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

Optimal. Leaf size=41 \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{2 a^5}+\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )}{8 a^5}+\frac{3 \log \left (\sin ^{-1}(a x)\right )}{8 a^5} \]

[Out]

-CosIntegral[2*ArcSin[a*x]]/(2*a^5) + CosIntegral[4*ArcSin[a*x]]/(8*a^5) + (3*Log[ArcSin[a*x]])/(8*a^5)

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Rubi [A]  time = 0.158949, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4723, 3312, 3302} \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{2 a^5}+\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )}{8 a^5}+\frac{3 \log \left (\sin ^{-1}(a x)\right )}{8 a^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(Sqrt[1 - a^2*x^2]*ArcSin[a*x]),x]

[Out]

-CosIntegral[2*ArcSin[a*x]]/(2*a^5) + CosIntegral[4*ArcSin[a*x]]/(8*a^5) + (3*Log[ArcSin[a*x]])/(8*a^5)

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^(
m + 1), Subst[Int[(a + b*x)^n*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

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

Mathematica [A]  time = 0.0718342, size = 31, normalized size = 0.76 \[ \frac{-4 \text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )+\text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )+3 \log \left (\sin ^{-1}(a x)\right )}{8 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(Sqrt[1 - a^2*x^2]*ArcSin[a*x]),x]

[Out]

(-4*CosIntegral[2*ArcSin[a*x]] + CosIntegral[4*ArcSin[a*x]] + 3*Log[ArcSin[a*x]])/(8*a^5)

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Maple [A]  time = 0.055, size = 36, normalized size = 0.9 \begin{align*} -{\frac{{\it Ci} \left ( 2\,\arcsin \left ( ax \right ) \right ) }{2\,{a}^{5}}}+{\frac{{\it Ci} \left ( 4\,\arcsin \left ( ax \right ) \right ) }{8\,{a}^{5}}}+{\frac{3\,\ln \left ( \arcsin \left ( ax \right ) \right ) }{8\,{a}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/2*Ci(2*arcsin(a*x))/a^5+1/8*Ci(4*arcsin(a*x))/a^5+3/8*ln(arcsin(a*x))/a^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/(sqrt(-a^2*x^2 + 1)*arcsin(a*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} x^{4}}{{\left (a^{2} x^{2} - 1\right )} \arcsin \left (a x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^4/((a^2*x^2 - 1)*arcsin(a*x)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname{asin}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/asin(a*x)/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**4/(sqrt(-(a*x - 1)*(a*x + 1))*asin(a*x)), x)

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Giac [A]  time = 1.31518, size = 47, normalized size = 1.15 \begin{align*} \frac{\operatorname{Ci}\left (4 \, \arcsin \left (a x\right )\right )}{8 \, a^{5}} - \frac{\operatorname{Ci}\left (2 \, \arcsin \left (a x\right )\right )}{2 \, a^{5}} + \frac{3 \, \log \left (\arcsin \left (a x\right )\right )}{8 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arcsin(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/8*cos_integral(4*arcsin(a*x))/a^5 - 1/2*cos_integral(2*arcsin(a*x))/a^5 + 3/8*log(arcsin(a*x))/a^5