3.318 \(\int \frac{x^2 \sqrt{1-c^2 x^2}}{a+b \sin ^{-1}(c x)} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{8 b c^3}+\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^3} \]

[Out]

-(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^3) - (Sin[(4*
a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(8*b*c^3)

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Rubi [A]  time = 0.250688, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4723, 4406, 3303, 3299, 3302} \[ -\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{8 b c^3}+\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{8 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3303

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.181625, size = 66, normalized size = 0.8 \[ -\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-\log \left (8 \left (a+b \sin ^{-1}(c x)\right )\right )}{8 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 77, normalized size = 0.9 \begin{align*} -{\frac{1}{8\,{c}^{3}b}{\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) }-{\frac{1}{8\,{c}^{3}b}{\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) }+{\frac{\ln \left ( a+b\arcsin \left ( cx \right ) \right ) }{8\,{c}^{3}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{b \arcsin \left (c x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{b \arcsin \left (c x\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.37644, size = 228, normalized size = 2.78 \begin{align*} -\frac{\cos \left (\frac{a}{b}\right )^{4} \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{3}} - \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^{3}} + \frac{\cos \left (\frac{a}{b}\right )^{2} \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{3}} + \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^{3}} - \frac{\operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{3}} + \frac{\log \left (b \arcsin \left (c x\right ) + a\right )}{8 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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