\(\int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx\) [322]

Optimal result

Integrand size = 28, antiderivative size = 121 \[ \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=-\frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b c^4}+\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{4 b c^4}-\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{4 b c^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int \frac {x^3}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=-\frac {3 \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )-\operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )-3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{4 b c^4} \] Input:

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

Output:

-1/4*(3*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - CosIntegral[3*(a/b + Arc 
Sin[c*x])]*Sin[(3*a)/b] - 3*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] + Cos[ 
(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])])/(b*c^4)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5224, 25, 3042, 3793, 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

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((-3*CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b])/4 + (CosIntegral[(3*(a + 
 b*ArcSin[c*x]))/b]*Sin[(3*a)/b])/4 + (3*Cos[a/b]*SinIntegral[(a + b*ArcSi 
n[c*x])/b])/4 - (Cos[(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/4)/( 
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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 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 = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )-3 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )+3 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{4 c^{4} b}\)	\(93\)

Input:

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

Output:

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

Fricas [F]

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

Sympy [F]

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

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

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

Output:

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

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

Output:

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

Reduce [F]

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

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

Output:

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