\(\int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx\) [63]

Optimal result

Integrand size = 27, antiderivative size = 183 \[ \int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {2 b x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {b x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}-\frac {b c x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4 d}+\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4 d^2} \] Output:

2/15*b*x*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*x^2+1)^(1/2)+1/45*b*x^3*(-c^2*d*x^ 
2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/25*b*c*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^ 
2+1)^(1/2)-1/3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))/c^4/d+1/5*(-c^2*d*x^ 
2+d)^(5/2)*(a+b*arcsin(c*x))/c^4/d^2
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.73 \[ \int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {\sqrt {d-c^2 d x^2} \left (15 a \sqrt {1-c^2 x^2} \left (-2-c^2 x^2+3 c^4 x^4\right )+b \left (30 c x+5 c^3 x^3-9 c^5 x^5\right )+15 b \sqrt {1-c^2 x^2} \left (-2-c^2 x^2+3 c^4 x^4\right ) \arcsin (c x)\right )}{225 c^4 \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(Sqrt[d - c^2*d*x^2]*(15*a*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 + 3*c^4*x^4) + 
b*(30*c*x + 5*c^3*x^3 - 9*c^5*x^5) + 15*b*Sqrt[1 - c^2*x^2]*(-2 - c^2*x^2 
+ 3*c^4*x^4)*ArcSin[c*x]))/(225*c^4*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5194, 27, 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.

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

\(\Big \downarrow \) 5194

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {-3 c^4 x^4+c^2 x^2+2}{15 c^4}dx}{\sqrt {1-c^2 x^2}}+\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \sqrt {d-c^2 d x^2} \int \left (-3 c^4 x^4+c^2 x^2+2\right )dx}{15 c^3 \sqrt {1-c^2 x^2}}+\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c^4 d^2}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c^4 d}+\frac {b \left (-\frac {3}{5} c^4 x^5+\frac {c^2 x^3}{3}+2 x\right ) \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}\)

Input:

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

Output:

(b*Sqrt[d - c^2*d*x^2]*(2*x + (c^2*x^3)/3 - (3*c^4*x^5)/5))/(15*c^3*Sqrt[1 
 - c^2*x^2]) - ((d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*c^4*d) + ((d 
 - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c^4*d^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 5194

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_) 
, x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin 
[c*x])   u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]]   Int[Sim 
plifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] 
&& EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 
1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [A] (verified)

Time = 0.67 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.98

method	result	size
orering	\(\frac {\left (81 c^{6} x^{6}-107 c^{4} x^{4}-120 c^{2} x^{2}+120\right ) \sqrt {-c^{2} d \,x^{2}+d}\, \left (a +b \arcsin \left (c x \right )\right )}{225 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\left (9 c^{4} x^{4}-5 c^{2} x^{2}-30\right ) \left (3 x^{2} \sqrt {-c^{2} d \,x^{2}+d}\, \left (a +b \arcsin \left (c x \right )\right )-\frac {x^{4} \left (a +b \arcsin \left (c x \right )\right ) c^{2} d}{\sqrt {-c^{2} d \,x^{2}+d}}+\frac {b c \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}{\sqrt {-c^{2} x^{2}+1}}\right )}{225 x^{2} c^{4}}\)	\(179\)
default	\(a \left (-\frac {x^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{5 c^{2} d}-\frac {2 \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{15 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (16 c^{6} x^{6}-28 c^{4} x^{4}-16 i \sqrt {-c^{2} x^{2}+1}\, x^{5} c^{5}+13 c^{2} x^{2}+20 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}-5 i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (i+5 \arcsin \left (c x \right )\right )}{800 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}+4 c^{4} x^{4}-3 i \sqrt {-c^{2} x^{2}+1}\, c x -5 c^{2} x^{2}+1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{288 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (17 i+15 \arcsin \left (c x \right )\right ) \cos \left (4 \arcsin \left (c x \right )\right )}{3600 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (2 i+15 \arcsin \left (c x \right )\right ) \sin \left (4 \arcsin \left (c x \right )\right )}{900 c^{4} \left (c^{2} x^{2}-1\right )}\right )\)	\(544\)
parts	\(a \left (-\frac {x^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{5 c^{2} d}-\frac {2 \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{15 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (16 c^{6} x^{6}-28 c^{4} x^{4}-16 i \sqrt {-c^{2} x^{2}+1}\, x^{5} c^{5}+13 c^{2} x^{2}+20 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}-5 i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (i+5 \arcsin \left (c x \right )\right )}{800 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{16 c^{4} \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 i \sqrt {-c^{2} x^{2}+1}\, x^{3} c^{3}+4 c^{4} x^{4}-3 i \sqrt {-c^{2} x^{2}+1}\, c x -5 c^{2} x^{2}+1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{288 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (17 i+15 \arcsin \left (c x \right )\right ) \cos \left (4 \arcsin \left (c x \right )\right )}{3600 c^{4} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i c^{2} x^{2}-c x \sqrt {-c^{2} x^{2}+1}-i\right ) \left (2 i+15 \arcsin \left (c x \right )\right ) \sin \left (4 \arcsin \left (c x \right )\right )}{900 c^{4} \left (c^{2} x^{2}-1\right )}\right )\)	\(544\)

Input:

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

Output:

1/225*(81*c^6*x^6-107*c^4*x^4-120*c^2*x^2+120)/c^4/(c^2*x^2-1)*(-c^2*d*x^2 
+d)^(1/2)*(a+b*arcsin(c*x))-1/225/x^2*(9*c^4*x^4-5*c^2*x^2-30)/c^4*(3*x^2* 
(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))-x^4/(-c^2*d*x^2+d)^(1/2)*(a+b*arcsi 
n(c*x))*c^2*d+b*c*x^3*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {{\left (9 \, b c^{5} x^{5} - 5 \, b c^{3} x^{3} - 30 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 15 \, {\left (3 \, a c^{6} x^{6} - 4 \, a c^{4} x^{4} - a c^{2} x^{2} + {\left (3 \, b c^{6} x^{6} - 4 \, b c^{4} x^{4} - b c^{2} x^{2} + 2 \, b\right )} \arcsin \left (c x\right ) + 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{225 \, {\left (c^{6} x^{2} - c^{4}\right )}} \] Input:

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

Output:

1/225*((9*b*c^5*x^5 - 5*b*c^3*x^3 - 30*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c 
^2*x^2 + 1) + 15*(3*a*c^6*x^6 - 4*a*c^4*x^4 - a*c^2*x^2 + (3*b*c^6*x^6 - 4 
*b*c^4*x^4 - b*c^2*x^2 + 2*b)*arcsin(c*x) + 2*a)*sqrt(-c^2*d*x^2 + d))/(c^ 
6*x^2 - c^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75 \[ \int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=-\frac {1}{15} \, b {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} \arcsin \left (c x\right ) - \frac {1}{15} \, a {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} - \frac {{\left (9 \, c^{4} \sqrt {d} x^{5} - 5 \, c^{2} \sqrt {d} x^{3} - 30 \, \sqrt {d} x\right )} b}{225 \, c^{3}} \] Input:

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

Output:

-1/15*b*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(3/2)/( 
c^4*d))*arcsin(c*x) - 1/15*a*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d) + 2*(-c 
^2*d*x^2 + d)^(3/2)/(c^4*d)) - 1/225*(9*c^4*sqrt(d)*x^5 - 5*c^2*sqrt(d)*x^ 
3 - 30*sqrt(d)*x)*b/c^3
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x)) \, dx=\frac {\sqrt {d}\, \left (3 \sqrt {-c^{2} x^{2}+1}\, a \,c^{4} x^{4}-\sqrt {-c^{2} x^{2}+1}\, a \,c^{2} x^{2}-2 \sqrt {-c^{2} x^{2}+1}\, a +15 \left (\int \sqrt {-c^{2} x^{2}+1}\, \mathit {asin} \left (c x \right ) x^{3}d x \right ) b \,c^{4}\right )}{15 c^{4}} \] Input:

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

Output:

(sqrt(d)*(3*sqrt( - c**2*x**2 + 1)*a*c**4*x**4 - sqrt( - c**2*x**2 + 1)*a* 
c**2*x**2 - 2*sqrt( - c**2*x**2 + 1)*a + 15*int(sqrt( - c**2*x**2 + 1)*asi 
n(c*x)*x**3,x)*b*c**4))/(15*c**4)