\(\int x^2 (d+e x^2) (a+b \arcsin (c x)) \, dx\) [425]

Optimal result

Integrand size = 19, antiderivative size = 120 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b \left (5 c^2 d+3 e\right ) \sqrt {1-c^2 x^2}}{15 c^5}-\frac {b \left (5 c^2 d+6 e\right ) \left (1-c^2 x^2\right )^{3/2}}{45 c^5}+\frac {b e \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+\frac {1}{3} d x^3 (a+b \arcsin (c x))+\frac {1}{5} e x^5 (a+b \arcsin (c x)) \] Output:

1/15*b*(5*c^2*d+3*e)*(-c^2*x^2+1)^(1/2)/c^5-1/45*b*(5*c^2*d+6*e)*(-c^2*x^2 
+1)^(3/2)/c^5+1/25*b*e*(-c^2*x^2+1)^(5/2)/c^5+1/3*d*x^3*(a+b*arcsin(c*x))+ 
1/5*e*x^5*(a+b*arcsin(c*x))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.80 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{225} \left (15 a x^3 \left (5 d+3 e x^2\right )+\frac {b \sqrt {1-c^2 x^2} \left (24 e+2 c^2 \left (25 d+6 e x^2\right )+c^4 \left (25 d x^2+9 e x^4\right )\right )}{c^5}+15 b x^3 \left (5 d+3 e x^2\right ) \arcsin (c x)\right ) \] Input:

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

Output:

(15*a*x^3*(5*d + 3*e*x^2) + (b*Sqrt[1 - c^2*x^2]*(24*e + 2*c^2*(25*d + 6*e 
*x^2) + c^4*(25*d*x^2 + 9*e*x^4)))/c^5 + 15*b*x^3*(5*d + 3*e*x^2)*ArcSin[c 
*x])/225
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {5230, 27, 354, 86, 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.

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

\(\Big \downarrow \) 5230

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/30*(b*c*((-2*(5*c^2*d + 3*e)*Sqrt[1 - c^2*x^2])/c^6 + (2*(5*c^2*d + 6*e 
)*(1 - c^2*x^2)^(3/2))/(3*c^6) - (6*e*(1 - c^2*x^2)^(5/2))/(5*c^6))) + (d* 
x^3*(a + b*ArcSin[c*x]))/3 + (e*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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

rule 2009

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

rule 5230

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.28

method	result	size
parts	\(a \left (\frac {1}{5} e \,x^{5}+\frac {1}{3} x^{3} d \right )+\frac {b \left (\frac {c^{3} \arcsin \left (c x \right ) e \,x^{5}}{5}+\frac {\arcsin \left (c x \right ) c^{3} x^{3} d}{3}-\frac {3 e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )+5 d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{15 c^{2}}\right )}{c^{3}}\)	\(154\)
derivativedivides	\(\frac {\frac {a \left (\frac {1}{3} d \,c^{5} x^{3}+\frac {1}{5} e \,c^{5} x^{5}\right )}{c^{2}}+\frac {b \left (\frac {\arcsin \left (c x \right ) d \,c^{5} x^{3}}{3}+\frac {\arcsin \left (c x \right ) e \,c^{5} x^{5}}{5}-\frac {e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}-\frac {d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c^{3}}\)	\(161\)
default	\(\frac {\frac {a \left (\frac {1}{3} d \,c^{5} x^{3}+\frac {1}{5} e \,c^{5} x^{5}\right )}{c^{2}}+\frac {b \left (\frac {\arcsin \left (c x \right ) d \,c^{5} x^{3}}{3}+\frac {\arcsin \left (c x \right ) e \,c^{5} x^{5}}{5}-\frac {e \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}-\frac {d \,c^{2} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c^{3}}\)	\(161\)
orering	\(\frac {\left (81 x^{8} e^{2} c^{6}+238 x^{6} e \,c^{6} d +125 c^{6} d^{2} x^{4}+12 e^{2} x^{6} c^{4}+106 c^{4} d e \,x^{4}+50 c^{4} d^{2} x^{2}+48 c^{2} e^{2} x^{4}-176 c^{2} d e \,x^{2}-100 c^{2} d^{2}-96 e^{2} x^{2}-48 d e \right ) \left (a +b \arcsin \left (c x \right )\right )}{225 x \left (e \,x^{2}+d \right ) c^{6}}-\frac {\left (9 c^{4} e \,x^{4}+25 c^{4} d \,x^{2}+12 c^{2} e \,x^{2}+50 c^{2} d +24 e \right ) \left (c x -1\right ) \left (c x +1\right ) \left (2 x \left (e \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )+2 e \,x^{3} \left (a +b \arcsin \left (c x \right )\right )+\frac {x^{2} \left (e \,x^{2}+d \right ) b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{225 c^{6} \left (e \,x^{2}+d \right ) x^{2}}\)	\(255\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.89 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {45 \, a c^{5} e x^{5} + 75 \, a c^{5} d x^{3} + 15 \, {\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3}\right )} \arcsin \left (c x\right ) + {\left (9 \, b c^{4} e x^{4} + 50 \, b c^{2} d + {\left (25 \, b c^{4} d + 12 \, b c^{2} e\right )} x^{2} + 24 \, b e\right )} \sqrt {-c^{2} x^{2} + 1}}{225 \, c^{5}} \] Input:

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

Output:

1/225*(45*a*c^5*e*x^5 + 75*a*c^5*d*x^3 + 15*(3*b*c^5*e*x^5 + 5*b*c^5*d*x^3 
)*arcsin(c*x) + (9*b*c^4*e*x^4 + 50*b*c^2*d + (25*b*c^4*d + 12*b*c^2*e)*x^ 
2 + 24*b*e)*sqrt(-c^2*x^2 + 1))/c^5
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.43 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a d x^{3}}{3} + \frac {a e x^{5}}{5} + \frac {b d x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b e x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {b d x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {b e x^{4} \sqrt {- c^{2} x^{2} + 1}}{25 c} + \frac {2 b d \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + \frac {4 b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{3}} + \frac {8 b e \sqrt {- c^{2} x^{2} + 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\a \left (\frac {d x^{3}}{3} + \frac {e x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*d*x**3/3 + a*e*x**5/5 + b*d*x**3*asin(c*x)/3 + b*e*x**5*asin( 
c*x)/5 + b*d*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + b*e*x**4*sqrt(-c**2*x**2 + 
1)/(25*c) + 2*b*d*sqrt(-c**2*x**2 + 1)/(9*c**3) + 4*b*e*x**2*sqrt(-c**2*x* 
*2 + 1)/(75*c**3) + 8*b*e*sqrt(-c**2*x**2 + 1)/(75*c**5), Ne(c, 0)), (a*(d 
*x**3/3 + e*x**5/5), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.18 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d + \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b e \] Input:

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

Output:

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/9*(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*d + 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*e
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (104) = 208\).

Time = 0.13 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.75 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {b d x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b e x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b e x \arcsin \left (c x\right )}{5 \, c^{4}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d}{9 \, c^{3}} + \frac {b e x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d}{3 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b e}{25 \, c^{5}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e}{15 \, c^{5}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e}{5 \, c^{5}} \] Input:

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

Output:

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/3*(c^2*x^2 - 1)*b*d*x*arcsin(c*x)/c^2 + 1/3* 
b*d*x*arcsin(c*x)/c^2 + 1/5*(c^2*x^2 - 1)^2*b*e*x*arcsin(c*x)/c^4 + 2/5*(c 
^2*x^2 - 1)*b*e*x*arcsin(c*x)/c^4 - 1/9*(-c^2*x^2 + 1)^(3/2)*b*d/c^3 + 1/5 
*b*e*x*arcsin(c*x)/c^4 + 1/3*sqrt(-c^2*x^2 + 1)*b*d/c^3 + 1/25*(c^2*x^2 - 
1)^2*sqrt(-c^2*x^2 + 1)*b*e/c^5 - 2/15*(-c^2*x^2 + 1)^(3/2)*b*e/c^5 + 1/5* 
sqrt(-c^2*x^2 + 1)*b*e/c^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.25 \[ \int x^2 \left (d+e x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {75 \mathit {asin} \left (c x \right ) b \,c^{5} d \,x^{3}+45 \mathit {asin} \left (c x \right ) b \,c^{5} e \,x^{5}+25 \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} d \,x^{2}+9 \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} e \,x^{4}+50 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} d +12 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} e \,x^{2}+24 \sqrt {-c^{2} x^{2}+1}\, b e +75 a \,c^{5} d \,x^{3}+45 a \,c^{5} e \,x^{5}}{225 c^{5}} \] Input:

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

Output:

(75*asin(c*x)*b*c**5*d*x**3 + 45*asin(c*x)*b*c**5*e*x**5 + 25*sqrt( - c**2 
*x**2 + 1)*b*c**4*d*x**2 + 9*sqrt( - c**2*x**2 + 1)*b*c**4*e*x**4 + 50*sqr 
t( - c**2*x**2 + 1)*b*c**2*d + 12*sqrt( - c**2*x**2 + 1)*b*c**2*e*x**2 + 2 
4*sqrt( - c**2*x**2 + 1)*b*e + 75*a*c**5*d*x**3 + 45*a*c**5*e*x**5)/(225*c 
**5)