3.4.28 \(\int \frac {(1-c^2 x^2)^{3/2}}{a+b \text {ArcSin}(c x)} \, dx\) [328]

Optimal. Leaf size=144 \[ \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c}+\frac {3 \log (a+b \text {ArcSin}(c x))}{8 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c} \]

[Out]

1/2*Ci(2*(a+b*arcsin(c*x))/b)*cos(2*a/b)/b/c+1/8*Ci(4*(a+b*arcsin(c*x))/b)*cos(4*a/b)/b/c+3/8*ln(a+b*arcsin(c*
x))/b/c+1/2*Si(2*(a+b*arcsin(c*x))/b)*sin(2*a/b)/b/c+1/8*Si(4*(a+b*arcsin(c*x))/b)*sin(4*a/b)/b/c

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Rubi [A]
time = 0.15, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4753, 3393, 3384, 3380, 3383} \begin {gather*} \frac {\cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c}+\frac {\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \text {ArcSin}(c x))}{b}\right )}{2 b c}+\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \text {ArcSin}(c x))}{b}\right )}{8 b c}+\frac {3 \log (a+b \text {ArcSin}(c x))}{8 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(2*b*c) + (Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[c*x
]))/b])/(8*b*c) + (3*Log[a + b*ArcSin[c*x]])/(8*b*c) + (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(
2*b*c) + (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(8*b*c)

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 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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - 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]

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 121, normalized size = 0.84 \begin {gather*} \frac {4 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\cos \left (\frac {4 a}{b}\right ) \text {CosIntegral}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+4 \log (a+b \text {ArcSin}(c x))-\log (8 (a+b \text {ArcSin}(c x)))+4 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )+\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\text {ArcSin}(c x)\right )\right )}{8 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])] + 4*Log[a
+ b*ArcSin[c*x]] - Log[8*(a + b*ArcSin[c*x])] + 4*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + Sin[(4*a)
/b]*SinIntegral[4*(a/b + ArcSin[c*x])])/(8*b*c)

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Maple [A]
time = 0.10, size = 111, normalized size = 0.77

method	result	size
default	\(\frac {\sinIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\cosineIntegral \left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )+4 \sinIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+4 \cosineIntegral \left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )+3 \ln \left (a +b \arcsin \left (c x \right )\right )}{8 c b}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/8/c*(Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)+Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)+4*Si(2*arcsin(c*x)+2*a/b)*sin(2*a
/b)+4*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)+3*ln(a+b*arcsin(c*x)))/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((-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate((-c^2*x^2 + 1)^(3/2)/(b*arcsin(c*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((-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{a + b \operatorname {asin}{\left (c x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 252, normalized size = 1.75 \begin {gather*} \frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c} + \frac {\cos \left (\frac {a}{b}\right )^{3} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c} - \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{2 \, b c} + \frac {\cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{b c} + \frac {\operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c} - \frac {\operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c} + \frac {3 \, \log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) + cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))
/(b*c) - cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) + cos(a/b)^2*cos_integral(2*a/b + 2*arcsin(c*x))
/(b*c) - 1/2*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c) + cos(a/b)*sin(a/b)*sin_integral(2*a/
b + 2*arcsin(c*x))/(b*c) + 1/8*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c) - 1/2*cos_integral(2*a/b + 2*arcsin(c
*x))/(b*c) + 3/8*log(b*arcsin(c*x) + a)/(b*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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