\(\int \frac {(1-c^2 x^2)^{3/2}}{(a+b \arcsin (c x))^2} \, dx\) [361]

Optimal result

Integrand size = 25, antiderivative size = 150 \[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}+\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c}+\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b^2 c}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.81 \[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\frac {-\frac {2 b \left (-1+c^2 x^2\right )^2}{a+b \arcsin (c x)}+2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )-2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{2 b^2 c} \] Input:

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

Output:

((-2*b*(-1 + c^2*x^2)^2)/(a + b*ArcSin[c*x]) + 2*CosIntegral[2*(a/b + ArcS 
in[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]*SinInteg 
ral[4*(a/b + ArcSin[c*x])])/(2*b^2*c)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5166, 5224, 25, 4906, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx\)

\(\Big \downarrow \) 5166

\(\displaystyle -\frac {4 c \int \frac {x \left (1-c^2 x^2\right )}{a+b \arcsin (c x)}dx}{b}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 5224

\(\displaystyle -\frac {4 \int -\frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 \int \frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle \frac {4 \int \left (\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{4 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )\right )}{b^2 c}-\frac {\left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

Input:

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

Output:

-((1 - c^2*x^2)^2/(b*c*(a + b*ArcSin[c*x]))) - (4*(-1/4*(CosIntegral[(2*(a 
 + b*ArcSin[c*x]))/b]*Sin[(2*a)/b]) - (CosIntegral[(4*(a + b*ArcSin[c*x])) 
/b]*Sin[(4*a)/b])/8 + (Cos[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b] 
)/4 + (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/8))/(b^2*c)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

rule 5166

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] + Simp[c*((2*p + 1)/(b*(n + 1)))*Simp[(d + e*x^2)^p/(1 
 - c^2*x^2)^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 5224

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x 
^2)^p]   Subst[Int[x^n*Sin[-a/b + x/b]^m*Cos[-a/b + x/b]^(2*p + 1), x], x, 
a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] 
 && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {4 \arcsin \left (c x \right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) b +8 \arcsin \left (c x \right ) \operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b -8 \arcsin \left (c x \right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) a +8 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a -8 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b +4 \cos \left (2 \arcsin \left (c x \right )\right ) b +3 b}{8 c \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(250\)

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

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]

\[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int { \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}} \,d x } \] Input:

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

Output:

-(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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 747 vs. \(2 (145) = 290\).

Time = 0.23 (sec) , antiderivative size = 747, normalized size of antiderivative = 4.98 \[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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*arcsin(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_inte 
gral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c*arcsin(c*x) + a*b^2*c) - 4*a*c 
os(a/b)^4*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)*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_integral(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*arcsin(c*x))/(b^3*c*arcsin(c*x) + a*b^2*c) - 2*a*c 
os(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)*si 
n_integral(4*a/b + 4*arcsin(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)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {{\left (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {\sqrt {-c^{2} x^{2}+1}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x -\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{2}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \right ) c^{2} \] Input:

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

Output:

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