\(\int (d-c^2 d x^2)^2 (a+b \arcsin (c x)) \, dx\) [2]

Optimal result

Integrand size = 22, antiderivative size = 131 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {8 b d^2 \sqrt {1-c^2 x^2}}{15 c}+\frac {4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{45 c}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{25 c}+d^2 x (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+\frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x)) \] Output:

8/15*b*d^2*(-c^2*x^2+1)^(1/2)/c+4/45*b*d^2*(-c^2*x^2+1)^(3/2)/c+1/25*b*d^2 
*(-c^2*x^2+1)^(5/2)/c+d^2*x*(a+b*arcsin(c*x))-2/3*c^2*d^2*x^3*(a+b*arcsin( 
c*x))+1/5*c^4*d^2*x^5*(a+b*arcsin(c*x))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.73 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {d^2 \left (15 a c x \left (15-10 c^2 x^2+3 c^4 x^4\right )+b \sqrt {1-c^2 x^2} \left (149-38 c^2 x^2+9 c^4 x^4\right )+15 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right ) \arcsin (c x)\right )}{225 c} \] Input:

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

Output:

(d^2*(15*a*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) + b*Sqrt[1 - c^2*x^2]*(149 - 
38*c^2*x^2 + 9*c^4*x^4) + 15*b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4)*ArcSin[c* 
x]))/(225*c)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {5154, 27, 1576, 1140, 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.

\(\displaystyle \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx\)

\(\Big \downarrow \) 5154

\(\displaystyle -b c \int \frac {d^2 x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{15 \sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+d^2 x (a+b \arcsin (c x))\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{15} b c d^2 \int \frac {x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{\sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+d^2 x (a+b \arcsin (c x))\)

\(\Big \downarrow \) 1576

\(\displaystyle -\frac {1}{30} b c d^2 \int \frac {3 c^4 x^4-10 c^2 x^2+15}{\sqrt {1-c^2 x^2}}dx^2+\frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+d^2 x (a+b \arcsin (c x))\)

\(\Big \downarrow \) 1140

\(\displaystyle -\frac {1}{30} b c d^2 \int \left (3 \left (1-c^2 x^2\right )^{3/2}+4 \sqrt {1-c^2 x^2}+\frac {8}{\sqrt {1-c^2 x^2}}\right )dx^2+\frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+d^2 x (a+b \arcsin (c x))\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} c^4 d^2 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 d^2 x^3 (a+b \arcsin (c x))+d^2 x (a+b \arcsin (c x))-\frac {1}{30} b c d^2 \left (-\frac {6 \left (1-c^2 x^2\right )^{5/2}}{5 c^2}-\frac {8 \left (1-c^2 x^2\right )^{3/2}}{3 c^2}-\frac {16 \sqrt {1-c^2 x^2}}{c^2}\right )\)

Input:

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

Output:

-1/30*(b*c*d^2*((-16*Sqrt[1 - c^2*x^2])/c^2 - (8*(1 - c^2*x^2)^(3/2))/(3*c 
^2) - (6*(1 - c^2*x^2)^(5/2))/(5*c^2))) + d^2*x*(a + b*ArcSin[c*x]) - (2*c 
^2*d^2*x^3*(a + b*ArcSin[c*x]))/3 + (c^4*d^2*x^5*(a + b*ArcSin[c*x]))/5
 

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 1140

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x 
_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; 
FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
 

rule 1576

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( 
p_.), x_Symbol] :> Simp[1/2   Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] 
, x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
 

rule 2009

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

rule 5154

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbo 
l] :> With[{u = IntHide[(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x])   u, x 
] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; Fr 
eeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.91

method	result	size
parts	\(d^{2} a \left (\frac {1}{5} c^{4} x^{5}-\frac {2}{3} c^{2} x^{3}+x \right )+\frac {d^{2} b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}-\frac {2 c^{3} x^{3} \arcsin \left (c x \right )}{3}+c x \arcsin \left (c x \right )+\frac {149 \sqrt {-c^{2} x^{2}+1}}{225}-\frac {38 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}\right )}{c}\)	\(119\)
derivativedivides	\(\frac {d^{2} a \left (\frac {1}{5} c^{5} x^{5}-\frac {2}{3} c^{3} x^{3}+c x \right )+d^{2} b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}-\frac {2 c^{3} x^{3} \arcsin \left (c x \right )}{3}+c x \arcsin \left (c x \right )+\frac {149 \sqrt {-c^{2} x^{2}+1}}{225}-\frac {38 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}\right )}{c}\)	\(122\)
default	\(\frac {d^{2} a \left (\frac {1}{5} c^{5} x^{5}-\frac {2}{3} c^{3} x^{3}+c x \right )+d^{2} b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}-\frac {2 c^{3} x^{3} \arcsin \left (c x \right )}{3}+c x \arcsin \left (c x \right )+\frac {149 \sqrt {-c^{2} x^{2}+1}}{225}-\frac {38 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{225}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}\right )}{c}\)	\(122\)
orering	\(\frac {x \left (81 c^{4} x^{4}-302 c^{2} x^{2}+821\right ) \left (-c^{2} d \,x^{2}+d \right )^{2} \left (a +b \arcsin \left (c x \right )\right )}{225 \left (c x -1\right ) \left (c x +1\right ) \left (c^{2} x^{2}-1\right )}-\frac {\left (9 c^{4} x^{4}-38 c^{2} x^{2}+149\right ) \left (-4 \left (-c^{2} d \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right ) c^{2} d x +\frac {\left (-c^{2} d \,x^{2}+d \right )^{2} b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{225 c^{2} \left (c x -1\right ) \left (c x +1\right )}\)	\(161\)

Input:

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

Output:

d^2*a*(1/5*c^4*x^5-2/3*c^2*x^3+x)+d^2*b/c*(1/5*c^5*x^5*arcsin(c*x)-2/3*c^3 
*x^3*arcsin(c*x)+c*x*arcsin(c*x)+149/225*(-c^2*x^2+1)^(1/2)-38/225*c^2*x^2 
*(-c^2*x^2+1)^(1/2)+1/25*c^4*x^4*(-c^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {45 \, a c^{5} d^{2} x^{5} - 150 \, a c^{3} d^{2} x^{3} + 225 \, a c d^{2} x + 15 \, {\left (3 \, b c^{5} d^{2} x^{5} - 10 \, b c^{3} d^{2} x^{3} + 15 \, b c d^{2} x\right )} \arcsin \left (c x\right ) + {\left (9 \, b c^{4} d^{2} x^{4} - 38 \, b c^{2} d^{2} x^{2} + 149 \, b d^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{225 \, c} \] Input:

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

Output:

1/225*(45*a*c^5*d^2*x^5 - 150*a*c^3*d^2*x^3 + 225*a*c*d^2*x + 15*(3*b*c^5* 
d^2*x^5 - 10*b*c^3*d^2*x^3 + 15*b*c*d^2*x)*arcsin(c*x) + (9*b*c^4*d^2*x^4 
- 38*b*c^2*d^2*x^2 + 149*b*d^2)*sqrt(-c^2*x^2 + 1))/c
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.26 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a c^{4} d^{2} x^{5}}{5} - \frac {2 a c^{2} d^{2} x^{3}}{3} + a d^{2} x + \frac {b c^{4} d^{2} x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {b c^{3} d^{2} x^{4} \sqrt {- c^{2} x^{2} + 1}}{25} - \frac {2 b c^{2} d^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} - \frac {38 b c d^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{225} + b d^{2} x \operatorname {asin}{\left (c x \right )} + \frac {149 b d^{2} \sqrt {- c^{2} x^{2} + 1}}{225 c} & \text {for}\: c \neq 0 \\a d^{2} x & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*c**4*d**2*x**5/5 - 2*a*c**2*d**2*x**3/3 + a*d**2*x + b*c**4*d 
**2*x**5*asin(c*x)/5 + b*c**3*d**2*x**4*sqrt(-c**2*x**2 + 1)/25 - 2*b*c**2 
*d**2*x**3*asin(c*x)/3 - 38*b*c*d**2*x**2*sqrt(-c**2*x**2 + 1)/225 + b*d** 
2*x*asin(c*x) + 149*b*d**2*sqrt(-c**2*x**2 + 1)/(225*c), Ne(c, 0)), (a*d** 
2*x, True))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.50 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a c^{4} d^{2} x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{4} d^{2} - \frac {2}{3} \, a c^{2} d^{2} x^{3} - \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{2} d^{2} + a d^{2} x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d^{2}}{c} \] Input:

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

Output:

1/5*a*c^4*d^2*x^5 + 1/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/c 
^2 + 4*sqrt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(-c^2*x^2 + 1)/c^6)*c)*b*c^4*d^2 
 - 2/3*a*c^2*d^2*x^3 - 2/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/ 
c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*c^2*d^2 + a*d^2*x + (c*x*arcsin(c*x) + 
sqrt(-c^2*x^2 + 1))*b*d^2/c
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.21 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a c^{4} d^{2} x^{5} - \frac {2}{3} \, a c^{2} d^{2} x^{3} + \frac {1}{5} \, {\left (c^{2} x^{2} - 1\right )}^{2} b d^{2} x \arcsin \left (c x\right ) - \frac {4}{15} \, {\left (c^{2} x^{2} - 1\right )} b d^{2} x \arcsin \left (c x\right ) + \frac {8}{15} \, b d^{2} x \arcsin \left (c x\right ) + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d^{2}}{25 \, c} + a d^{2} x + \frac {4 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d^{2}}{45 \, c} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1} b d^{2}}{15 \, c} \] Input:

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

Output:

1/5*a*c^4*d^2*x^5 - 2/3*a*c^2*d^2*x^3 + 1/5*(c^2*x^2 - 1)^2*b*d^2*x*arcsin 
(c*x) - 4/15*(c^2*x^2 - 1)*b*d^2*x*arcsin(c*x) + 8/15*b*d^2*x*arcsin(c*x) 
+ 1/25*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2/c + a*d^2*x + 4/45*(-c^2*x 
^2 + 1)^(3/2)*b*d^2/c + 8/15*sqrt(-c^2*x^2 + 1)*b*d^2/c
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {d^{2} \left (45 \mathit {asin} \left (c x \right ) b \,c^{5} x^{5}-150 \mathit {asin} \left (c x \right ) b \,c^{3} x^{3}+225 \mathit {asin} \left (c x \right ) b c x +9 \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} x^{4}-38 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} x^{2}+149 \sqrt {-c^{2} x^{2}+1}\, b +45 a \,c^{5} x^{5}-150 a \,c^{3} x^{3}+225 a c x \right )}{225 c} \] Input:

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

Output:

(d**2*(45*asin(c*x)*b*c**5*x**5 - 150*asin(c*x)*b*c**3*x**3 + 225*asin(c*x 
)*b*c*x + 9*sqrt( - c**2*x**2 + 1)*b*c**4*x**4 - 38*sqrt( - c**2*x**2 + 1) 
*b*c**2*x**2 + 149*sqrt( - c**2*x**2 + 1)*b + 45*a*c**5*x**5 - 150*a*c**3* 
x**3 + 225*a*c*x))/(225*c)