\(\int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx\) [291]

Optimal result

Integrand size = 28, antiderivative size = 183 \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx=-\frac {\operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b c^4}-\frac {\operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b c^4}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b c^4}+\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b c^4}-\frac {\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b c^4} \] Output:

-1/8*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b/c^4-1/16*Ci(3*(a+b*arcsin(c*x))/b) 
*sin(3*a/b)/b/c^4+1/16*Ci(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b/c^4+1/8*cos( 
a/b)*Si((a+b*arcsin(c*x))/b)/b/c^4+1/16*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/ 
b)/b/c^4-1/16*cos(5*a/b)*Si(5*(a+b*arcsin(c*x))/b)/b/c^4
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx=\frac {-2 \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )-\operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+\operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )+2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{16 b c^4} \] Input:

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

Output:

(-2*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - CosIntegral[3*(a/b + ArcSin[ 
c*x])]*Sin[(3*a)/b] + CosIntegral[5*(a/b + ArcSin[c*x])]*Sin[(5*a)/b] + 2* 
Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] + Cos[(3*a)/b]*SinIntegral[3*(a/b 
+ ArcSin[c*x])] - Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b*c 
^4)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.84, 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.

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (-\frac {\sin \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{8} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {1}{16} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {1}{16} \sin \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{16} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{b c^4}\)

Input:

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

Output:

(-1/8*(CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b]) - (CosIntegral[(3*(a + 
 b*ArcSin[c*x]))/b]*Sin[(3*a)/b])/16 + (CosIntegral[(5*(a + b*ArcSin[c*x]) 
)/b]*Sin[(5*a)/b])/16 + (Cos[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/8 + 
(Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/16 - (Cos[(5*a)/b]*S 
inIntegral[(5*(a + b*ArcSin[c*x]))/b])/16)/(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 = 0.18 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75

method	result	size
default	\(-\frac {2 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )-2 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )-\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )+\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 c^{4} b}\)	\(138\)

Input:

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

Output:

-1/16/c^4*(2*Ci(arcsin(c*x)+a/b)*sin(a/b)-2*Si(arcsin(c*x)+a/b)*cos(a/b)+S 
i(5*arcsin(c*x)+5*a/b)*cos(5*a/b)-Ci(5*arcsin(c*x)+5*a/b)*sin(5*a/b)-Si(3* 
arcsin(c*x)+3*a/b)*cos(3*a/b)+Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b))/b
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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