3.326 \(\int \frac {(1-a^2-2 a b x-b^2 x^2)^{3/2}}{\sin ^{-1}(a+b x)^4} \, dx\)

Optimal. Leaf size=155 \[ \frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac {4 \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}-\frac {8 \left (1-(a+b x)^2\right ) (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {2 \left (1-(a+b x)^2\right )^{3/2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3} \]

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4807, 4659, 4721, 4635, 4406, 12, 3299} \[ \frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac {4 \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}-\frac {8 \left (1-(a+b x)^2\right ) (a+b x)^2}{3 b \sin ^{-1}(a+b x)}+\frac {2 \left (1-(a+b x)^2\right )^{3/2} (a+b x)}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - (a + b*x)^2)^2/(3*b*ArcSin[a + b*x]^3) + (2*(a + b*x)*(1 - (a + b*x)^2)^(3/2))/(3*b*ArcSin[a + b*x]^2) +
(2*(1 - (a + b*x)^2))/(3*b*ArcSin[a + b*x]) - (8*(a + b*x)^2*(1 - (a + b*x)^2))/(3*b*ArcSin[a + b*x]) + (2*Si
nIntegral[2*ArcSin[a + b*x]])/(3*b) + (4*SinIntegral[4*ArcSin[a + b*x]])/(3*b)

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 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 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 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 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]

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)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2}}{\sin ^{-1}(x)^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}-\frac {4 \operatorname {Subst}\left (\int \frac {x \left (1-x^2\right )}{\sin ^{-1}(x)^3} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}-\frac {2 \operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}+\frac {8 \operatorname {Subst}\left (\int \frac {x^2 \sqrt {1-x^2}}{\sin ^{-1}(x)^2} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{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 {16 \operatorname {Subst}\left (\int \frac {x}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}-\frac {32 \operatorname {Subst}\left (\int \frac {x^3}{\sin ^{-1}(x)} \, dx,x,a+b x\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{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 {16 \operatorname {Subst}\left (\int \frac {\cos (x) \sin (x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac {32 \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^3(x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{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 {16 \operatorname {Subst}\left (\int \frac {\sin (2 x)}{2 x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}-\frac {32 \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 )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{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 {4 \operatorname {Subst}\left (\int \frac {\sin (4 x)}{x} \, dx,x,\sin ^{-1}(a+b x)\right )}{3 b}\\ &=-\frac {\left (1-(a+b x)^2\right )^2}{3 b \sin ^{-1}(a+b x)^3}+\frac {2 (a+b x) \left (1-(a+b x)^2\right )^{3/2}}{3 b \sin ^{-1}(a+b x)^2}+\frac {2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}-\frac {8 (a+b x)^2 \left (1-(a+b x)^2\right )}{3 b \sin ^{-1}(a+b x)}+\frac {2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )}{3 b}+\frac {4 \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 143, normalized size = 0.92 \[ \frac {\frac {\left (a^2+2 a b x+b^2 x^2-1\right ) \left (2 \left (4 a^2+8 a b x+4 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2-2 (a+b x) \sqrt {-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)-a^2-2 a b x-b^2 x^2+1\right )}{\sin ^{-1}(a+b x)^3}+2 \text {Si}\left (2 \sin ^{-1}(a+b x)\right )+4 \text {Si}\left (4 \sin ^{-1}(a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-1 + a^2 + 2*a*b*x + b^2*x^2)*(1 - a^2 - 2*a*b*x - b^2*x^2 - 2*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*
ArcSin[a + b*x] + 2*(-1 + 4*a^2 + 8*a*b*x + 4*b^2*x^2)*ArcSin[a + b*x]^2))/ArcSin[a + b*x]^3 + 2*SinIntegral[2
*ArcSin[a + b*x]] + 4*SinIntegral[4*ArcSin[a + b*x]])/(3*b)

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{\arcsin \left (b x + a\right )^{4}}, x\right ) \]

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)^4,x, algorithm="fricas")

[Out]

integral((-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/arcsin(b*x + a)^4, x)

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giac [A]  time = 2.97, size = 163, normalized size = 1.05 \[ \frac {8 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )} + \frac {4 \, \operatorname {Si}\left (4 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac {2 \, \operatorname {Si}\left (2 \, \arcsin \left (b x + a\right )\right )}{3 \, b} + \frac {2 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )}}{3 \, b \arcsin \left (b x + a\right )^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{b \arcsin \left (b x + a\right )} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}^{2}}{3 \, b \arcsin \left (b x + a\right )^{3}} \]

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)^4,x, algorithm="giac")

[Out]

8/3*(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)) + 4/3*sin_integral(4*arcsin(b*x + a))/b + 2/3*sin_inte
gral(2*arcsin(b*x + a))/b + 2/3*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*(b*x + a)/(b*arcsin(b*x + a)^2) + 2*(b^2*
x^2 + 2*a*b*x + a^2 - 1)/(b*arcsin(b*x + a)) - 1/3*(b^2*x^2 + 2*a*b*x + a^2 - 1)^2/(b*arcsin(b*x + a)^3)

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maple [A]  time = 0.17, size = 148, normalized size = 0.95 \[ \frac {16 \Si \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+32 \Si \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+8 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+8 \cos \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+4 \sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )+2 \sin \left (4 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-4 \cos \left (2 \arcsin \left (b x +a \right )\right )-\cos \left (4 \arcsin \left (b x +a \right )\right )-3}{24 b \arcsin \left (b x +a \right )^{3}} \]

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)^4,x)

[Out]

1/24/b*(16*Si(2*arcsin(b*x+a))*arcsin(b*x+a)^3+32*Si(4*arcsin(b*x+a))*arcsin(b*x+a)^3+8*cos(2*arcsin(b*x+a))*a
rcsin(b*x+a)^2+8*cos(4*arcsin(b*x+a))*arcsin(b*x+a)^2+4*sin(2*arcsin(b*x+a))*arcsin(b*x+a)+2*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)^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asin}\left (a+b\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - b^2*x^2 - 2*a*b*x - a^2)^(3/2)/asin(a + b*x)^4,x)

[Out]

int((1 - b^2*x^2 - 2*a*b*x - a^2)^(3/2)/asin(a + b*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{\operatorname {asin}^{4}{\left (a + b x \right )}}\, dx \]

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)**4,x)

[Out]

Integral((-(a + b*x - 1)*(a + b*x + 1))**(3/2)/asin(a + b*x)**4, x)

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