\(\int \frac {x^3 (1-c^2 x^2)^{3/2}}{a+b \arcsin (c x)} \, dx\) [298]

Optimal result

Integrand size = 28, antiderivative size = 245 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arcsin (c x)} \, dx=-\frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{64 b c^4}-\frac {3 \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{64 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{64 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {7 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {7 a}{b}\right )}{64 b c^4}+\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{64 b c^4}+\frac {3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{64 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{64 b c^4}-\frac {\cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )}{64 b c^4} \] Output:

-3/64*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b/c^4-3/64*Ci(3*(a+b*arcsin(c*x))/b 
)*sin(3*a/b)/b/c^4+1/64*Ci(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b/c^4+1/64*Ci 
(7*(a+b*arcsin(c*x))/b)*sin(7*a/b)/b/c^4+3/64*cos(a/b)*Si((a+b*arcsin(c*x) 
)/b)/b/c^4+3/64*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b/c^4-1/64*cos(5*a/b) 
*Si(5*(a+b*arcsin(c*x))/b)/b/c^4-1/64*cos(7*a/b)*Si(7*(a+b*arcsin(c*x))/b) 
/b/c^4
 

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arcsin (c x)} \, dx=\frac {-3 \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )-3 \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )+\operatorname {CosIntegral}\left (7 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {7 a}{b}\right )+3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+3 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {7 a}{b}\right ) \text {Si}\left (7 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{64 b c^4} \] Input:

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

Output:

(-3*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - 3*CosIntegral[3*(a/b + ArcSi 
n[c*x])]*Sin[(3*a)/b] + CosIntegral[5*(a/b + ArcSin[c*x])]*Sin[(5*a)/b] + 
CosIntegral[7*(a/b + ArcSin[c*x])]*Sin[(7*a)/b] + 3*Cos[a/b]*SinIntegral[a 
/b + ArcSin[c*x]] + 3*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])] - Co 
s[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])] - Cos[(7*a)/b]*SinIntegral[7 
*(a/b + ArcSin[c*x])])/(64*b*c^4)
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {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.

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (-\frac {\sin \left (\frac {7 a}{b}-\frac {7 (a+b \arcsin (c x))}{b}\right )}{64 (a+b \arcsin (c x))}-\frac {\sin \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right )}{64 (a+b \arcsin (c x))}+\frac {3 \sin \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{64 (a+b \arcsin (c x))}+\frac {3 \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{64 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {3}{64} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {3}{64} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {1}{64} \sin \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )+\frac {1}{64} \sin \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )+\frac {3}{64} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {3}{64} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {1}{64} \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )-\frac {1}{64} \cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )}{b c^4}\)

Input:

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

Output:

((-3*CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b])/64 - (3*CosIntegral[(3*( 
a + b*ArcSin[c*x]))/b]*Sin[(3*a)/b])/64 + (CosIntegral[(5*(a + b*ArcSin[c* 
x]))/b]*Sin[(5*a)/b])/64 + (CosIntegral[(7*(a + b*ArcSin[c*x]))/b]*Sin[(7* 
a)/b])/64 + (3*Cos[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/64 + (3*Cos[(3 
*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/64 - (Cos[(5*a)/b]*SinInteg 
ral[(5*(a + b*ArcSin[c*x]))/b])/64 - (Cos[(7*a)/b]*SinIntegral[(7*(a + b*A 
rcSin[c*x]))/b])/64)/(b*c^4)
 

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 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 = 0.19 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {3 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-3 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+3 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-3 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )-\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )+\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-\operatorname {Si}\left (7 \arcsin \left (c x \right )+\frac {7 a}{b}\right ) \cos \left (\frac {7 a}{b}\right )+\operatorname {Ci}\left (7 \arcsin \left (c x \right )+\frac {7 a}{b}\right ) \sin \left (\frac {7 a}{b}\right )}{64 c^{4} b}\)	\(184\)

Input:

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

Output:

1/64/c^4*(3*Si(3*arcsin(c*x)+3*a/b)*cos(3*a/b)-3*Ci(3*arcsin(c*x)+3*a/b)*s 
in(3*a/b)+3*Si(arcsin(c*x)+a/b)*cos(a/b)-3*Ci(arcsin(c*x)+a/b)*sin(a/b)-Si 
(5*arcsin(c*x)+5*a/b)*cos(5*a/b)+Ci(5*arcsin(c*x)+5*a/b)*sin(5*a/b)-Si(7*a 
rcsin(c*x)+7*a/b)*cos(7*a/b)+Ci(7*arcsin(c*x)+7*a/b)*sin(7*a/b))/b
 

Fricas [F]

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

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

Output:

integral(-(c^2*x^5 - x^3)*sqrt(-c^2*x^2 + 1)/(b*arcsin(c*x) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 614 vs. \(2 (229) = 458\).

Time = 0.18 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.51 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arcsin (c x)} \, dx =\text {Too large to display} \] Input:

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

Output:

cos(a/b)^6*cos_integral(7*a/b + 7*arcsin(c*x))*sin(a/b)/(b*c^4) - cos(a/b) 
^7*sin_integral(7*a/b + 7*arcsin(c*x))/(b*c^4) - 5/4*cos(a/b)^4*cos_integr 
al(7*a/b + 7*arcsin(c*x))*sin(a/b)/(b*c^4) + 1/4*cos(a/b)^4*cos_integral(5 
*a/b + 5*arcsin(c*x))*sin(a/b)/(b*c^4) + 7/4*cos(a/b)^5*sin_integral(7*a/b 
 + 7*arcsin(c*x))/(b*c^4) - 1/4*cos(a/b)^5*sin_integral(5*a/b + 5*arcsin(c 
*x))/(b*c^4) + 3/8*cos(a/b)^2*cos_integral(7*a/b + 7*arcsin(c*x))*sin(a/b) 
/(b*c^4) - 3/16*cos(a/b)^2*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b 
*c^4) - 3/16*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^ 
4) - 7/8*cos(a/b)^3*sin_integral(7*a/b + 7*arcsin(c*x))/(b*c^4) + 5/16*cos 
(a/b)^3*sin_integral(5*a/b + 5*arcsin(c*x))/(b*c^4) + 3/16*cos(a/b)^3*sin_ 
integral(3*a/b + 3*arcsin(c*x))/(b*c^4) - 1/64*cos_integral(7*a/b + 7*arcs 
in(c*x))*sin(a/b)/(b*c^4) + 1/64*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a 
/b)/(b*c^4) + 3/64*cos_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^4) - 
3/64*cos_integral(a/b + arcsin(c*x))*sin(a/b)/(b*c^4) + 7/64*cos(a/b)*sin_ 
integral(7*a/b + 7*arcsin(c*x))/(b*c^4) - 5/64*cos(a/b)*sin_integral(5*a/b 
 + 5*arcsin(c*x))/(b*c^4) - 9/64*cos(a/b)*sin_integral(3*a/b + 3*arcsin(c* 
x))/(b*c^4) + 3/64*cos(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 \left (1-c^2 x^2\right )^{3/2}}{a+b \arcsin (c x)} \, dx=-\left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{5}}{\mathit {asin} \left (c x \right ) b +a}d x \right ) c^{2}+\int \frac {\sqrt {-c^{2} x^{2}+1}\, x^{3}}{\mathit {asin} \left (c x \right ) b +a}d x \] Input:

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

Output:

 - int((sqrt( - c**2*x**2 + 1)*x**5)/(asin(c*x)*b + a),x)*c**2 + int((sqrt 
( - c**2*x**2 + 1)*x**3)/(asin(c*x)*b + a),x)