3.4.19 \(\int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\text {ArcSin}(a+b x)^4} \, dx\) [319]

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

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 = {4891, 4751, 4729, 4807, 4731, 4491, 12, 3380, 4737} \begin {gather*} \frac {2 \text {Si}(2 \text {ArcSin}(a+b x))}{3 b}-\frac {2 (a+b x)^2}{3 b \text {ArcSin}(a+b x)}+\frac {\sqrt {1-(a+b x)^2} (a+b x)}{3 b \text {ArcSin}(a+b x)^2}+\frac {1}{3 b \text {ArcSin}(a+b x)}-\frac {1-(a+b x)^2}{3 b \text {ArcSin}(a+b x)^3} \end {gather*}

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]

-1/3*(1 - (a + b*x)^2)/(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 12

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

Rule 3380

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 4491

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 4729

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 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sin[-
a/b + x/b]^m*Cos[-a/b + x/b], x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4737

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

Rule 4751

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)/(b*(n + 1)))*Simp[(d + e*x^2)
^p/(1 - c^2*x^2)^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 4807

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] - Dist[f*(m/(b
*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], 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]

Rule 4891

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

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Mathematica [A]
time = 0.07, size = 117, normalized size = 1.02 \begin {gather*} \frac {-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} \text {ArcSin}(a+b x)-\left (-1+2 a^2+4 a b x+2 b^2 x^2\right ) \text {ArcSin}(a+b x)^2+2 \text {ArcSin}(a+b x)^3 \text {Si}(2 \text {ArcSin}(a+b x))}{3 b \text {ArcSin}(a+b x)^3} \end {gather*}

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 = 0.46, size = 81, normalized size = 0.70

method	result	size
default	\(\frac {4 \sinIntegral \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{3}+2 \cos \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )^{2}+\sin \left (2 \arcsin \left (b x +a \right )\right ) \arcsin \left (b x +a \right )-\cos \left (2 \arcsin \left (b x +a \right )\right )-1}{6 b \arcsin \left (b x +a \right )^{3}}\)	\(81\)

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,method=_RETURNVERBOSE)

[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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.48, size = 128, normalized size = 1.11 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{{\mathrm {asin}\left (a+b\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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