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

Optimal. Leaf size=138 \[ 3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-6 \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+6 \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-6 i \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )+6 i \text{PolyLog}\left (4,e^{i \sin ^{-1}(a x)}\right )-2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

[Out]

-2*ArcSin[a*x]^3*ArcTanh[E^(I*ArcSin[a*x])] + (3*I)*ArcSin[a*x]^2*PolyLog[2, -E^(I*ArcSin[a*x])] - (3*I)*ArcSi
n[a*x]^2*PolyLog[2, E^(I*ArcSin[a*x])] - 6*ArcSin[a*x]*PolyLog[3, -E^(I*ArcSin[a*x])] + 6*ArcSin[a*x]*PolyLog[
3, E^(I*ArcSin[a*x])] - (6*I)*PolyLog[4, -E^(I*ArcSin[a*x])] + (6*I)*PolyLog[4, E^(I*ArcSin[a*x])]

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Rubi [A]  time = 0.160109, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4709, 4183, 2531, 6609, 2282, 6589} \[ 3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-6 \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+6 \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-6 i \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )+6 i \text{PolyLog}\left (4,e^{i \sin ^{-1}(a x)}\right )-2 \sin ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcSin[a*x]^3*ArcTanh[E^(I*ArcSin[a*x])] + (3*I)*ArcSin[a*x]^2*PolyLog[2, -E^(I*ArcSin[a*x])] - (3*I)*ArcSi
n[a*x]^2*PolyLog[2, E^(I*ArcSin[a*x])] - 6*ArcSin[a*x]*PolyLog[3, -E^(I*ArcSin[a*x])] + 6*ArcSin[a*x]*PolyLog[
3, E^(I*ArcSin[a*x])] - (6*I)*PolyLog[4, -E^(I*ArcSin[a*x])] + (6*I)*PolyLog[4, E^(I*ArcSin[a*x])]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.14999, size = 180, normalized size = 1.3 \[ -\frac{1}{8} i \left (-24 \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{-i \sin ^{-1}(a x)}\right )-24 \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )+48 i \sin ^{-1}(a x) \text{PolyLog}\left (3,e^{-i \sin ^{-1}(a x)}\right )-48 i \sin ^{-1}(a x) \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,e^{-i \sin ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,-e^{i \sin ^{-1}(a x)}\right )-2 \sin ^{-1}(a x)^4+8 i \sin ^{-1}(a x)^3 \log \left (1-e^{-i \sin ^{-1}(a x)}\right )-8 i \sin ^{-1}(a x)^3 \log \left (1+e^{i \sin ^{-1}(a x)}\right )+\pi ^4\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-I/8)*(Pi^4 - 2*ArcSin[a*x]^4 + (8*I)*ArcSin[a*x]^3*Log[1 - E^((-I)*ArcSin[a*x])] - (8*I)*ArcSin[a*x]^3*Log[1
+ E^(I*ArcSin[a*x])] - 24*ArcSin[a*x]^2*PolyLog[2, E^((-I)*ArcSin[a*x])] - 24*ArcSin[a*x]^2*PolyLog[2, -E^(I*
ArcSin[a*x])] + (48*I)*ArcSin[a*x]*PolyLog[3, E^((-I)*ArcSin[a*x])] - (48*I)*ArcSin[a*x]*PolyLog[3, -E^(I*ArcS
in[a*x])] + 48*PolyLog[4, E^((-I)*ArcSin[a*x])] + 48*PolyLog[4, -E^(I*ArcSin[a*x])])

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Maple [A]  time = 0.066, size = 221, normalized size = 1.6 \begin{align*} \left ( \arcsin \left ( ax \right ) \right ) ^{3}\ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) - \left ( \arcsin \left ( ax \right ) \right ) ^{3}\ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +6\,\arcsin \left ( ax \right ){\it polylog} \left ( 3,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) -6\,\arcsin \left ( ax \right ){\it polylog} \left ( 3,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -3\,i \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +3\,i \left ( \arcsin \left ( ax \right ) \right ) ^{2}{\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -6\,i{\it polylog} \left ( 4,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +6\,i{\it polylog} \left ( 4,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (a x\right )^{3}}{\sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1} \arcsin \left (a x\right )^{3}}{a^{2} x^{3} - x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (a x\right )^{3}}{\sqrt{-a^{2} x^{2} + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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