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

Optimal. Leaf size=264 \[ \frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {Li}_4\left (-e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {Li}_4\left (e^{i \sin ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {3 a \sin ^{-1}(a x)^2}{2 x} \]

[Out]

-3/2*a*arcsin(a*x)^2/x-6*a^2*arcsin(a*x)*arctanh(I*a*x+(-a^2*x^2+1)^(1/2))-a^2*arcsin(a*x)^3*arctanh(I*a*x+(-a
^2*x^2+1)^(1/2))+3*I*a^2*polylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))+3/2*I*a^2*arcsin(a*x)^2*polylog(2,-I*a*x-(-a^2*x
^2+1)^(1/2))-3*I*a^2*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))-3/2*I*a^2*arcsin(a*x)^2*polylog(2,I*a*x+(-a^2*x^2+1)^
(1/2))-3*a^2*arcsin(a*x)*polylog(3,-I*a*x-(-a^2*x^2+1)^(1/2))+3*a^2*arcsin(a*x)*polylog(3,I*a*x+(-a^2*x^2+1)^(
1/2))-3*I*a^2*polylog(4,-I*a*x-(-a^2*x^2+1)^(1/2))+3*I*a^2*polylog(4,I*a*x+(-a^2*x^2+1)^(1/2))-1/2*arcsin(a*x)
^3*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.36, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4701, 4709, 4183, 2531, 6609, 2282, 6589, 4627, 2279, 2391} \[ \frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-3 a^2 \sin ^{-1}(a x) \text {PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+3 a^2 \sin ^{-1}(a x) \text {PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {PolyLog}\left (4,e^{i \sin ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {3 a \sin ^{-1}(a x)^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/(x^3*Sqrt[1 - a^2*x^2]),x]

[Out]

(-3*a*ArcSin[a*x]^2)/(2*x) - (Sqrt[1 - a^2*x^2]*ArcSin[a*x]^3)/(2*x^2) - 6*a^2*ArcSin[a*x]*ArcTanh[E^(I*ArcSin
[a*x])] - a^2*ArcSin[a*x]^3*ArcTanh[E^(I*ArcSin[a*x])] + (3*I)*a^2*PolyLog[2, -E^(I*ArcSin[a*x])] + ((3*I)/2)*
a^2*ArcSin[a*x]^2*PolyLog[2, -E^(I*ArcSin[a*x])] - (3*I)*a^2*PolyLog[2, E^(I*ArcSin[a*x])] - ((3*I)/2)*a^2*Arc
Sin[a*x]^2*PolyLog[2, E^(I*ArcSin[a*x])] - 3*a^2*ArcSin[a*x]*PolyLog[3, -E^(I*ArcSin[a*x])] + 3*a^2*ArcSin[a*x
]*PolyLog[3, E^(I*ArcSin[a*x])] - (3*I)*a^2*PolyLog[4, -E^(I*ArcSin[a*x])] + (3*I)*a^2*PolyLog[4, E^(I*ArcSin[
a*x])]

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 4183

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

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4701

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

Rule 4709

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

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}{x^3 \sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}+\frac {1}{2} (3 a) \int \frac {\sin ^{-1}(a x)^2}{x^2} \, dx+\frac {1}{2} a^2 \int \frac {\sin ^{-1}(a x)^3}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}+\frac {1}{2} a^2 \operatorname {Subst}\left (\int x^3 \csc (x) \, dx,x,\sin ^{-1}(a x)\right )+\left (3 a^2\right ) \int \frac {\sin ^{-1}(a x)}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {1}{2} \left (3 a^2\right ) \operatorname {Subst}\left (\int x^2 \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\frac {1}{2} \left (3 a^2\right ) \operatorname {Subst}\left (\int x^2 \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\left (3 a^2\right ) \operatorname {Subst}\left (\int x \csc (x) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-\left (3 i a^2\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\left (3 i a^2\right ) \operatorname {Subst}\left (\int x \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )-\left (3 a^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\left (3 a^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )+\left (3 i a^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )-\left (3 i a^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )+\left (3 a^2\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )-\left (3 a^2\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )+\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-\left (3 i a^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )+\left (3 i a^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )\\ &=-\frac {3 a \sin ^{-1}(a x)^2}{2 x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{2 x^2}-6 a^2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )+\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-\frac {3}{2} i a^2 \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+3 a^2 \sin ^{-1}(a x) \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-3 i a^2 \text {Li}_4\left (-e^{i \sin ^{-1}(a x)}\right )+3 i a^2 \text {Li}_4\left (e^{i \sin ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]  time = 5.04, size = 317, normalized size = 1.20 \[ \frac {1}{16} a^2 \left (24 i \sin ^{-1}(a x)^2 \text {Li}_2\left (e^{-i \sin ^{-1}(a x)}\right )+48 \sin ^{-1}(a x) \text {Li}_3\left (e^{-i \sin ^{-1}(a x)}\right )-48 \sin ^{-1}(a x) \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+24 i \left (\sin ^{-1}(a x)^2+2\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-48 i \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-48 i \text {Li}_4\left (e^{-i \sin ^{-1}(a x)}\right )-48 i \text {Li}_4\left (-e^{i \sin ^{-1}(a x)}\right )+2 i \sin ^{-1}(a x)^4+8 \sin ^{-1}(a x)^3 \log \left (1-e^{-i \sin ^{-1}(a x)}\right )-8 \sin ^{-1}(a x)^3 \log \left (1+e^{i \sin ^{-1}(a x)}\right )+48 \sin ^{-1}(a x) \log \left (1-e^{i \sin ^{-1}(a x)}\right )-48 \sin ^{-1}(a x) \log \left (1+e^{i \sin ^{-1}(a x)}\right )-12 \sin ^{-1}(a x)^2 \tan \left (\frac {1}{2} \sin ^{-1}(a x)\right )-12 \sin ^{-1}(a x)^2 \cot \left (\frac {1}{2} \sin ^{-1}(a x)\right )-2 \sin ^{-1}(a x)^3 \csc ^2\left (\frac {1}{2} \sin ^{-1}(a x)\right )+2 \sin ^{-1}(a x)^3 \sec ^2\left (\frac {1}{2} \sin ^{-1}(a x)\right )-i \pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^3/(x^3*Sqrt[1 - a^2*x^2]),x]

[Out]

(a^2*((-I)*Pi^4 + (2*I)*ArcSin[a*x]^4 - 12*ArcSin[a*x]^2*Cot[ArcSin[a*x]/2] - 2*ArcSin[a*x]^3*Csc[ArcSin[a*x]/
2]^2 + 8*ArcSin[a*x]^3*Log[1 - E^((-I)*ArcSin[a*x])] + 48*ArcSin[a*x]*Log[1 - E^(I*ArcSin[a*x])] - 48*ArcSin[a
*x]*Log[1 + E^(I*ArcSin[a*x])] - 8*ArcSin[a*x]^3*Log[1 + E^(I*ArcSin[a*x])] + (24*I)*ArcSin[a*x]^2*PolyLog[2,
E^((-I)*ArcSin[a*x])] + (24*I)*(2 + ArcSin[a*x]^2)*PolyLog[2, -E^(I*ArcSin[a*x])] - (48*I)*PolyLog[2, E^(I*Arc
Sin[a*x])] + 48*ArcSin[a*x]*PolyLog[3, E^((-I)*ArcSin[a*x])] - 48*ArcSin[a*x]*PolyLog[3, -E^(I*ArcSin[a*x])] -
(48*I)*PolyLog[4, E^((-I)*ArcSin[a*x])] - (48*I)*PolyLog[4, -E^(I*ArcSin[a*x])] + 2*ArcSin[a*x]^3*Sec[ArcSin[
a*x]/2]^2 - 12*ArcSin[a*x]^2*Tan[ArcSin[a*x]/2]))/16

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )^{3}}{a^{2} x^{5} - x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*arcsin(a*x)^3/(a^2*x^5 - x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^3/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(arcsin(a*x)^3/(sqrt(-a^2*x^2 + 1)*x^3), x)

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maple [A]  time = 0.35, size = 428, normalized size = 1.62 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \arcsin \left (a x \right )^{2} \left (a^{2} x^{2} \arcsin \left (a x \right )-3 a x \sqrt {-a^{2} x^{2}+1}-\arcsin \left (a x \right )\right )}{2 \left (a^{2} x^{2}-1\right ) x^{2}}-\frac {\ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right ) \arcsin \left (a x \right )^{3} a^{2}}{2}+\frac {\ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right ) \arcsin \left (a x \right )^{3} a^{2}}{2}-3 a^{2} \arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )+\frac {3 i a^{2} \arcsin \left (a x \right )^{2} \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )}{2}-3 a^{2} \arcsin \left (a x \right ) \polylog \left (3, -i a x -\sqrt {-a^{2} x^{2}+1}\right )+3 a^{2} \arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-\frac {3 i a^{2} \arcsin \left (a x \right )^{2} \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )}{2}+3 a^{2} \arcsin \left (a x \right ) \polylog \left (3, i a x +\sqrt {-a^{2} x^{2}+1}\right )+3 i a^{2} \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )-3 i a^{2} \polylog \left (4, -i a x -\sqrt {-a^{2} x^{2}+1}\right )-3 i a^{2} \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )+3 i a^{2} \polylog \left (4, i a x +\sqrt {-a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)^3/x^3/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/2*(-a^2*x^2+1)^(1/2)/(a^2*x^2-1)/x^2*arcsin(a*x)^2*(a^2*x^2*arcsin(a*x)-3*a*x*(-a^2*x^2+1)^(1/2)-arcsin(a*x
))-1/2*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))*arcsin(a*x)^3*a^2+1/2*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))*arcsin(a*x)^3*a^2-3
*a^2*arcsin(a*x)*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))+3/2*I*a^2*arcsin(a*x)^2*polylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))-3
*a^2*arcsin(a*x)*polylog(3,-I*a*x-(-a^2*x^2+1)^(1/2))+3*a^2*arcsin(a*x)*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-3/2*I*a
^2*arcsin(a*x)^2*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))+3*a^2*arcsin(a*x)*polylog(3,I*a*x+(-a^2*x^2+1)^(1/2))+3*I
*a^2*polylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))-3*I*a^2*polylog(4,-I*a*x-(-a^2*x^2+1)^(1/2))-3*I*a^2*polylog(2,I*a*x
+(-a^2*x^2+1)^(1/2))+3*I*a^2*polylog(4,I*a*x+(-a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (a x\right )^{3}}{\sqrt {-a^{2} x^{2} + 1} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^3/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arcsin(a*x)^3/(sqrt(-a^2*x^2 + 1)*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a*x)**3/(x**3*sqrt(-(a*x - 1)*(a*x + 1))), x)

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