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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 0.07, size = 65, normalized size = 1.00 \[ -\frac {\text {Ci}\left (\sin ^{-1}(a+b x)\right )}{2 b}+\frac {a+b x}{2 b \sin ^{-1}(a+b x)}-\frac {\sqrt {1-(a+b x)^2}}{2 b \sin ^{-1}(a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^(-3),x]

[Out]

-1/2*Sqrt[1 - (a + b*x)^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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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\arcsin \left (b x + a\right )^{3}}, x\right ) \]

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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giac [A]  time = 0.39, size = 57, normalized size = 0.88 \[ -\frac {\operatorname {Ci}\left (\arcsin \left (b x + a\right )\right )}{2 \, b} + \frac {b x + a}{2 \, b \arcsin \left (b x + a\right )} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{2 \, b \arcsin \left (b x + a\right )^{2}} \]

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)

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maple [A]  time = 0.03, size = 53, normalized size = 0.82 \[ \frac {-\frac {\sqrt {1-\left (b x +a \right )^{2}}}{2 \arcsin \left (b x +a \right )^{2}}+\frac {b x +a}{2 \arcsin \left (b x +a \right )}-\frac {\Ci \left (\arcsin \left (b x +a \right )\right )}{2}}{b} \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2} \int \frac {1}{\arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}\,{d x} - {\left (b x + a\right )} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right ) + \sqrt {b x + a + 1} \sqrt {-b x - a + 1}}{2 \, b \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2*integrate(1/arctan2(b*x + a, sqrt(b*x + a + 1
)*sqrt(-b*x - a + 1)), x) - (b*x + a)*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)) + sqrt(b*x + a +
1)*sqrt(-b*x - a + 1))/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {asin}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asin(a + b*x)^3,x)

[Out]

int(1/asin(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {asin}^{3}{\left (a + b x \right )}}\, dx \]

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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