3.324 \(\int \frac{(1-a^2-2 a b x-b^2 x^2)^{3/2}}{\sin ^{-1}(a+b x)^2} \, dx\)

Optimal. Leaf size=57 \[ -\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac{\text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.152061, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4807, 4659, 4723, 4406, 3299} \[ -\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac{\text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - (a + b*x)^2)^2/(b*ArcSin[a + b*x])) - SinIntegral[2*ArcSin[a + b*x]]/b - SinIntegral[4*ArcSin[a + b*x]]
/(2*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 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 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 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{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}}{\sin ^{-1}(a+b x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac{4 \operatorname{Subst}\left (\int \frac{x \left (1-x^2\right )}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac{4 \operatorname{Subst}\left (\int \frac{\cos ^3(x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac{4 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 x}+\frac{\sin (4 x)}{8 x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac{\operatorname{Subst}\left (\int \frac{\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\sin (2 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{\left (1-(a+b x)^2\right )^2}{b \sin ^{-1}(a+b x)}-\frac{\text{Si}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac{\text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.313519, size = 70, normalized size = 1.23 \[ -\frac{2 \left (a^2+2 a b x+b^2 x^2-1\right )^2+2 \sin ^{-1}(a+b x) \text{Si}\left (2 \sin ^{-1}(a+b x)\right )+\sin ^{-1}(a+b x) \text{Si}\left (4 \sin ^{-1}(a+b x)\right )}{2 b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 70, normalized size = 1.2 \begin{align*} -{\frac{8\,{\it Si} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +4\,{\it Si} \left ( 4\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) +4\,\cos \left ( 2\,\arcsin \left ( bx+a \right ) \right ) +\cos \left ( 4\,\arcsin \left ( bx+a \right ) \right ) +3}{8\,b\arcsin \left ( bx+a \right ) }} \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)^(3/2)/arcsin(b*x+a)^2,x)

[Out]

-1/8/b*(8*Si(2*arcsin(b*x+a))*arcsin(b*x+a)+4*Si(4*arcsin(b*x+a))*arcsin(b*x+a)+4*cos(2*arcsin(b*x+a))+cos(4*a
rcsin(b*x+a))+3)/arcsin(b*x+a)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \,{\left (3 \, a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 4 \, b \arctan \left (b x + a, \sqrt{b x + a + 1} \sqrt{-b x - a + 1}\right ) \int \frac{b^{3} x^{3} + 3 \, a b^{2} x^{2} + a^{3} +{\left (3 \, a^{2} - 1\right )} b x - a}{\arctan \left (b x + a, \sqrt{b x + a + 1} \sqrt{-b x - a + 1}\right )}\,{d x} - 2 \, a^{2} + 1}{b \arctan \left (b x + a, \sqrt{b x + a + 1} \sqrt{-b x - a + 1}\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)^(3/2)/arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

-(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 + a^4 + 4*(a^3 - a)*b*x - b*arctan2(b*x + a, sqrt(b*x + a + 1)
*sqrt(-b*x - a + 1))*integrate(4*(b^3*x^3 + 3*a*b^2*x^2 + a^3 + (3*a^2 - 1)*b*x - a)/arctan2(b*x + a, sqrt(b*x
+ a + 1)*sqrt(-b*x - a + 1)), x) - 2*a^2 + 1)/(b*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1)))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{\arcsin \left (b x + a\right )^{2}}, 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)^(3/2)/arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

integral((-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/arcsin(b*x + a)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}{\operatorname{asin}^{2}{\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)**(3/2)/asin(b*x+a)**2,x)

[Out]

Integral((-(a + b*x - 1)*(a + b*x + 1))**(3/2)/asin(a + b*x)**2, x)

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Giac [A]  time = 1.27461, size = 82, normalized size = 1.44 \begin{align*} -\frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{b \arcsin \left (b x + a\right )} - \frac{\operatorname{Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{2 \, b} - \frac{\operatorname{Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} \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)^(3/2)/arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

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