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

Optimal. Leaf size=11 \[ \frac{\log \left (\sin ^{-1}(a+b x)\right )}{b} \]

[Out]

Log[ArcSin[a + b*x]]/b

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Rubi [A]  time = 0.0761958, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {4807, 4639} \[ \frac{\log \left (\sin ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[ArcSin[a + b*x]]/b

Rule 4807

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-(C/d^2) + (C*x^2)/d^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 4639

Int[1/(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Simp[Log[a + b*ArcSin[c*x]]
/(b*c*Sqrt[d]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]

Rubi steps

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

Mathematica [A]  time = 0.033133, size = 11, normalized size = 1. \[ \frac{\log \left (\sin ^{-1}(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[ArcSin[a + b*x]]/b

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Maple [A]  time = 0.046, size = 12, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( \arcsin \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(arcsin(b*x+a))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.23768, size = 34, normalized size = 3.09 \begin{align*} \frac{\log \left (-\arcsin \left (b x + a\right )\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-arcsin(b*x + a))/b

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Sympy [A]  time = 1.04593, size = 22, normalized size = 2. \begin{align*} \begin{cases} \frac{\log{\left (\operatorname{asin}{\left (a + b x \right )} \right )}}{b} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{1 - a^{2}} \operatorname{asin}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asin(b*x+a)/(-b**2*x**2-2*a*b*x-a**2+1)**(1/2),x)

[Out]

Piecewise((log(asin(a + b*x))/b, Ne(b, 0)), (x/(sqrt(1 - a**2)*asin(a)), True))

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Giac [A]  time = 1.25405, size = 16, normalized size = 1.45 \begin{align*} \frac{\log \left ({\left | \arcsin \left (b x + a\right ) \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(arcsin(b*x + a)))/b