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

Optimal. Leaf size=98 \[ -\frac{\text{CosIntegral}\left (\sin ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{CosIntegral}\left (3 \sin ^{-1}(a x)\right )}{32 a^5}-\frac{25 \text{CosIntegral}\left (5 \sin ^{-1}(a x)\right )}{32 a^5}-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)} \]

[Out]

-(x^4*Sqrt[1 - a^2*x^2])/(2*a*ArcSin[a*x]^2) - (2*x^3)/(a^2*ArcSin[a*x]) + (5*x^5)/(2*ArcSin[a*x]) - CosIntegr
al[ArcSin[a*x]]/(16*a^5) + (27*CosIntegral[3*ArcSin[a*x]])/(32*a^5) - (25*CosIntegral[5*ArcSin[a*x]])/(32*a^5)

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Rubi [A]  time = 0.342807, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4633, 4719, 4635, 4406, 3302} \[ -\frac{\text{CosIntegral}\left (\sin ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{CosIntegral}\left (3 \sin ^{-1}(a x)\right )}{32 a^5}-\frac{25 \text{CosIntegral}\left (5 \sin ^{-1}(a x)\right )}{32 a^5}-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^4*Sqrt[1 - a^2*x^2])/(2*a*ArcSin[a*x]^2) - (2*x^3)/(a^2*ArcSin[a*x]) + (5*x^5)/(2*ArcSin[a*x]) - CosIntegr
al[ArcSin[a*x]]/(16*a^5) + (27*CosIntegral[3*ArcSin[a*x]])/(32*a^5) - (25*CosIntegral[5*ArcSin[a*x]])/(32*a^5)

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 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}{\sin ^{-1}(a x)^3} \, dx &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}+\frac{2 \int \frac{x^3}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2} \, dx}{a}-\frac{1}{2} (5 a) \int \frac{x^5}{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2} \, dx\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)}-\frac{25}{2} \int \frac{x^4}{\sin ^{-1}(a x)} \, dx+\frac{6 \int \frac{x^2}{\sin ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)}+\frac{6 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{a^5}-\frac{25 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^4(x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)}+\frac{6 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{4 x}-\frac{\cos (3 x)}{4 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a^5}-\frac{25 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{8 x}-\frac{3 \cos (3 x)}{16 x}+\frac{\cos (5 x)}{16 x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{2 a^5}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)}-\frac{25 \operatorname{Subst}\left (\int \frac{\cos (5 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{32 a^5}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{2 a^5}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{2 a^5}-\frac{25 \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{16 a^5}+\frac{75 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\sin ^{-1}(a x)\right )}{32 a^5}\\ &=-\frac{x^4 \sqrt{1-a^2 x^2}}{2 a \sin ^{-1}(a x)^2}-\frac{2 x^3}{a^2 \sin ^{-1}(a x)}+\frac{5 x^5}{2 \sin ^{-1}(a x)}-\frac{\text{Ci}\left (\sin ^{-1}(a x)\right )}{16 a^5}+\frac{27 \text{Ci}\left (3 \sin ^{-1}(a x)\right )}{32 a^5}-\frac{25 \text{Ci}\left (5 \sin ^{-1}(a x)\right )}{32 a^5}\\ \end{align*}

Mathematica [A]  time = 0.184825, size = 103, normalized size = 1.05 \[ -\frac{16 a^4 x^4 \sqrt{1-a^2 x^2}-80 a^5 x^5 \sin ^{-1}(a x)+64 a^3 x^3 \sin ^{-1}(a x)+2 \sin ^{-1}(a x)^2 \text{CosIntegral}\left (\sin ^{-1}(a x)\right )-27 \sin ^{-1}(a x)^2 \text{CosIntegral}\left (3 \sin ^{-1}(a x)\right )+25 \sin ^{-1}(a x)^2 \text{CosIntegral}\left (5 \sin ^{-1}(a x)\right )}{32 a^5 \sin ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16*a^4*x^4*Sqrt[1 - a^2*x^2] + 64*a^3*x^3*ArcSin[a*x] - 80*a^5*x^5*ArcSin[a*x] + 2*ArcSin[a*x]^2*CosIntegral
[ArcSin[a*x]] - 27*ArcSin[a*x]^2*CosIntegral[3*ArcSin[a*x]] + 25*ArcSin[a*x]^2*CosIntegral[5*ArcSin[a*x]])/(32
*a^5*ArcSin[a*x]^2)

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Maple [A]  time = 0.042, size = 121, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ( -{\frac{1}{16\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{ax}{16\,\arcsin \left ( ax \right ) }}-{\frac{{\it Ci} \left ( \arcsin \left ( ax \right ) \right ) }{16}}+{\frac{3\,\cos \left ( 3\,\arcsin \left ( ax \right ) \right ) }{32\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}-{\frac{9\,\sin \left ( 3\,\arcsin \left ( ax \right ) \right ) }{32\,\arcsin \left ( ax \right ) }}+{\frac{27\,{\it Ci} \left ( 3\,\arcsin \left ( ax \right ) \right ) }{32}}-{\frac{\cos \left ( 5\,\arcsin \left ( ax \right ) \right ) }{32\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}}}+{\frac{5\,\sin \left ( 5\,\arcsin \left ( ax \right ) \right ) }{32\,\arcsin \left ( ax \right ) }}-{\frac{25\,{\it Ci} \left ( 5\,\arcsin \left ( ax \right ) \right ) }{32}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(-1/16/arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)+1/16*a*x/arcsin(a*x)-1/16*Ci(arcsin(a*x))+3/32/arcsin(a*x)^2*cos
(3*arcsin(a*x))-9/32/arcsin(a*x)*sin(3*arcsin(a*x))+27/32*Ci(3*arcsin(a*x))-1/32/arcsin(a*x)^2*cos(5*arcsin(a*
x))+5/32/arcsin(a*x)*sin(5*arcsin(a*x))-25/32*Ci(5*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^{4} + \arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )^{2} \int \frac{{\left (25 \, a^{2} x^{2} - 12\right )} x^{2}}{\arctan \left (a x, \sqrt{a x + 1} \sqrt{-a x + 1}\right )}\,{d x} -{\left (5 \, a^{2} x^{5} - 4 \, x^{3}\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^4/arcsin(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^4 + arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2*integrate((25*a^2*x^4
- 12*x^2)/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)), x) - (5*a^2*x^5 - 4*x^3)*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^{4}}{\arcsin \left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4/arcsin(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/asin(a*x)**3, x)

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Giac [A]  time = 1.3283, size = 230, normalized size = 2.35 \begin{align*} \frac{5 \,{\left (a^{2} x^{2} - 1\right )}^{2} x}{2 \, a^{4} \arcsin \left (a x\right )} + \frac{3 \,{\left (a^{2} x^{2} - 1\right )} x}{a^{4} \arcsin \left (a x\right )} + \frac{x}{2 \, a^{4} \arcsin \left (a x\right )} - \frac{25 \, \operatorname{Ci}\left (5 \, \arcsin \left (a x\right )\right )}{32 \, a^{5}} + \frac{27 \, \operatorname{Ci}\left (3 \, \arcsin \left (a x\right )\right )}{32 \, a^{5}} - \frac{\operatorname{Ci}\left (\arcsin \left (a x\right )\right )}{16 \, a^{5}} - \frac{{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt{-a^{2} x^{2} + 1}}{2 \, a^{5} \arcsin \left (a x\right )^{2}} + \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{a^{5} \arcsin \left (a x\right )^{2}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{2 \, a^{5} \arcsin \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*(a^2*x^2 - 1)^2*x/(a^4*arcsin(a*x)) + 3*(a^2*x^2 - 1)*x/(a^4*arcsin(a*x)) + 1/2*x/(a^4*arcsin(a*x)) - 25/3
2*cos_integral(5*arcsin(a*x))/a^5 + 27/32*cos_integral(3*arcsin(a*x))/a^5 - 1/16*cos_integral(arcsin(a*x))/a^5
- 1/2*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)/(a^5*arcsin(a*x)^2) + (-a^2*x^2 + 1)^(3/2)/(a^5*arcsin(a*x)^2) - 1/2
*sqrt(-a^2*x^2 + 1)/(a^5*arcsin(a*x)^2)