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

Optimal. Leaf size=150 \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c}+\frac{\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{b^2 c}-\frac{\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b^2 c}-\frac{\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.272936, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4659, 4723, 4406, 3303, 3299, 3302} \[ \frac{\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c}+\frac{\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c}-\frac{\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{b^2 c}-\frac{\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c x)\right )}{2 b^2 c}-\frac{\left (1-c^2 x^2\right )^2}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4659

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(Sqrt[1 - c^2*x^2]*
(d + e*x^2)^p*(a + b*ArcSin[c*x])^(n + 1))/(b*c*(n + 1)), x] + Dist[(c*(2*p + 1)*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(b*(n + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n + 1), x],
x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

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

Mathematica [A]  time = 0.623514, size = 122, normalized size = 0.81 \[ \frac{-\frac{2 b \left (c^2 x^2-1\right )^2}{a+b \sin ^{-1}(c x)}+2 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-2 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{2 b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.048, size = 250, normalized size = 1.7 \begin{align*} -{\frac{1}{8\,c \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+8\,\arcsin \left ( cx \right ){\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) b-8\,\arcsin \left ( cx \right ){\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\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+8\,{\it Si} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-8\,{\it Ci} \left ( 2\,\arcsin \left ( cx \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a+\cos \left ( 4\,\arcsin \left ( cx \right ) \right ) b+4\,\cos \left ( 2\,\arcsin \left ( cx \right ) \right ) b+3\,b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{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((-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.6512, size = 1008, normalized size = 6.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(3/2)/(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*arcsin(c*x) + a*b^2*c) - 4*b*ar
csin(c*x)*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + 4*a*cos(a/b)^3*cos_in
tegral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 4*a*cos(a/b)^4*sin_integral(4*a/b + 4*a
rcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a
/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 2*b*arcsin(c*x)*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3
*c*arcsin(c*x) + a*b^2*c) + 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x)
+ a*b^2*c) - 2*b*arcsin(c*x)*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*
a*cos(a/b)*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 2*a*cos(a/b)*cos_integ
ral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) + 4*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcs
in(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*a*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x)
+ a*b^2*c) - (c^2*x^2 - 1)^2*b/(b^3*c*arcsin(c*x) + a*b^2*c) - 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*arcsi
n(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + b*arcsin(c*x)*sin_integral(2*a/b + 2*arcsin(c*x))/(b^3*c*arcsin(c*x) +
a*b^2*c) - 1/2*a*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) + a*sin_integral(2*a/b + 2
*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c)