\(\int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [109]

Optimal result

Integrand size = 27, antiderivative size = 148 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {2 b x \sqrt {1-c^2 x^2}}{3 c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^4 d}-\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^2 d} \] Output:

2/3*b*x*(-c^2*x^2+1)^(1/2)/c^3/(-c^2*d*x^2+d)^(1/2)+1/9*b*x^3*(-c^2*x^2+1) 
^(1/2)/c/(-c^2*d*x^2+d)^(1/2)-2/3*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/c 
^4/d-1/3*x^2*(-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))/c^2/d
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=\frac {b c x \sqrt {1-c^2 x^2} \left (6+c^2 x^2\right )+3 a \left (-2+c^2 x^2+c^4 x^4\right )+3 b \left (-2+c^2 x^2+c^4 x^4\right ) \arcsin (c x)}{9 c^4 \sqrt {d-c^2 d x^2}} \] Input:

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

Output:

(b*c*x*Sqrt[1 - c^2*x^2]*(6 + c^2*x^2) + 3*a*(-2 + c^2*x^2 + c^4*x^4) + 3* 
b*(-2 + c^2*x^2 + c^4*x^4)*ArcSin[c*x])/(9*c^4*Sqrt[d - c^2*d*x^2])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5210, 15, 5182, 24}

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.

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

\(\Big \downarrow \) 5210

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

\(\Big \downarrow \) 15

\(\displaystyle \frac {2 \int \frac {x (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}}dx}{3 c^2}-\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^2 d}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 5182

\(\displaystyle \frac {2 \left (\frac {b \sqrt {1-c^2 x^2} \int 1dx}{c \sqrt {d-c^2 d x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{c^2 d}\right )}{3 c^2}-\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^2 d}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{3 c^2 d}+\frac {2 \left (\frac {b x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}}-\frac {\sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{c^2 d}\right )}{3 c^2}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 5182

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_ 
.), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 
1))), x] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   I 
nt[(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] && GtQ[n, 0] && NeQ[p, -1]
 

rule 5210

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.08

method	result	size
orering	\(\frac {\left (5 c^{4} x^{4}+12 c^{2} x^{2}-24\right ) \left (a +b \arcsin \left (c x \right )\right )}{9 c^{4} \sqrt {-c^{2} d \,x^{2}+d}}-\frac {\left (c^{2} x^{2}+6\right ) \left (c x -1\right ) \left (c x +1\right ) \left (\frac {3 x^{2} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {-c^{2} d \,x^{2}+d}}+\frac {x^{3} b c}{\sqrt {-c^{2} x^{2}+1}\, \sqrt {-c^{2} d \,x^{2}+d}}+\frac {x^{4} \left (a +b \arcsin \left (c x \right )\right ) c^{2} d}{\left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )}{9 x^{2} c^{4}}\)	\(160\)
default	\(a \left (-\frac {x^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 c^{2} d}-\frac {2 \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, c x +2 c^{2} x^{2}-1\right ) \left (i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \sqrt {-c^{2} x^{2}+1}\, c x +2 c^{2} x^{2}-1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{24 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (4 \arcsin \left (c x \right )\right )}{72 c^{4} d \left (c^{2} x^{2}-1\right )}\right )\)	\(407\)
parts	\(a \left (-\frac {x^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 c^{2} d}-\frac {2 \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-2 i \sqrt {-c^{2} x^{2}+1}\, c x +2 c^{2} x^{2}-1\right ) \left (i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (\arcsin \left (c x \right )+i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \sqrt {-c^{2} x^{2}+1}\, c x +2 c^{2} x^{2}-1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{24 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (4 \arcsin \left (c x \right )\right )}{72 c^{4} d \left (c^{2} x^{2}-1\right )}\right )\)	\(407\)

Input:

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

Output:

1/9*(5*c^4*x^4+12*c^2*x^2-24)/c^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2)-1 
/9/x^2*(c^2*x^2+6)/c^4*(c*x-1)*(c*x+1)*(3*x^2*(a+b*arcsin(c*x))/(-c^2*d*x^ 
2+d)^(1/2)+x^3*b*c/(-c^2*x^2+1)^(1/2)/(-c^2*d*x^2+d)^(1/2)+x^4*(a+b*arcsin 
(c*x))/(-c^2*d*x^2+d)^(3/2)*c^2*d)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {{\left (b c^{3} x^{3} + 6 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 3 \, {\left (a c^{4} x^{4} + a c^{2} x^{2} + {\left (b c^{4} x^{4} + b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{9 \, {\left (c^{6} d x^{2} - c^{4} d\right )}} \] Input:

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

Output:

-1/9*((b*c^3*x^3 + 6*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 3*(a 
*c^4*x^4 + a*c^2*x^2 + (b*c^4*x^4 + b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sq 
rt(-c^2*d*x^2 + d))/(c^6*d*x^2 - c^4*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}} \, dx=-\frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} + \frac {{\left (c^{2} x^{3} + 6 \, x\right )} b}{9 \, c^{3} \sqrt {d}} \] Input:

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

Output:

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

Giac [F(-2)]

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

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

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

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

Output:

( - sqrt( - c**2*x**2 + 1)*a*c**2*x**2 - 2*sqrt( - c**2*x**2 + 1)*a + 3*in 
t((asin(c*x)*x**3)/sqrt( - c**2*x**2 + 1),x)*b*c**4)/(3*sqrt(d)*c**4)