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

Optimal. Leaf size=94 \[ -\frac{\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}+\frac{\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}-\frac{x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-((x^2*(1 - c^2*x^2))/(b*c*(a + b*ArcSin[c*x]))) - (CosIntegral[(4*(a + b*ArcSin[c*x]))/b]*Sin[(4*a)/b])/(2*b^
2*c^3) + (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/(2*b^2*c^3)

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Rubi [A]  time = 0.468122, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4721, 4635, 4406, 12, 3303, 3299, 3302} \[ -\frac{\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c^3}+\frac{\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c^3}-\frac{x^2 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^2*(1 - c^2*x^2))/(b*c*(a + b*ArcSin[c*x]))) - (CosIntegral[(4*a)/b + 4*ArcSin[c*x]]*Sin[(4*a)/b])/(2*b^2*
c^3) + (Cos[(4*a)/b]*SinIntegral[(4*a)/b + 4*ArcSin[c*x]])/(2*b^2*c^3)

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 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 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 12

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

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

Mathematica [A]  time = 0.322192, size = 82, normalized size = 0.87 \[ \frac{\frac{2 b c^2 x^2 \left (c^2 x^2-1\right )}{a+b \sin ^{-1}(c x)}-\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{2 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c^2*x^2*(-1 + c^2*x^2))/(a + b*ArcSin[c*x]) - CosIntegral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] + Cos[(4*a
)/b]*SinIntegral[4*(a/b + ArcSin[c*x])])/(2*b^2*c^3)

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Maple [A]  time = 0.046, size = 136, normalized size = 1.5 \begin{align*}{\frac{1}{8\,{c}^{3} \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( 4\,\arcsin \left ( cx \right ){\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) b-4\,\arcsin \left ( cx \right ){\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) b+4\,{\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) a-4\,{\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) a+\cos \left ( 4\,\arcsin \left ( cx \right ) \right ) b-b \right ) } \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))^2,x)

[Out]

1/8/c^3*(4*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-4*arcsin(c*x)*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*b
+4*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*a-4*Ci(4*arcsin(c*x)+4*a/b)*sin(4*a/b)*a+cos(4*arcsin(c*x))*b-b)/(a+b*ar
csin(c*x))/b^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{c^{2} x^{4} - x^{2} - 2 \,{\left (b^{2} c \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c\right )} \int \frac{2 \, c^{2} x^{3} - x}{b^{2} c \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c}\,{d x}}{b^{2} c \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c} \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))^2,x, algorithm="maxima")

[Out]

(c^2*x^4 - x^2 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(2*(2*c^2*x^3 - x)/(b^2*c
*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c), x))/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) +
a*b*c)

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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^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*x^2 + 1)*x^2/(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2), 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 )}}{\left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

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Giac [B]  time = 1.52963, size = 760, normalized size = 8.09 \begin{align*} \text{result too large to display} \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))^2,x, algorithm="giac")

[Out]

-4*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 4
*b*arcsin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 4*a*cos(a/b)
^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 4*a*cos(a/b)^4*sin_integra
l(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*a
rcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arc
sin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 2*a*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c
^3*arcsin(c*x) + a*b^2*c^3) - 4*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*
c^3) + (c^2*x^2 - 1)^2*b/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*arcsin(c
*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + (c^2*x^2 - 1)*b/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/2*a*sin_integra
l(4*a/b + 4*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3)