3.4.16 \(\int \frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{\text {ArcSin}(a+b x)} \, dx\) [316]

Optimal. Leaf size=31 \[ \frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{2 b}+\frac {\log (\text {ArcSin}(a+b x))}{2 b} \]

[Out]

1/2*Ci(2*arcsin(b*x+a))/b+1/2*ln(arcsin(b*x+a))/b

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Rubi [A]
time = 0.08, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4891, 4753, 3393, 3383} \begin {gather*} \frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))}{2 b}+\frac {\log (\text {ArcSin}(a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

CosIntegral[2*ArcSin[a + b*x]]/(2*b) + Log[ArcSin[a + b*x]]/(2*b)

Rule 3383

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 3393

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 4753

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

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

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Mathematica [A]
time = 0.05, size = 24, normalized size = 0.77 \begin {gather*} \frac {\text {CosIntegral}(2 \text {ArcSin}(a+b x))+\log (\text {ArcSin}(a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(CosIntegral[2*ArcSin[a + b*x]] + Log[ArcSin[a + b*x]])/(2*b)

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Maple [A]
time = 0.49, size = 23, normalized size = 0.74

method	result	size
default	\(\frac {\ln \left (\arcsin \left (b x +a \right )\right )+\cosineIntegral \left (2 \arcsin \left (b x +a \right )\right )}{2 b}\)	\(23\)

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),x,method=_RETURNVERBOSE)

[Out]

1/2*(ln(arcsin(b*x+a))+Ci(2*arcsin(b*x+a)))/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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),x, algorithm="maxima")

[Out]

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

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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),x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [A]
time = 0.45, size = 27, normalized size = 0.87 \begin {gather*} \frac {\operatorname {Ci}\left (2 \, \arcsin \left (b x + a\right )\right )}{2 \, b} + \frac {\log \left (\arcsin \left (b x + a\right )\right )}{2 \, b} \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),x, algorithm="giac")

[Out]

1/2*cos_integral(2*arcsin(b*x + a))/b + 1/2*log(arcsin(b*x + a))/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{\mathrm {asin}\left (a+b\,x\right )} \,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),x)

[Out]

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

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