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

Optimal. Leaf size=142 \[ \frac{3 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2 c^4}-\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^4}+\frac{3 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2 c^4}-\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.340451, antiderivative size = 138, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4719, 4635, 4406, 3303, 3299, 3302} \[ \frac{3 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 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 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^3}{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \int \frac{x^2}{a+b \sin ^{-1}(c x)} \, dx}{b c}\\ &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{4 (a+b x)}-\frac{\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{\left (3 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{\left (3 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}+\frac{\left (3 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{\left (3 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}\\ &=-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^4}+\frac{3 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^4}-\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.268754, size = 113, normalized size = 0.8 \[ \frac{3 \left (\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-\cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )-\sin \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )\right )}{4 b^2 c^4}-\frac{x^3}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 227, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,{c}^{4} \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( 3\,\arcsin \left ( cx \right ){\it Si} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) b+3\,\arcsin \left ( cx \right ){\it Ci} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) b-3\,\arcsin \left ( cx \right ){\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b-3\,\arcsin \left ( cx \right ){\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b+3\,{\it Si} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) a+3\,{\it Ci} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a-3\,{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a-3\,{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a+3\,xbc-\sin \left ( 3\,\arcsin \left ( cx \right ) \right ) b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

-(x^3 - 3*(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integrate(x^2/(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^{3}}{a^{2} c^{2} x^{2} +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arcsin \left (c x\right )^{2} - a^{2} + 2 \,{\left (a b c^{2} x^{2} - a b\right )} \arcsin \left (c x\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\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**3/(a+b*asin(c*x))**2/(-c**2*x**2+1)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.55248, size = 859, normalized size = 6.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*arcsin(c*x)*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 3*b*arcsin
(c*x)*cos(a/b)^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 3*a*cos(a/b)
^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 3*a*cos(a/b)^2*sin(a/b)*sin_integra
l(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - (c^2*x^2 - 1)*b*c*x/(b^3*c^4*arcsin(c*x) + a*b^2*
c^4) + 9/4*b*arcsin(c*x)*cos(a/b)*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*
b*arcsin(c*x)*cos(a/b)*cos_integral(a/b + arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*b*arcsin(c*x)*s
in(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*b*arcsin(c*x)*sin(a/b)*sin
_integral(a/b + arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - b*c*x/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 9/4
*a*cos(a/b)*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*a*cos(a/b)*cos_integra
l(a/b + arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*a*sin(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b
^3*c^4*arcsin(c*x) + a*b^2*c^4) + 3/4*a*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*
c^4)