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

Optimal. Leaf size=83 \[ -\frac{\text{Si}\left (2 \sin ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \sin ^{-1}(a x)\right )}{a^4}-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)} \]

[Out]

-(x^3*Sqrt[1 - a^2*x^2])/(2*a*ArcSin[a*x]^2) - (3*x^2)/(2*a^2*ArcSin[a*x]) + (2*x^4)/ArcSin[a*x] - SinIntegral
[2*ArcSin[a*x]]/(2*a^4) + SinIntegral[4*ArcSin[a*x]]/a^4

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Rubi [A]  time = 0.299843, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4633, 4719, 4635, 4406, 3299, 12} \[ -\frac{\text{Si}\left (2 \sin ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \sin ^{-1}(a x)\right )}{a^4}-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^3*Sqrt[1 - a^2*x^2])/(2*a*ArcSin[a*x]^2) - (3*x^2)/(2*a^2*ArcSin[a*x]) + (2*x^4)/ArcSin[a*x] - SinIntegral
[2*ArcSin[a*x]]/(2*a^4) + SinIntegral[4*ArcSin[a*x]]/a^4

Rule 4633

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

Rule 4719

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

Rule 4635

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
in[x]^m*Cos[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int \frac{x^3}{\sin ^{-1}(a x)^3} \, dx &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}+\frac{3 \int \frac{x^2}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2} \, dx}{2 a}-(2 a) \int \frac{x^4}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2} \, dx\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)}-8 \int \frac{x^3}{\sin ^{-1}(a x)} \, dx+\frac{3 \int \frac{x}{\sin ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}-\frac{8 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}-\frac{8 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 x}-\frac{\sin (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a^4}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{2 a^4}-\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{3 x^2}{2 a^2 \sin ^{-1}(a x)}+\frac{2 x^4}{\sin ^{-1}(a x)}-\frac{\text{Si}\left (2 \sin ^{-1}(a x)\right )}{2 a^4}+\frac{\text{Si}\left (4 \sin ^{-1}(a x)\right )}{a^4}\\ \end{align*}

Mathematica [A]  time = 0.186956, size = 73, normalized size = 0.88 \[ \frac{\frac{a^2 x^2 \left (\left (4 a^2 x^2-3\right ) \sin ^{-1}(a x)-a x \sqrt{1-a^2 x^2}\right )}{\sin ^{-1}(a x)^2}-\text{Si}\left (2 \sin ^{-1}(a x)\right )+2 \text{Si}\left (4 \sin ^{-1}(a x)\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*x^2*(-(a*x*Sqrt[1 - a^2*x^2]) + (-3 + 4*a^2*x^2)*ArcSin[a*x]))/ArcSin[a*x]^2 - SinIntegral[2*ArcSin[a*x]
] + 2*SinIntegral[4*ArcSin[a*x]])/(2*a^4)

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Maple [A]  time = 0.034, size = 82, normalized size = 1. \begin{align*}{\frac{1}{{a}^{4}} \left ( -{\frac{\sin \left ( 2\,\arcsin \left ( ax \right ) \right ) }{8\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}-{\frac{\cos \left ( 2\,\arcsin \left ( ax \right ) \right ) }{4\,\arcsin \left ( ax \right ) }}-{\frac{{\it Si} \left ( 2\,\arcsin \left ( ax \right ) \right ) }{2}}+{\frac{\sin \left ( 4\,\arcsin \left ( ax \right ) \right ) }{16\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}+{\frac{\cos \left ( 4\,\arcsin \left ( ax \right ) \right ) }{4\,\arcsin \left ( ax \right ) }}+{\it Si} \left ( 4\,\arcsin \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsin(a*x)^3,x)

[Out]

1/a^4*(-1/8/arcsin(a*x)^2*sin(2*arcsin(a*x))-1/4/arcsin(a*x)*cos(2*arcsin(a*x))-1/2*Si(2*arcsin(a*x))+1/16/arc
sin(a*x)^2*sin(4*arcsin(a*x))+1/4/arcsin(a*x)*cos(4*arcsin(a*x))+Si(4*arcsin(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\sqrt{a x + 1} \sqrt{-a x + 1} a x^{3} + 2 \, \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2} \int \frac{{\left (8 \, a^{2} x^{2} - 3\right )} x}{\arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}\,{d x} -{\left (4 \, a^{2} x^{4} - 3 \, x^{2}\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}{2 \, a^{2} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^3 + 2*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2*integrate((8*a^2*x^3
- 3*x)/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)), x) - (4*a^2*x^4 - 3*x^2)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-
a*x + 1)))/(a^2*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3/arcsin(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/asin(a*x)**3, x)

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Giac [A]  time = 1.36199, size = 169, normalized size = 2.04 \begin{align*} \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{2 \, a^{3} \arcsin \left (a x\right )^{2}} + \frac{2 \,{\left (a^{2} x^{2} - 1\right )}^{2}}{a^{4} \arcsin \left (a x\right )} + \frac{\operatorname{Si}\left (4 \, \arcsin \left (a x\right )\right )}{a^{4}} - \frac{\operatorname{Si}\left (2 \, \arcsin \left (a x\right )\right )}{2 \, a^{4}} - \frac{\sqrt{-a^{2} x^{2} + 1} x}{2 \, a^{3} \arcsin \left (a x\right )^{2}} + \frac{5 \,{\left (a^{2} x^{2} - 1\right )}}{2 \, a^{4} \arcsin \left (a x\right )} + \frac{1}{2 \, a^{4} \arcsin \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-a^2*x^2 + 1)^(3/2)*x/(a^3*arcsin(a*x)^2) + 2*(a^2*x^2 - 1)^2/(a^4*arcsin(a*x)) + sin_integral(4*arcsin(a
*x))/a^4 - 1/2*sin_integral(2*arcsin(a*x))/a^4 - 1/2*sqrt(-a^2*x^2 + 1)*x/(a^3*arcsin(a*x)^2) + 5/2*(a^2*x^2 -
1)/(a^4*arcsin(a*x)) + 1/2/(a^4*arcsin(a*x))