3.326 \(\int \frac{x^2 (1-c^2 x^2)^{3/2}}{a+b \sin ^{-1}(c x)} \, dx\)

Optimal. Leaf size=206 \[ \frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^3}-\frac{\cos \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (\frac{6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b c^3}-\frac{\sin \left (\frac{6 a}{b}\right ) \text{Si}\left (\frac{6 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{32 b c^3}+\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3} \]

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) - (Cos[(4*a)/b]*CosIntegral[(4*(a + b*ArcSin[
c*x]))/b])/(16*b*c^3) - (Cos[(6*a)/b]*CosIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) + Log[a + b*ArcSin[c*
x]]/(16*b*c^3) + (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(32*b*c^3) - (Sin[(4*a)/b]*SinIntegral[
(4*(a + b*ArcSin[c*x]))/b])/(16*b*c^3) - (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[c*x]))/b])/(32*b*c^3)

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Rubi [A]  time = 0.423249, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac{\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac{\cos \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (\frac{6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}+\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac{\sin \left (\frac{6 a}{b}\right ) \text{Si}\left (\frac{6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}+\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSin[c*x]])/(32*b*c^3) - (Cos[(4*a)/b]*CosIntegral[(4*a)/b + 4*ArcSin[
c*x]])/(16*b*c^3) - (Cos[(6*a)/b]*CosIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*b*c^3) + Log[a + b*ArcSin[c*x]]/(1
6*b*c^3) + (Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(32*b*c^3) - (Sin[(4*a)/b]*SinIntegral[(4*a)/b
+ 4*ArcSin[c*x]])/(16*b*c^3) - (Sin[(6*a)/b]*SinIntegral[(6*a)/b + 6*ArcSin[c*x]])/(32*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 \left (1-c^2 x^2\right )^{3/2}}{a+b \sin ^{-1}(c x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\cos ^4(x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{16 (a+b x)}+\frac{\cos (2 x)}{32 (a+b x)}-\frac{\cos (4 x)}{16 (a+b x)}-\frac{\cos (6 x)}{32 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3}\\ &=\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac{\operatorname{Subst}\left (\int \frac{\cos (6 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}-\frac{\operatorname{Subst}\left (\int \frac{\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 c^3}\\ &=\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac{\cos \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 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 )}{16 c^3}-\frac{\cos \left (\frac{6 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}+\frac{\sin \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 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 )}{16 c^3}-\frac{\sin \left (\frac{6 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{32 c^3}\\ &=\frac{\cos \left (\frac{2 a}{b}\right ) \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac{\cos \left (\frac{4 a}{b}\right ) \text{Ci}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac{\cos \left (\frac{6 a}{b}\right ) \text{Ci}\left (\frac{6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}+\frac{\log \left (a+b \sin ^{-1}(c x)\right )}{16 b c^3}+\frac{\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{32 b c^3}-\frac{\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{16 b c^3}-\frac{\sin \left (\frac{6 a}{b}\right ) \text{Si}\left (\frac{6 a}{b}+6 \sin ^{-1}(c x)\right )}{32 b c^3}\\ \end{align*}

Mathematica [A]  time = 0.595416, size = 165, normalized size = 0.8 \[ -\frac{-\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 \cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\cos \left (\frac{6 a}{b}\right ) \text{CosIntegral}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 \sin \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{6 a}{b}\right ) \text{Si}\left (6 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 \log \left (a+b \sin ^{-1}(c x)\right )-4 \log \left (8 \left (a+b \sin ^{-1}(c x)\right )\right )}{32 b c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(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])]) + 2*Cos[(4*a)/b]*CosIntegral[4*(a/b + ArcSin[c*x])] + Cos
[(6*a)/b]*CosIntegral[6*(a/b + ArcSin[c*x])] + 2*Log[a + b*ArcSin[c*x]] - 4*Log[8*(a + b*ArcSin[c*x])] - Sin[(
2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] + 2*Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] + Sin[(6*a)/b]*
SinIntegral[6*(a/b + ArcSin[c*x])])/(32*b*c^3)

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Maple [A]  time = 0.05, size = 193, normalized size = 0.9 \begin{align*} -{\frac{1}{16\,b{c}^{3}}{\it Si} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) }-{\frac{1}{16\,b{c}^{3}}{\it Ci} \left ( 4\,\arcsin \left ( cx \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) }+{\frac{1}{32\,b{c}^{3}}{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) }+{\frac{1}{32\,b{c}^{3}}{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) }-{\frac{1}{32\,b{c}^{3}}{\it Si} \left ( 6\,\arcsin \left ( cx \right ) +6\,{\frac{a}{b}} \right ) \sin \left ( 6\,{\frac{a}{b}} \right ) }-{\frac{1}{32\,b{c}^{3}}{\it Ci} \left ( 6\,\arcsin \left ( cx \right ) +6\,{\frac{a}{b}} \right ) \cos \left ( 6\,{\frac{a}{b}} \right ) }+{\frac{\ln \left ( a+b\arcsin \left ( cx \right ) \right ) }{16\,b{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/c^3/b*Si(4*arcsin(c*x)+4*a/b)*sin(4*a/b)-1/16/c^3/b*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b)+1/32/c^3/b*Si(2*a
rcsin(c*x)+2*a/b)*sin(2*a/b)+1/32/c^3/b*Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)-1/32/c^3/b*Si(6*arcsin(c*x)+6*a/b)*
sin(6*a/b)-1/32/c^3/b*Ci(6*arcsin(c*x)+6*a/b)*cos(6*a/b)+1/16*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{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} 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)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate((-c^2*x^2 + 1)^(3/2)*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{{\left (c^{2} x^{4} - x^{2}\right )} \sqrt{-c^{2} x^{2} + 1}}{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)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral(-(c^2*x^4 - x^2)*sqrt(-c^2*x^2 + 1)/(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} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.37152, size = 639, normalized size = 3.1 \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)^(3/2)/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-cos(a/b)^6*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - cos(a/b)^5*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*
x))/(b*c^3) + 3/2*cos(a/b)^4*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/2*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(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/2*cos(a/b)^3*sin
(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) - 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3
) + 1/2*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)^2*cos_integral(2*a/b + 2*arcsin
(c*x))/(b*c^3) - 3/16*cos(a/b)*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) + 1/4*cos(a/b)*sin(a/b)*si
n_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) + 1/16*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^3)
+ 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^3) - 1/16*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^3) - 1/32*
cos_integral(2*a/b + 2*arcsin(c*x))/(b*c^3) + 1/16*log(b*arcsin(c*x) + a)/(b*c^3)