3.149 \(\int \frac{1}{\sin ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=41 \[ -\frac{\text{Si}\left (\sin ^{-1}(a+b x)\right )}{b}-\frac{\sqrt{1-(a+b x)^2}}{b \sin ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0789604, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4803, 4621, 4723, 3299} \[ -\frac{\text{Si}\left (\sin ^{-1}(a+b x)\right )}{b}-\frac{\sqrt{1-(a+b x)^2}}{b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^(-2),x]

[Out]

-(Sqrt[1 - (a + b*x)^2]/(b*ArcSin[a + b*x])) - SinIntegral[ArcSin[a + b*x]]/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 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 3299

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

Rubi steps

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

Mathematica [A]  time = 0.070635, size = 37, normalized size = 0.9 \[ -\frac{\text{Si}\left (\sin ^{-1}(a+b x)\right )+\frac{\sqrt{1-(a+b x)^2}}{\sin ^{-1}(a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^(-2),x]

[Out]

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

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Maple [A]  time = 0.03, size = 38, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{1}{\arcsin \left ( bx+a \right ) }\sqrt{1- \left ( bx+a \right ) ^{2}}}-{\it Si} \left ( \arcsin \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)-Si(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)^2,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 )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(b*x + a)^(-2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{asin}^{2}{\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)**2,x)

[Out]

Integral(asin(a + b*x)**(-2), x)

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Giac [A]  time = 1.15882, size = 53, normalized size = 1.29 \begin{align*} -\frac{\operatorname{Si}\left (\arcsin \left (b x + a\right )\right )}{b} - \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{b \arcsin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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