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

Optimal. Leaf size=200 \[ \frac{3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{6 \sin ^{-1}(a x) \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(a x)}\right )}{a c}+\frac{6 \sin ^{-1}(a x) \text{PolyLog}\left (3,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{6 i \text{PolyLog}\left (4,-i e^{i \sin ^{-1}(a x)}\right )}{a c}+\frac{6 i \text{PolyLog}\left (4,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{2 i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.13375, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4657, 4181, 2531, 6609, 2282, 6589} \[ \frac{3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{3 i \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{6 \sin ^{-1}(a x) \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(a x)}\right )}{a c}+\frac{6 \sin ^{-1}(a x) \text{PolyLog}\left (3,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{6 i \text{PolyLog}\left (4,-i e^{i \sin ^{-1}(a x)}\right )}{a c}+\frac{6 i \text{PolyLog}\left (4,i e^{i \sin ^{-1}(a x)}\right )}{a c}-\frac{2 i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/(c - a^2*c*x^2),x]

[Out]

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

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

Mathematica [A]  time = 0.181513, size = 162, normalized size = 0.81 \[ -\frac{i \left (-3 \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )+3 \sin ^{-1}(a x)^2 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )-6 i \sin ^{-1}(a x) \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(a x)}\right )+6 i \sin ^{-1}(a x) \text{PolyLog}\left (3,i e^{i \sin ^{-1}(a x)}\right )+6 \text{PolyLog}\left (4,-i e^{i \sin ^{-1}(a x)}\right )-6 \text{PolyLog}\left (4,i e^{i \sin ^{-1}(a x)}\right )+2 \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )\right )}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(2*ArcSin[a*x]^3*ArcTan[E^(I*ArcSin[a*x])] - 3*ArcSin[a*x]^2*PolyLog[2, (-I)*E^(I*ArcSin[a*x])] + 3*ArcS
in[a*x]^2*PolyLog[2, I*E^(I*ArcSin[a*x])] - (6*I)*ArcSin[a*x]*PolyLog[3, (-I)*E^(I*ArcSin[a*x])] + (6*I)*ArcSi
n[a*x]*PolyLog[3, I*E^(I*ArcSin[a*x])] + 6*PolyLog[4, (-I)*E^(I*ArcSin[a*x])] - 6*PolyLog[4, I*E^(I*ArcSin[a*x
])]))/(a*c)

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Maple [F]  time = 0.1, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{3}}{-{a}^{2}c{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.24713, size = 49, normalized size = 0.24 \begin{align*} \frac{1}{2} \,{\left (\frac{\log \left (a x + 1\right )}{a c} - \frac{\log \left (a x - 1\right )}{a c}\right )} \arcsin \left (a x\right )^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(asin(a*x)**3/(a**2*x**2 - 1), x)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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