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

Optimal. Leaf size=90 \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{b}-\frac{\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)} \]

[Out]

-(1 - (a + b*x)^2)^2/(2*b*ArcSin[a + b*x]^2) + (2*(a + b*x)*(1 - (a + b*x)^2)^(3/2))/(b*ArcSin[a + b*x]) - Cos
Integral[2*ArcSin[a + b*x]]/b - CosIntegral[4*ArcSin[a + b*x]]/b

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Rubi [A]  time = 0.283546, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4807, 4659, 4721, 4661, 3312, 3302, 4723, 4406} \[ -\frac{\text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )}{b}-\frac{\text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{b}-\frac{\left (1-(a+b x)^2\right )^2}{2 b \sin ^{-1}(a+b x)^2}+\frac{2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{b \sin ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - (a + b*x)^2)^2/(2*b*ArcSin[a + b*x]^2) + (2*(a + b*x)*(1 - (a + b*x)^2)^(3/2))/(b*ArcSin[a + b*x]) - Cos
Integral[2*ArcSin[a + b*x]]/b - CosIntegral[4*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 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 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^m*Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntPar
t[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p - 1/
2)*(a + b*ArcSin[c*x])^(n + 1), x], x] + Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(n +
1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x], x])
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

Rule 4661

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[(
a + b*x)^n*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] && I
GtQ[2*p, 0] && (IntegerQ[p] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 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]

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

Mathematica [A]  time = 0.43503, size = 110, normalized size = 1.22 \[ -\frac{\frac{\left (a^2+2 a b x+b^2 x^2-1\right ) \left (4 (a+b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+a^2+2 a b x+b^2 x^2-1\right )}{\sin ^{-1}(a+b x)^2}+2 \text{CosIntegral}\left (2 \sin ^{-1}(a+b x)\right )+2 \text{CosIntegral}\left (4 \sin ^{-1}(a+b x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 108, normalized size = 1.2 \begin{align*} -{\frac{16\,{\it Ci} \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}+16\,{\it Ci} \left ( 4\,\arcsin \left ( bx+a \right ) \right ) \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}-8\,\sin \left ( 2\,\arcsin \left ( bx+a \right ) \right ) \arcsin \left ( bx+a \right ) -4\,\sin \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}{16\,b \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}} \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)^3,x)

[Out]

-1/16/b*(16*Ci(2*arcsin(b*x+a))*arcsin(b*x+a)^2+16*Ci(4*arcsin(b*x+a))*arcsin(b*x+a)^2-8*sin(2*arcsin(b*x+a))*
arcsin(b*x+a)-4*sin(4*arcsin(b*x+a))*arcsin(b*x+a)+4*cos(2*arcsin(b*x+a))+cos(4*arcsin(b*x+a))+3)/arcsin(b*x+a
)^2

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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)^(3/2)/arcsin(b*x+a)^3,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{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{\arcsin \left (b x + a\right )^{3}}, 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)^3,x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [A]  time = 1.30654, size = 136, normalized size = 1.51 \begin{align*} \frac{2 \,{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}{\left (b x + a\right )}}{b \arcsin \left (b x + a\right )} - \frac{\operatorname{Ci}\left (4 \, \arcsin \left (b x + a\right )\right )}{b} - \frac{\operatorname{Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{b} - \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{2 \, b \arcsin \left (b x + a\right )^{2}} \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)^3,x, algorithm="giac")

[Out]

2*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*(b*x + a)/(b*arcsin(b*x + a)) - cos_integral(4*arcsin(b*x + a))/b - cos
_integral(2*arcsin(b*x + a))/b - 1/2*(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)^2)