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

Optimal. Leaf size=108 \[ 6 i a \sin ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-6 i a \sin ^{-1}(a x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-6 a \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+6 a \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^3}{x}-6 a \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.1632, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4627, 4709, 4183, 2531, 2282, 6589} \[ 6 i a \sin ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-6 i a \sin ^{-1}(a x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-6 a \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+6 a \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^3}{x}-6 a \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/x^2,x]

[Out]

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

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

Mathematica [A]  time = 0.12395, size = 133, normalized size = 1.23 \[ a \left (6 i \sin ^{-1}(a x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-6 i \sin ^{-1}(a x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-6 \text{PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+6 \text{PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^3}{a x}+3 \sin ^{-1}(a x)^2 \log \left (1-e^{i \sin ^{-1}(a x)}\right )-3 \sin ^{-1}(a x)^2 \log \left (1+e^{i \sin ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^3/x^2,x]

[Out]

a*(-(ArcSin[a*x]^3/(a*x)) + 3*ArcSin[a*x]^2*Log[1 - E^(I*ArcSin[a*x])] - 3*ArcSin[a*x]^2*Log[1 + E^(I*ArcSin[a
*x])] + (6*I)*ArcSin[a*x]*PolyLog[2, -E^(I*ArcSin[a*x])] - (6*I)*ArcSin[a*x]*PolyLog[2, E^(I*ArcSin[a*x])] - 6
*PolyLog[3, -E^(I*ArcSin[a*x])] + 6*PolyLog[3, E^(I*ArcSin[a*x])])

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Maple [A]  time = 0.074, size = 179, normalized size = 1.7 \begin{align*} -{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{3}}{x}}-3\,a \left ( \arcsin \left ( ax \right ) \right ) ^{2}\ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +6\,ia\arcsin \left ( ax \right ){\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -6\,a{\it polylog} \left ( 3,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +3\,a \left ( \arcsin \left ( ax \right ) \right ) ^{2}\ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -6\,ia\arcsin \left ( ax \right ){\it polylog} \left ( 2,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +6\,a{\it polylog} \left ( 3,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^2,x)

[Out]

-arcsin(a*x)^3/x-3*a*arcsin(a*x)^2*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))+6*I*a*arcsin(a*x)*polylog(2,-I*a*x-(-a^2*x^2
+1)^(1/2))-6*a*polylog(3,-I*a*x-(-a^2*x^2+1)^(1/2))+3*a*arcsin(a*x)^2*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-6*I*a*arc
sin(a*x)*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))+6*a*polylog(3,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*} -\frac{\arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3} + 3 \, a x \int \frac{\sqrt{-a x + 1} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2}}{\sqrt{a x + 1}{\left (a x - 1\right )} x}\,{d x}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^3 + 3*a*x*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(a*x, sqr
t(a*x + 1)*sqrt(-a*x + 1))^2/(a^2*x^3 - x), x))/x

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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}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(a*x)^3/x^2, 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^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**3/x**2,x)

[Out]

Integral(asin(a*x)**3/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(a*x)^3/x^2, x)