subst}\left (\int x \text{sech}(x) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}\\ &=-\frac{1}{4 a c^3 \sqrt{1+a^2 x^2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{i \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}-\frac{i \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int x \text{Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}\\ &=-\frac{1}{4 a c^3 \sqrt{1+a^2 x^2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int \text{Li}_3\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int \text{Li}_3\left (i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{4 a c^3}\\ &=-\frac{1}{4 a c^3 \sqrt{1+a^2 x^2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{5 i \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}-\frac{5 i \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{(9 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{(9 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}\\ &=-\frac{1}{4 a c^3 \sqrt{1+a^2 x^2}}-\frac{x \sinh ^{-1}(a x)}{4 c^3 \left (1+a^2 x^2\right )}+\frac{\sinh ^{-1}(a x)^2}{4 a c^3 \left (1+a^2 x^2\right )^{3/2}}+\frac{9 \sinh ^{-1}(a x)^2}{8 a c^3 \sqrt{1+a^2 x^2}}+\frac{x \sinh ^{-1}(a x)^3}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac{3 x \sinh ^{-1}(a x)^3}{8 c^3 \left (1+a^2 x^2\right )}-\frac{5 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^3}+\frac{3 \sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{5 i \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}-\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}-\frac{5 i \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^3}+\frac{9 i \sinh ^{-1}(a x)^2 \text{Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{8 a c^3}+\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \sinh ^{-1}(a x) \text{Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}-\frac{9 i \text{Li}_4\left (-i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}+\frac{9 i \text{Li}_4\left (i e^{\sinh ^{-1}(a x)}\right )}{4 a c^3}\\ \end{align*}

Mathematica [A]  time = 5.36398, size = 654, normalized size = 1.6 \[ -\frac{i \left (576 \sinh ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )+576 i \pi \sinh ^{-1}(a x) \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )+1152 \sinh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(a x)}\right )-1152 \sinh ^{-1}(a x) \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(a x)}\right )-16 \left (-36 \sinh ^{-1}(a x)^2-36 i \pi \sinh ^{-1}(a x)+9 \pi ^2+80\right ) \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(a x)}\right )+1280 \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(a x)}\right )-144 \pi ^2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )+576 i \pi \text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(a x)}\right )-576 i \pi \text{PolyLog}\left (3,i e^{\sinh ^{-1}(a x)}\right )+1152 \text{PolyLog}\left (4,-i e^{-\sinh ^{-1}(a x)}\right )+1152 \text{PolyLog}\left (4,-i e^{\sinh ^{-1}(a x)}\right )-\frac{128 i}{\sqrt{a^2 x^2+1}}+\frac{192 i a x \sinh ^{-1}(a x)^3}{a^2 x^2+1}+\frac{128 i a x \sinh ^{-1}(a x)^3}{\left (a^2 x^2+1\right )^2}+\frac{576 i \sinh ^{-1}(a x)^2}{\sqrt{a^2 x^2+1}}+\frac{128 i \sinh ^{-1}(a x)^2}{\left (a^2 x^2+1\right )^{3/2}}-\frac{128 i a x \sinh ^{-1}(a x)}{a^2 x^2+1}-48 \sinh ^{-1}(a x)^4-96 i \pi \sinh ^{-1}(a x)^3+72 \pi ^2 \sinh ^{-1}(a x)^2+24 i \pi ^3 \sinh ^{-1}(a x)-192 \sinh ^{-1}(a x)^3 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )+192 \sinh ^{-1}(a x)^3 \log \left (1+i e^{\sinh ^{-1}(a x)}\right )-288 i \pi \sinh ^{-1}(a x)^2 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )+288 i \pi \sinh ^{-1}(a x)^2 \log \left (1-i e^{\sinh ^{-1}(a x)}\right )-1280 \sinh ^{-1}(a x) \log \left (1-i e^{-\sinh ^{-1}(a x)}\right )+144 \pi ^2 \sinh ^{-1}(a x) \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )+1280 \sinh ^{-1}(a x) \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )-144 \pi ^2 \sinh ^{-1}(a x) \log \left (1-i e^{\sinh ^{-1}(a x)}\right )+24 i \pi ^3 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )-24 i \pi ^3 \log \left (1+i e^{\sinh ^{-1}(a x)}\right )+24 i \pi ^3 \log \left (\tan \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(a x)\right )\right )\right )+21 \pi ^4\right )}{512 a c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/512)*(21*Pi^4 - (128*I)/Sqrt[1 + a^2*x^2] + (24*I)*Pi^3*ArcSinh[a*x] - ((128*I)*a*x*ArcSinh[a*x])/(1 + a^
2*x^2) + 72*Pi^2*ArcSinh[a*x]^2 + ((128*I)*ArcSinh[a*x]^2)/(1 + a^2*x^2)^(3/2) + ((576*I)*ArcSinh[a*x]^2)/Sqrt
[1 + a^2*x^2] - (96*I)*Pi*ArcSinh[a*x]^3 + ((128*I)*a*x*ArcSinh[a*x]^3)/(1 + a^2*x^2)^2 + ((192*I)*a*x*ArcSinh
[a*x]^3)/(1 + a^2*x^2) - 48*ArcSinh[a*x]^4 - 1280*ArcSinh[a*x]*Log[1 - I/E^ArcSinh[a*x]] + (24*I)*Pi^3*Log[1 +
I/E^ArcSinh[a*x]] + 1280*ArcSinh[a*x]*Log[1 + I/E^ArcSinh[a*x]] + 144*Pi^2*ArcSinh[a*x]*Log[1 + I/E^ArcSinh[a
*x]] - (288*I)*Pi*ArcSinh[a*x]^2*Log[1 + I/E^ArcSinh[a*x]] - 192*ArcSinh[a*x]^3*Log[1 + I/E^ArcSinh[a*x]] - 14
4*Pi^2*ArcSinh[a*x]*Log[1 - I*E^ArcSinh[a*x]] + (288*I)*Pi*ArcSinh[a*x]^2*Log[1 - I*E^ArcSinh[a*x]] - (24*I)*P
i^3*Log[1 + I*E^ArcSinh[a*x]] + 192*ArcSinh[a*x]^3*Log[1 + I*E^ArcSinh[a*x]] + (24*I)*Pi^3*Log[Tan[(Pi + (2*I)
*ArcSinh[a*x])/4]] - 16*(80 + 9*Pi^2 - (36*I)*Pi*ArcSinh[a*x] - 36*ArcSinh[a*x]^2)*PolyLog[2, (-I)/E^ArcSinh[a
*x]] + 1280*PolyLog[2, I/E^ArcSinh[a*x]] + 576*ArcSinh[a*x]^2*PolyLog[2, (-I)*E^ArcSinh[a*x]] - 144*Pi^2*PolyL
og[2, I*E^ArcSinh[a*x]] + (576*I)*Pi*ArcSinh[a*x]*PolyLog[2, I*E^ArcSinh[a*x]] + (576*I)*Pi*PolyLog[3, (-I)/E^
ArcSinh[a*x]] + 1152*ArcSinh[a*x]*PolyLog[3, (-I)/E^ArcSinh[a*x]] - 1152*ArcSinh[a*x]*PolyLog[3, (-I)*E^ArcSin
h[a*x]] - (576*I)*Pi*PolyLog[3, I*E^ArcSinh[a*x]] + 1152*PolyLog[4, (-I)/E^ArcSinh[a*x]] + 1152*PolyLog[4, (-I
)*E^ArcSinh[a*x]]))/(a*c^3)

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Maple [F]  time = 0.227, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}{ \left ({a}^{2}c{x}^{2}+c \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsinh}\left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arsinh}\left (a x\right )^{3}}{a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

integral(arcsinh(a*x)^3/(a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**3/(a**6*x**6 + 3*a**4*x**4 + 3*a**2*x**2 + 1), x)/c**3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsinh}\left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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