3.294 \(\int \frac {\sin ^{-1}(a x)^3}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=455 \[ -\frac {1}{4 a c^3 \sqrt {1-a^2 x^2}}+\frac {3 x \sin ^{-1}(a x)^3}{8 c^3 \left (1-a^2 x^2\right )}+\frac {x \sin ^{-1}(a x)^3}{4 c^3 \left (1-a^2 x^2\right )^2}-\frac {9 \sin ^{-1}(a x)^2}{8 a c^3 \sqrt {1-a^2 x^2}}-\frac {\sin ^{-1}(a x)^2}{4 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \sin ^{-1}(a x)}{4 c^3 \left (1-a^2 x^2\right )}+\frac {9 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{8 a c^3}-\frac {9 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{8 a c^3}-\frac {9 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {9 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {5 i \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^3}-\frac {5 i \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^3}-\frac {9 i \text {Li}_4\left (-i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {9 i \text {Li}_4\left (i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}-\frac {3 i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}-\frac {5 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^3} \]

[Out]

1/4*x*arcsin(a*x)/c^3/(-a^2*x^2+1)-1/4*arcsin(a*x)^2/a/c^3/(-a^2*x^2+1)^(3/2)+1/4*x*arcsin(a*x)^3/c^3/(-a^2*x^
2+1)^2+3/8*x*arcsin(a*x)^3/c^3/(-a^2*x^2+1)+9/8*I*arcsin(a*x)^2*polylog(2,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3
+5/2*I*polylog(2,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-3/4*I*arcsin(a*x)^3*arctan(I*a*x+(-a^2*x^2+1)^(1/2))/a/c
^3-9/4*I*polylog(4,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-9/8*I*arcsin(a*x)^2*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(1
/2)))/a/c^3+9/4*I*polylog(4,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-9/4*arcsin(a*x)*polylog(3,-I*(I*a*x+(-a^2*x^2+
1)^(1/2)))/a/c^3+9/4*arcsin(a*x)*polylog(3,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-5*I*arcsin(a*x)*arctan(I*a*x+(-
a^2*x^2+1)^(1/2))/a/c^3-5/2*I*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-1/4/a/c^3/(-a^2*x^2+1)^(1/2)-9/8*a
rcsin(a*x)^2/a/c^3/(-a^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.51, antiderivative size = 455, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {4655, 4657, 4181, 2531, 6609, 2282, 6589, 4677, 2279, 2391, 261} \[ \frac {9 i \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{8 a c^3}-\frac {9 i \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{8 a c^3}-\frac {9 \sin ^{-1}(a x) \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {9 \sin ^{-1}(a x) \text {PolyLog}\left (3,i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {5 i \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^3}-\frac {5 i \text {PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{2 a c^3}-\frac {9 i \text {PolyLog}\left (4,-i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}+\frac {9 i \text {PolyLog}\left (4,i e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}-\frac {1}{4 a c^3 \sqrt {1-a^2 x^2}}+\frac {3 x \sin ^{-1}(a x)^3}{8 c^3 \left (1-a^2 x^2\right )}+\frac {x \sin ^{-1}(a x)^3}{4 c^3 \left (1-a^2 x^2\right )^2}-\frac {9 \sin ^{-1}(a x)^2}{8 a c^3 \sqrt {1-a^2 x^2}}-\frac {\sin ^{-1}(a x)^2}{4 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {x \sin ^{-1}(a x)}{4 c^3 \left (1-a^2 x^2\right )}-\frac {3 i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{4 a c^3}-\frac {5 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/(c - a^2*c*x^2)^3,x]

[Out]

-1/(4*a*c^3*Sqrt[1 - a^2*x^2]) + (x*ArcSin[a*x])/(4*c^3*(1 - a^2*x^2)) - ArcSin[a*x]^2/(4*a*c^3*(1 - a^2*x^2)^
(3/2)) - (9*ArcSin[a*x]^2)/(8*a*c^3*Sqrt[1 - a^2*x^2]) + (x*ArcSin[a*x]^3)/(4*c^3*(1 - a^2*x^2)^2) + (3*x*ArcS
in[a*x]^3)/(8*c^3*(1 - a^2*x^2)) - ((5*I)*ArcSin[a*x]*ArcTan[E^(I*ArcSin[a*x])])/(a*c^3) - (((3*I)/4)*ArcSin[a
*x]^3*ArcTan[E^(I*ArcSin[a*x])])/(a*c^3) + (((5*I)/2)*PolyLog[2, (-I)*E^(I*ArcSin[a*x])])/(a*c^3) + (((9*I)/8)
*ArcSin[a*x]^2*PolyLog[2, (-I)*E^(I*ArcSin[a*x])])/(a*c^3) - (((5*I)/2)*PolyLog[2, I*E^(I*ArcSin[a*x])])/(a*c^
3) - (((9*I)/8)*ArcSin[a*x]^2*PolyLog[2, I*E^(I*ArcSin[a*x])])/(a*c^3) - (9*ArcSin[a*x]*PolyLog[3, (-I)*E^(I*A
rcSin[a*x])])/(4*a*c^3) + (9*ArcSin[a*x]*PolyLog[3, I*E^(I*ArcSin[a*x])])/(4*a*c^3) - (((9*I)/4)*PolyLog[4, (-
I)*E^(I*ArcSin[a*x])])/(a*c^3) + (((9*I)/4)*PolyLog[4, I*E^(I*ArcSin[a*x])])/(a*c^3)

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]

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 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 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 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 4181

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

Rule 4655

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

Rule 4657

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

Rule 4677

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

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 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]

Rubi steps

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

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Mathematica [B]  time = 12.52, size = 1544, normalized size = 3.39 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^3/(c - a^2*c*x^2)^3,x]

[Out]

-(((1 + 5*ArcSin[a*x]^2)/4 - (5*(ArcSin[a*x]*(Log[1 - I*E^(I*ArcSin[a*x])] - Log[1 + I*E^(I*ArcSin[a*x])]) + I
*(PolyLog[2, (-I)*E^(I*ArcSin[a*x])] - PolyLog[2, I*E^(I*ArcSin[a*x])])))/2 - (3*((Pi^3*Log[Cot[(Pi/2 - ArcSin
[a*x])/2]])/8 + (3*Pi^2*((Pi/2 - ArcSin[a*x])*(Log[1 - E^(I*(Pi/2 - ArcSin[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcS
in[a*x]))]) + I*(PolyLog[2, -E^(I*(Pi/2 - ArcSin[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcSin[a*x]))])))/4 - (3*Pi
*((Pi/2 - ArcSin[a*x])^2*(Log[1 - E^(I*(Pi/2 - ArcSin[a*x]))] - Log[1 + E^(I*(Pi/2 - ArcSin[a*x]))]) + (2*I)*(
Pi/2 - ArcSin[a*x])*(PolyLog[2, -E^(I*(Pi/2 - ArcSin[a*x]))] - PolyLog[2, E^(I*(Pi/2 - ArcSin[a*x]))]) + 2*(-P
olyLog[3, -E^(I*(Pi/2 - ArcSin[a*x]))] + PolyLog[3, E^(I*(Pi/2 - ArcSin[a*x]))])))/2 + 8*((I/64)*(Pi/2 - ArcSi
n[a*x])^4 + (I/4)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)^4 - ((Pi/2 - ArcSin[a*x])^3*Log[1 + E^(I*(Pi/2 - ArcSin[a
*x]))])/8 - (Pi^3*(I*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2) - Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))
]))/8 - (Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)^3*Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))] + ((3*I)/8)
*(Pi/2 - ArcSin[a*x])^2*PolyLog[2, -E^(I*(Pi/2 - ArcSin[a*x]))] + (3*Pi^2*((I/2)*(Pi/2 + (-1/2*Pi + ArcSin[a*x
])/2)^2 - (Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)*Log[1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))] + (I/2)*Pol
yLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))]))/4 + ((3*I)/2)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)^2*Po
lyLog[2, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))] - (3*(Pi/2 - ArcSin[a*x])*PolyLog[3, -E^(I*(Pi/2 - Arc
Sin[a*x]))])/4 - (3*Pi*((I/3)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)^3 - (Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)^2*Log[
1 + E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))] + I*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2)*PolyLog[2, -E^((2*I)*
(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))] - PolyLog[3, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))]/2))/2 - (3*(P
i/2 + (-1/2*Pi + ArcSin[a*x])/2)*PolyLog[3, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))])/2 - ((3*I)/4)*Poly
Log[4, -E^(I*(Pi/2 - ArcSin[a*x]))] - ((3*I)/4)*PolyLog[4, -E^((2*I)*(Pi/2 + (-1/2*Pi + ArcSin[a*x])/2))])))/8
- ArcSin[a*x]^3/(16*(Cos[ArcSin[a*x]/2] - Sin[ArcSin[a*x]/2])^4) - (2*ArcSin[a*x] - ArcSin[a*x]^2 + 3*ArcSin[
a*x]^3)/(16*(Cos[ArcSin[a*x]/2] - Sin[ArcSin[a*x]/2])^2) + (ArcSin[a*x]^2*Sin[ArcSin[a*x]/2])/(8*(Cos[ArcSin[a
*x]/2] - Sin[ArcSin[a*x]/2])^3) + ArcSin[a*x]^3/(16*(Cos[ArcSin[a*x]/2] + Sin[ArcSin[a*x]/2])^4) - (ArcSin[a*x
]^2*Sin[ArcSin[a*x]/2])/(8*(Cos[ArcSin[a*x]/2] + Sin[ArcSin[a*x]/2])^3) - (-2*ArcSin[a*x] - ArcSin[a*x]^2 - 3*
ArcSin[a*x]^3)/(16*(Cos[ArcSin[a*x]/2] + Sin[ArcSin[a*x]/2])^2) - (-Sin[ArcSin[a*x]/2] - 5*ArcSin[a*x]^2*Sin[A
rcSin[a*x]/2])/(4*(Cos[ArcSin[a*x]/2] - Sin[ArcSin[a*x]/2])) - (Sin[ArcSin[a*x]/2] + 5*ArcSin[a*x]^2*Sin[ArcSi
n[a*x]/2])/(4*(Cos[ArcSin[a*x]/2] + Sin[ArcSin[a*x]/2])))/(a*c^3))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\arcsin \left (a x\right )^{3}}{a^{6} c^{3} x^{6} - 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} - c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-arcsin(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\arcsin \left (a x\right )^{3}}{{\left (a^{2} c x^{2} - c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-arcsin(a*x)^3/(a^2*c*x^2 - c)^3, x)

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maple [A]  time = 0.34, size = 726, normalized size = 1.60 \[ -\frac {3 a^{2} \arcsin \left (a x \right )^{3} x^{3}}{8 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}+\frac {9 a \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}\, x^{2}}{8 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}-\frac {a^{2} \arcsin \left (a x \right ) x^{3}}{4 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}+\frac {a \,x^{2} \sqrt {-a^{2} x^{2}+1}}{4 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}+\frac {5 \arcsin \left (a x \right )^{3} x}{8 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}-\frac {11 \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{8 a \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}+\frac {\arcsin \left (a x \right ) x}{4 \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}-\frac {\sqrt {-a^{2} x^{2}+1}}{4 a \left (a^{4} x^{4}-2 a^{2} x^{2}+1\right ) c^{3}}+\frac {3 \arcsin \left (a x \right )^{3} \ln \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{8 a \,c^{3}}-\frac {9 i \polylog \left (4, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{4 a \,c^{3}}+\frac {9 \arcsin \left (a x \right ) \polylog \left (3, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{4 a \,c^{3}}-\frac {5 i \dilog \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{3}}-\frac {3 \arcsin \left (a x \right )^{3} \ln \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{8 a \,c^{3}}+\frac {9 i \polylog \left (4, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{4 a \,c^{3}}-\frac {9 \arcsin \left (a x \right ) \polylog \left (3, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{4 a \,c^{3}}+\frac {9 i \arcsin \left (a x \right )^{2} \polylog \left (2, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{8 a \,c^{3}}-\frac {5 \arcsin \left (a x \right ) \ln \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{3}}+\frac {5 \arcsin \left (a x \right ) \ln \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{3}}-\frac {9 i \arcsin \left (a x \right )^{2} \polylog \left (2, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{8 a \,c^{3}}+\frac {5 i \dilog \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*a^2/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)^3*x^3+9/8*a/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)^2*(-a^2*x^2+1
)^(1/2)*x^2-1/4*a^2/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)*x^3+1/4*a/(a^4*x^4-2*a^2*x^2+1)/c^3*x^2*(-a^2*x^2+1)
^(1/2)+5/8/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)^3*x-11/8/a/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)^2*(-a^2*x^2+
1)^(1/2)+1/4/(a^4*x^4-2*a^2*x^2+1)/c^3*arcsin(a*x)*x-1/4/a/(a^4*x^4-2*a^2*x^2+1)/c^3*(-a^2*x^2+1)^(1/2)+3/8/a/
c^3*arcsin(a*x)^3*ln(1-I*(I*a*x+(-a^2*x^2+1)^(1/2)))+9/4*I*polylog(4,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3+9/4*a
rcsin(a*x)*polylog(3,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-9/8*I*arcsin(a*x)^2*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(
1/2)))/a/c^3-3/8/a/c^3*arcsin(a*x)^3*ln(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))-5/2*I/a/c^3*dilog(1-I*(I*a*x+(-a^2*x^2
+1)^(1/2)))-9/4*arcsin(a*x)*polylog(3,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^3-9/4*I*polylog(4,-I*(I*a*x+(-a^2*x^2
+1)^(1/2)))/a/c^3-5/2/a/c^3*arcsin(a*x)*ln(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))+5/2/a/c^3*arcsin(a*x)*ln(1-I*(I*a*x
+(-a^2*x^2+1)^(1/2)))+5/2*I/a/c^3*dilog(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))+9/8*I*arcsin(a*x)^2*polylog(2,-I*(I*a*
x+(-a^2*x^2+1)^(1/2)))/a/c^3

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maxima [A]  time = 1.04, size = 78, normalized size = 0.17 \[ -\frac {1}{16} \, {\left (\frac {2 \, {\left (3 \, a^{2} x^{3} - 5 \, x\right )}}{a^{4} c^{3} x^{4} - 2 \, a^{2} c^{3} x^{2} + c^{3}} - \frac {3 \, \log \left (a x + 1\right )}{a c^{3}} + \frac {3 \, \log \left (a x - 1\right )}{a c^{3}}\right )} \arcsin \left (a x\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(3*a^2*x^3 - 5*x)/(a^4*c^3*x^4 - 2*a^2*c^3*x^2 + c^3) - 3*log(a*x + 1)/(a*c^3) + 3*log(a*x - 1)/(a*c^
3))*arcsin(a*x)^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {asin}\left (a\,x\right )}^3}{{\left (c-a^2\,c\,x^2\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)^3/(c - a^2*c*x^2)^3,x)

[Out]

int(asin(a*x)^3/(c - a^2*c*x^2)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\operatorname {asin}^{3}{\left (a x \right )}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(asin(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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