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

Optimal. Leaf size=65 \[ -\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b}+\frac{a+b x}{2 b \sin ^{-1}(a+b x)}-\frac{\sqrt{1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0855354, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {4803, 4621, 4719, 4623, 3302} \[ -\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b}+\frac{a+b x}{2 b \sin ^{-1}(a+b x)}-\frac{\sqrt{1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^(-3),x]

[Out]

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

Rule 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rule 4621

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

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 4623

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

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

Mathematica [A]  time = 0.0675659, size = 65, normalized size = 1. \[ -\frac{\text{CosIntegral}\left (\sin ^{-1}(a+b x)\right )}{2 b}+\frac{a+b x}{2 b \sin ^{-1}(a+b x)}-\frac{\sqrt{1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^(-3),x]

[Out]

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

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Maple [A]  time = 0.027, size = 53, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{2\, \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}\sqrt{1- \left ( bx+a \right ) ^{2}}}+{\frac{bx+a}{2\,\arcsin \left ( bx+a \right ) }}-{\frac{{\it Ci} \left ( \arcsin \left ( bx+a \right ) \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsin(b*x+a)^3,x)

[Out]

1/b*(-1/2/arcsin(b*x+a)^2*(1-(b*x+a)^2)^(1/2)+1/2*(b*x+a)/arcsin(b*x+a)-1/2*Ci(arcsin(b*x+a)))

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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(1/arcsin(b*x+a)^3,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{1}{\arcsin \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsin(b*x+a)^3,x, algorithm="fricas")

[Out]

integral(arcsin(b*x + a)^(-3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{asin}^{3}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asin(b*x+a)**3,x)

[Out]

Integral(asin(a + b*x)**(-3), x)

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Giac [A]  time = 1.15734, size = 77, normalized size = 1.18 \begin{align*} -\frac{\operatorname{Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b} + \frac{b x + a}{2 \, b \arcsin \left (b x + a\right )} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{2 \, b \arcsin \left (b x + a\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/2*cos_integral(arcsin(b*x + a))/b + 1/2*(b*x + a)/(b*arcsin(b*x + a)) - 1/2*sqrt(-(b*x + a)^2 + 1)/(b*arcsi
n(b*x + a)^2)