3.68 \(\int \frac{x^3}{\sin ^{-1}(a x)^4} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{3 a^4}+\frac{4 \text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )}{3 a^4}+\frac{8 x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}-\frac{x \sqrt{1-a^2 x^2}}{a^3 \sin ^{-1}(a x)}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2} \]

[Out]

-(x^3*Sqrt[1 - a^2*x^2])/(3*a*ArcSin[a*x]^3) - x^2/(2*a^2*ArcSin[a*x]^2) + (2*x^4)/(3*ArcSin[a*x]^2) - (x*Sqrt
[1 - a^2*x^2])/(a^3*ArcSin[a*x]) + (8*x^3*Sqrt[1 - a^2*x^2])/(3*a*ArcSin[a*x]) - CosIntegral[2*ArcSin[a*x]]/(3
*a^4) + (4*CosIntegral[4*ArcSin[a*x]])/(3*a^4)

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Rubi [A]  time = 0.281678, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4633, 4719, 4631, 3302} \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )}{3 a^4}+\frac{4 \text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )}{3 a^4}+\frac{8 x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}-\frac{x \sqrt{1-a^2 x^2}}{a^3 \sin ^{-1}(a x)}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcSin[a*x]^4,x]

[Out]

-(x^3*Sqrt[1 - a^2*x^2])/(3*a*ArcSin[a*x]^3) - x^2/(2*a^2*ArcSin[a*x]^2) + (2*x^4)/(3*ArcSin[a*x]^2) - (x*Sqrt
[1 - a^2*x^2])/(a^3*ArcSin[a*x]) + (8*x^3*Sqrt[1 - a^2*x^2])/(3*a*ArcSin[a*x]) - CosIntegral[2*ArcSin[a*x]]/(3
*a^4) + (4*CosIntegral[4*ArcSin[a*x]])/(3*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 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 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^3}{\sin ^{-1}(a x)^4} \, dx &=-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}+\frac{\int \frac{x^2}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^3} \, dx}{a}-\frac{1}{3} (4 a) \int \frac{x^4}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^3} \, dx\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2}-\frac{8}{3} \int \frac{x^3}{\sin ^{-1}(a x)^2} \, dx+\frac{\int \frac{x}{\sin ^{-1}(a x)^2} \, dx}{a^2}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2}-\frac{x \sqrt{1-a^2 x^2}}{a^3 \sin ^{-1}(a x)}+\frac{8 x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^4}-\frac{8 \operatorname{Subst}\left (\int \left (\frac{\cos (2 x)}{2 x}-\frac{\cos (4 x)}{2 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2}-\frac{x \sqrt{1-a^2 x^2}}{a^3 \sin ^{-1}(a x)}+\frac{8 x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}+\frac{\text{Ci}\left (2 \sin ^{-1}(a x)\right )}{a^4}-\frac{4 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{3 a^4}+\frac{4 \operatorname{Subst}\left (\int \frac{\cos (4 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{3 a^4}\\ &=-\frac{x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)^3}-\frac{x^2}{2 a^2 \sin ^{-1}(a x)^2}+\frac{2 x^4}{3 \sin ^{-1}(a x)^2}-\frac{x \sqrt{1-a^2 x^2}}{a^3 \sin ^{-1}(a x)}+\frac{8 x^3 \sqrt{1-a^2 x^2}}{3 a \sin ^{-1}(a x)}-\frac{\text{Ci}\left (2 \sin ^{-1}(a x)\right )}{3 a^4}+\frac{4 \text{Ci}\left (4 \sin ^{-1}(a x)\right )}{3 a^4}\\ \end{align*}

Mathematica [A]  time = 0.359611, size = 107, normalized size = 0.74 \[ \frac{\frac{a x \left (-2 a^2 x^2 \sqrt{1-a^2 x^2}+a x \left (4 a^2 x^2-3\right ) \sin ^{-1}(a x)+2 \sqrt{1-a^2 x^2} \left (8 a^2 x^2-3\right ) \sin ^{-1}(a x)^2\right )}{\sin ^{-1}(a x)^3}-2 \text{CosIntegral}\left (2 \sin ^{-1}(a x)\right )+8 \text{CosIntegral}\left (4 \sin ^{-1}(a x)\right )}{6 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcSin[a*x]^4,x]

[Out]

((a*x*(-2*a^2*x^2*Sqrt[1 - a^2*x^2] + a*x*(-3 + 4*a^2*x^2)*ArcSin[a*x] + 2*Sqrt[1 - a^2*x^2]*(-3 + 8*a^2*x^2)*
ArcSin[a*x]^2))/ArcSin[a*x]^3 - 2*CosIntegral[2*ArcSin[a*x]] + 8*CosIntegral[4*ArcSin[a*x]])/(6*a^4)

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Maple [A]  time = 0.03, size = 114, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{4}} \left ( -{\frac{\sin \left ( 2\,\arcsin \left ( ax \right ) \right ) }{12\, \left ( \arcsin \left ( ax \right ) \right ) ^{3}}}-{\frac{\cos \left ( 2\,\arcsin \left ( ax \right ) \right ) }{12\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}+{\frac{\sin \left ( 2\,\arcsin \left ( ax \right ) \right ) }{6\,\arcsin \left ( ax \right ) }}-{\frac{{\it Ci} \left ( 2\,\arcsin \left ( ax \right ) \right ) }{3}}+{\frac{\sin \left ( 4\,\arcsin \left ( ax \right ) \right ) }{24\, \left ( \arcsin \left ( ax \right ) \right ) ^{3}}}+{\frac{\cos \left ( 4\,\arcsin \left ( ax \right ) \right ) }{12\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}-{\frac{\sin \left ( 4\,\arcsin \left ( ax \right ) \right ) }{3\,\arcsin \left ( ax \right ) }}+{\frac{4\,{\it Ci} \left ( 4\,\arcsin \left ( ax \right ) \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsin(a*x)^4,x)

[Out]

1/a^4*(-1/12/arcsin(a*x)^3*sin(2*arcsin(a*x))-1/12/arcsin(a*x)^2*cos(2*arcsin(a*x))+1/6/arcsin(a*x)*sin(2*arcs
in(a*x))-1/3*Ci(2*arcsin(a*x))+1/24/arcsin(a*x)^3*sin(4*arcsin(a*x))+1/12/arcsin(a*x)^2*cos(4*arcsin(a*x))-1/3
/arcsin(a*x)*sin(4*arcsin(a*x))+4/3*Ci(4*arcsin(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, a^{3} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3} \int \frac{{\left (32 \, a^{4} x^{4} - 30 \, a^{2} x^{2} + 3\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{{\left (a^{5} x^{2} - a^{3}\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}\,{d x} + 2 \,{\left (a^{2} x^{3} -{\left (8 \, a^{2} x^{3} - 3 \, x\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2}\right )} \sqrt{a x + 1} \sqrt{-a x + 1} -{\left (4 \, a^{3} x^{4} - 3 \, a x^{2}\right )} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}{6 \, a^{3} \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsin(a*x)^4,x, algorithm="maxima")

[Out]

-1/6*(6*a^3*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^3*integrate(1/3*(32*a^4*x^4 - 30*a^2*x^2 + 3)*sqrt(a*x
+ 1)*sqrt(-a*x + 1)/((a^5*x^2 - a^3)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))), x) + 2*(a^2*x^3 - (8*a^2*x^3
 - 3*x)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)*sqrt(a*x + 1)*sqrt(-a*x + 1) - (4*a^3*x^4 - 3*a*x^2)*arc
tan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))/(a^3*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^3)

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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 )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsin(a*x)^4,x, algorithm="fricas")

[Out]

integral(x^3/arcsin(a*x)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\operatorname{asin}^{4}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/asin(a*x)**4,x)

[Out]

Integral(x**3/asin(a*x)**4, x)

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Giac [A]  time = 1.37685, size = 235, normalized size = 1.63 \begin{align*} -\frac{8 \,{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{3 \, a^{3} \arcsin \left (a x\right )} + \frac{5 \, \sqrt{-a^{2} x^{2} + 1} x}{3 \, a^{3} \arcsin \left (a x\right )} + \frac{4 \, \operatorname{Ci}\left (4 \, \arcsin \left (a x\right )\right )}{3 \, a^{4}} - \frac{\operatorname{Ci}\left (2 \, \arcsin \left (a x\right )\right )}{3 \, a^{4}} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{3 \, a^{3} \arcsin \left (a x\right )^{3}} + \frac{2 \,{\left (a^{2} x^{2} - 1\right )}^{2}}{3 \, a^{4} \arcsin \left (a x\right )^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1} x}{3 \, a^{3} \arcsin \left (a x\right )^{3}} + \frac{5 \,{\left (a^{2} x^{2} - 1\right )}}{6 \, a^{4} \arcsin \left (a x\right )^{2}} + \frac{1}{6 \, a^{4} \arcsin \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsin(a*x)^4,x, algorithm="giac")

[Out]

-8/3*(-a^2*x^2 + 1)^(3/2)*x/(a^3*arcsin(a*x)) + 5/3*sqrt(-a^2*x^2 + 1)*x/(a^3*arcsin(a*x)) + 4/3*cos_integral(
4*arcsin(a*x))/a^4 - 1/3*cos_integral(2*arcsin(a*x))/a^4 + 1/3*(-a^2*x^2 + 1)^(3/2)*x/(a^3*arcsin(a*x)^3) + 2/
3*(a^2*x^2 - 1)^2/(a^4*arcsin(a*x)^2) - 1/3*sqrt(-a^2*x^2 + 1)*x/(a^3*arcsin(a*x)^3) + 5/6*(a^2*x^2 - 1)/(a^4*
arcsin(a*x)^2) + 1/6/(a^4*arcsin(a*x)^2)