3.319 \(\int \frac{\sqrt{1-a^2-2 a b x-b^2 x^2}}{\sin ^{-1}(a+b x)^4} \, dx\)

Optimal. Leaf size=115 \[ \frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{\sqrt{1-(a+b x)^2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3} \]

[Out]

-(1 - (a + b*x)^2)/(3*b*ArcSin[a + b*x]^3) + ((a + b*x)*Sqrt[1 - (a + b*x)^2])/(3*b*ArcSin[a + b*x]^2) + 1/(3*
b*ArcSin[a + b*x]) - (2*(a + b*x)^2)/(3*b*ArcSin[a + b*x]) + (2*SinIntegral[2*ArcSin[a + b*x]])/(3*b)

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Rubi [A]  time = 0.224037, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4807, 4659, 4633, 4719, 4635, 4406, 12, 3299, 4641} \[ \frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{\sqrt{1-(a+b x)^2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]/ArcSin[a + b*x]^4,x]

[Out]

-(1 - (a + b*x)^2)/(3*b*ArcSin[a + b*x]^3) + ((a + b*x)*Sqrt[1 - (a + b*x)^2])/(3*b*ArcSin[a + b*x]^2) + 1/(3*
b*ArcSin[a + b*x]) - (2*(a + b*x)^2)/(3*b*ArcSin[a + b*x]) + (2*SinIntegral[2*ArcSin[a + b*x]])/(3*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 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x],
x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 4641

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-a^2-2 a b x-b^2 x^2}}{\sin ^{-1}(a+b x)^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{\sin ^{-1}(x)^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}-\frac{2 \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}+\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{2 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac{1-(a+b x)^2}{3 b \sin ^{-1}(a+b x)^3}+\frac{(a+b x) \sqrt{1-(a+b x)^2}}{3 b \sin ^{-1}(a+b x)^2}+\frac{1}{3 b \sin ^{-1}(a+b x)}-\frac{2 (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac{2 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.110668, size = 117, normalized size = 1.02 \[ \frac{-\left (2 a^2+4 a b x+2 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2+(a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+a^2+2 \sin ^{-1}(a+b x)^3 \text{Si}\left (2 \sin ^{-1}(a+b x)\right )+2 a b x+b^2 x^2-1}{3 b \sin ^{-1}(a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]/ArcSin[a + b*x]^4,x]

[Out]

(-1 + a^2 + 2*a*b*x + b^2*x^2 + (a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x] - (-1 + 2*a^2 + 4*
a*b*x + 2*b^2*x^2)*ArcSin[a + b*x]^2 + 2*ArcSin[a + b*x]^3*SinIntegral[2*ArcSin[a + b*x]])/(3*b*ArcSin[a + b*x
]^3)

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Maple [A]  time = 0.053, size = 81, normalized size = 0.7 \begin{align*}{\frac{4\,{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}+2\,\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) -\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) -1}{6\,b \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b^2*x^2-2*a*b*x-a^2+1)^(1/2)/arcsin(b*x+a)^4,x)

[Out]

1/6/b*(4*Si(2*arcsin(b*x+a))*arcsin(b*x+a)^3+2*cos(2*arcsin(b*x+a))*arcsin(b*x+a)^2+sin(2*arcsin(b*x+a))*arcsi
n(b*x+a)-cos(2*arcsin(b*x+a))-1)/arcsin(b*x+a)^3

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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((-b^2*x^2-2*a*b*x-a^2+1)^(1/2)/arcsin(b*x+a)^4,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{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\arcsin \left (b x + a\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2-2*a*b*x-a^2+1)^(1/2)/arcsin(b*x+a)^4,x, algorithm="fricas")

[Out]

integral(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/arcsin(b*x + a)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{\operatorname{asin}^{4}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b**2*x**2-2*a*b*x-a**2+1)**(1/2)/asin(b*x+a)**4,x)

[Out]

Integral(sqrt(-(a + b*x - 1)*(a + b*x + 1))/asin(a + b*x)**4, x)

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Giac [A]  time = 1.3426, size = 173, normalized size = 1.5 \begin{align*} \frac{2 \, \operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} - \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{3 \, b \arcsin \left (b x + a\right )} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} - \frac{1}{3 \, b \arcsin \left (b x + a\right )} + \frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{3 \, b \arcsin \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b^2*x^2-2*a*b*x-a^2+1)^(1/2)/arcsin(b*x+a)^4,x, algorithm="giac")

[Out]

2/3*sin_integral(2*arcsin(b*x + a))/b - 2/3*(b^2*x^2 + 2*a*b*x + a^2 - 1)/(b*arcsin(b*x + a)) + 1/3*sqrt(-b^2*
x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b*arcsin(b*x + a)^2) - 1/3/(b*arcsin(b*x + a)) + 1/3*(b^2*x^2 + 2*a*b*x +
a^2 - 1)/(b*arcsin(b*x + a)^3)