3.53 \(\int \frac{x^4}{\sin ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=69 \[ -\frac{\text{Si}\left (\sin ^{-1}(a x)\right )}{8 a^5}+\frac{9 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a^5}-\frac{x^4 \sqrt{1-a^2 x^2}}{a \sin ^{-1}(a x)} \]

[Out]

-((x^4*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x])) - SinIntegral[ArcSin[a*x]]/(8*a^5) + (9*SinIntegral[3*ArcSin[a*x]])
/(16*a^5) - (5*SinIntegral[5*ArcSin[a*x]])/(16*a^5)

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Rubi [A]  time = 0.0576204, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4631, 3299} \[ -\frac{\text{Si}\left (\sin ^{-1}(a x)\right )}{8 a^5}+\frac{9 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a^5}-\frac{x^4 \sqrt{1-a^2 x^2}}{a \sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcSin[a*x]^2,x]

[Out]

-((x^4*Sqrt[1 - a^2*x^2])/(a*ArcSin[a*x])) - SinIntegral[ArcSin[a*x]]/(8*a^5) + (9*SinIntegral[3*ArcSin[a*x]])
/(16*a^5) - (5*SinIntegral[5*ArcSin[a*x]])/(16*a^5)

Rule 4631

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

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]

Rubi steps

\begin{align*} \int \frac{x^4}{\sin ^{-1}(a x)^2} \, dx &=-\frac{x^4 \sqrt{1-a^2 x^2}}{a \sin ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \left (-\frac{\sin (x)}{8 x}+\frac{9 \sin (3 x)}{16 x}-\frac{5 \sin (5 x)}{16 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a^5}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{a \sin ^{-1}(a x)}-\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{8 a^5}-\frac{5 \operatorname{Subst}\left (\int \frac{\sin (5 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a^5}+\frac{9 \operatorname{Subst}\left (\int \frac{\sin (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a^5}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{a \sin ^{-1}(a x)}-\frac{\text{Si}\left (\sin ^{-1}(a x)\right )}{8 a^5}+\frac{9 \text{Si}\left (3 \sin ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a^5}\\ \end{align*}

Mathematica [A]  time = 0.208683, size = 61, normalized size = 0.88 \[ -\frac{\frac{16 a^4 x^4 \sqrt{1-a^2 x^2}}{\sin ^{-1}(a x)}+2 \text{Si}\left (\sin ^{-1}(a x)\right )-9 \text{Si}\left (3 \sin ^{-1}(a x)\right )+5 \text{Si}\left (5 \sin ^{-1}(a x)\right )}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcSin[a*x]^2,x]

[Out]

-((16*a^4*x^4*Sqrt[1 - a^2*x^2])/ArcSin[a*x] + 2*SinIntegral[ArcSin[a*x]] - 9*SinIntegral[3*ArcSin[a*x]] + 5*S
inIntegral[5*ArcSin[a*x]])/(16*a^5)

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Maple [A]  time = 0.026, size = 81, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ( -{\frac{1}{8\,\arcsin \left ( ax \right ) }\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{\it Si} \left ( \arcsin \left ( ax \right ) \right ) }{8}}+{\frac{3\,\cos \left ( 3\,\arcsin \left ( ax \right ) \right ) }{16\,\arcsin \left ( ax \right ) }}+{\frac{9\,{\it Si} \left ( 3\,\arcsin \left ( ax \right ) \right ) }{16}}-{\frac{\cos \left ( 5\,\arcsin \left ( ax \right ) \right ) }{16\,\arcsin \left ( ax \right ) }}-{\frac{5\,{\it Si} \left ( 5\,\arcsin \left ( ax \right ) \right ) }{16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsin(a*x)^2,x)

[Out]

1/a^5*(-1/8/arcsin(a*x)*(-a^2*x^2+1)^(1/2)-1/8*Si(arcsin(a*x))+3/16/arcsin(a*x)*cos(3*arcsin(a*x))+9/16*Si(3*a
rcsin(a*x))-1/16/arcsin(a*x)*cos(5*arcsin(a*x))-5/16*Si(5*arcsin(a*x)))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{4}}{\arcsin \left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^4/arcsin(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{asin}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/asin(a*x)**2,x)

[Out]

Integral(x**4/asin(a*x)**2, x)

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Giac [A]  time = 1.30518, size = 155, normalized size = 2.25 \begin{align*} -\frac{{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1}}{a^{5} \arcsin \left (a x\right )} - \frac{5 \, \operatorname{Si}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a^{5}} + \frac{9 \, \operatorname{Si}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a^{5}} - \frac{\operatorname{Si}\left (\arcsin \left (a x\right )\right )}{8 \, a^{5}} + \frac{2 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{5} \arcsin \left (a x\right )} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a^{5} \arcsin \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)^2,x, algorithm="giac")

[Out]

-(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)/(a^5*arcsin(a*x)) - 5/16*sin_integral(5*arcsin(a*x))/a^5 + 9/16*sin_integr
al(3*arcsin(a*x))/a^5 - 1/8*sin_integral(arcsin(a*x))/a^5 + 2*(-a^2*x^2 + 1)^(3/2)/(a^5*arcsin(a*x)) - sqrt(-a
^2*x^2 + 1)/(a^5*arcsin(a*x))