Integrand size = 18, antiderivative size = 150 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt {1-c^2 x^2}}{15 c^5}-\frac {2 b e \left (5 c^2 d+3 e\right ) \left (1-c^2 x^2\right )^{3/2}}{45 c^5}+\frac {b e^2 \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x)) \]
-2/45*b*e*(5*c^2*d+3*e)*(-c^2*x^2+1)^(3/2)/c^5+1/25*b*e^2*(-c^2*x^2+1)^(5/ 2)/c^5+d^2*x*(a+b*arcsin(c*x))+2/3*d*e*x^3*(a+b*arcsin(c*x))+1/5*e^2*x^5*( a+b*arcsin(c*x))+1/15*b*(15*c^4*d^2+10*c^2*d*e+3*e^2)*(-c^2*x^2+1)^(1/2)/c ^5
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{225} \left (15 a x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )+\frac {b \sqrt {1-c^2 x^2} \left (24 e^2+4 c^2 e \left (25 d+3 e x^2\right )+c^4 \left (225 d^2+50 d e x^2+9 e^2 x^4\right )\right )}{c^5}+15 b x \left (15 d^2+10 d e x^2+3 e^2 x^4\right ) \arcsin (c x)\right ) \]
(15*a*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4) + (b*Sqrt[1 - c^2*x^2]*(24*e^2 + 4*c^2*e*(25*d + 3*e*x^2) + c^4*(225*d^2 + 50*d*e*x^2 + 9*e^2*x^4)))/c^5 + 15*b*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4)*ArcSin[c*x])/225
Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5170, 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.
| \(\displaystyle \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5170 |
| \(\displaystyle -b c \int \frac {x \left (3 e^2 x^4+10 d e x^2+15 d^2\right )}{15 \sqrt {1-c^2 x^2}}dx+d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x))\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{15} b c \int \frac {x \left (3 e^2 x^4+10 d e x^2+15 d^2\right )}{\sqrt {1-c^2 x^2}}dx+d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x))\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle -\frac {1}{30} b c \int \frac {3 e^2 x^4+10 d e x^2+15 d^2}{\sqrt {1-c^2 x^2}}dx^2+d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x))\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle -\frac {1}{30} b c \int \left (\frac {3 \left (1-c^2 x^2\right )^{3/2} e^2}{c^4}-\frac {2 \left (5 d c^2+3 e\right ) \sqrt {1-c^2 x^2} e}{c^4}+\frac {15 d^2 c^4+10 d e c^2+3 e^2}{c^4 \sqrt {1-c^2 x^2}}\right )dx^2+d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle d^2 x (a+b \arcsin (c x))+\frac {2}{3} d e x^3 (a+b \arcsin (c x))+\frac {1}{5} e^2 x^5 (a+b \arcsin (c x))-\frac {1}{30} b c \left (\frac {4 e \left (1-c^2 x^2\right )^{3/2} \left (5 c^2 d+3 e\right )}{3 c^6}-\frac {6 e^2 \left (1-c^2 x^2\right )^{5/2}}{5 c^6}-\frac {2 \sqrt {1-c^2 x^2} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right )}{c^6}\right )\) |
-1/30*(b*c*((-2*(15*c^4*d^2 + 10*c^2*d*e + 3*e^2)*Sqrt[1 - c^2*x^2])/c^6 + (4*e*(5*c^2*d + 3*e)*(1 - c^2*x^2)^(3/2))/(3*c^6) - (6*e^2*(1 - c^2*x^2)^ (5/2))/(5*c^6))) + d^2*x*(a + b*ArcSin[c*x]) + (2*d*e*x^3*(a + b*ArcSin[c* x]))/3 + (e^2*x^5*(a + b*ArcSin[c*x]))/5
3.7.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2 , 0])
Time = 0.19 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.29
| method | result | size |
| parts | \(a \left (\frac {1}{5} e^{2} x^{5}+\frac {2}{3} d e \,x^{3}+d^{2} x \right )+\frac {b \left (\frac {c \arcsin \left (c x \right ) e^{2} x^{5}}{5}+\frac {2 c \arcsin \left (c x \right ) e d \,x^{3}}{3}+\arcsin \left (c x \right ) d^{2} c x -\frac {3 e^{2} \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )-15 d^{2} c^{4} \sqrt {-c^{2} x^{2}+1}+10 d \,c^{2} e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{15 c^{4}}\right )}{c}\) | \(194\) |
| derivativedivides | \(\frac {\frac {a \left (d^{2} c^{5} x +\frac {2}{3} d \,c^{5} e \,x^{3}+\frac {1}{5} e^{2} c^{5} x^{5}\right )}{c^{4}}+\frac {b \left (\arcsin \left (c x \right ) d^{2} c^{5} x +\frac {2 \arcsin \left (c x \right ) d \,c^{5} e \,x^{3}}{3}+\frac {\arcsin \left (c x \right ) e^{2} c^{5} x^{5}}{5}-\frac {e^{2} \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}+d^{2} c^{4} \sqrt {-c^{2} x^{2}+1}-\frac {2 d \,c^{2} e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{4}}}{c}\) | \(209\) |
| default | \(\frac {\frac {a \left (d^{2} c^{5} x +\frac {2}{3} d \,c^{5} e \,x^{3}+\frac {1}{5} e^{2} c^{5} x^{5}\right )}{c^{4}}+\frac {b \left (\arcsin \left (c x \right ) d^{2} c^{5} x +\frac {2 \arcsin \left (c x \right ) d \,c^{5} e \,x^{3}}{3}+\frac {\arcsin \left (c x \right ) e^{2} c^{5} x^{5}}{5}-\frac {e^{2} \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}+d^{2} c^{4} \sqrt {-c^{2} x^{2}+1}-\frac {2 d \,c^{2} e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{4}}}{c}\) | \(209\) |
a*(1/5*e^2*x^5+2/3*d*e*x^3+d^2*x)+b/c*(1/5*c*arcsin(c*x)*e^2*x^5+2/3*c*arc sin(c*x)*e*d*x^3+arcsin(c*x)*d^2*c*x-1/15/c^4*(3*e^2*(-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))-15*d^ 2*c^4*(-c^2*x^2+1)^(1/2)+10*d*c^2*e*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*( -c^2*x^2+1)^(1/2))))
Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {45 \, a c^{5} e^{2} x^{5} + 150 \, a c^{5} d e x^{3} + 225 \, a c^{5} d^{2} x + 15 \, {\left (3 \, b c^{5} e^{2} x^{5} + 10 \, b c^{5} d e x^{3} + 15 \, b c^{5} d^{2} x\right )} \arcsin \left (c x\right ) + {\left (9 \, b c^{4} e^{2} x^{4} + 225 \, b c^{4} d^{2} + 100 \, b c^{2} d e + 24 \, b e^{2} + 2 \, {\left (25 \, b c^{4} d e + 6 \, b c^{2} e^{2}\right )} x^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{225 \, c^{5}} \]
1/225*(45*a*c^5*e^2*x^5 + 150*a*c^5*d*e*x^3 + 225*a*c^5*d^2*x + 15*(3*b*c^ 5*e^2*x^5 + 10*b*c^5*d*e*x^3 + 15*b*c^5*d^2*x)*arcsin(c*x) + (9*b*c^4*e^2* x^4 + 225*b*c^4*d^2 + 100*b*c^2*d*e + 24*b*e^2 + 2*(25*b*c^4*d*e + 6*b*c^2 *e^2)*x^2)*sqrt(-c^2*x^2 + 1))/c^5
Time = 0.41 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.60 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\begin {cases} a d^{2} x + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{5}}{5} + b d^{2} x \operatorname {asin}{\left (c x \right )} + \frac {2 b d e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b e^{2} x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {b d^{2} \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {2 b d e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {b e^{2} x^{4} \sqrt {- c^{2} x^{2} + 1}}{25 c} + \frac {4 b d e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + \frac {4 b e^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{3}} + \frac {8 b e^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{2} x + \frac {2 d e x^{3}}{3} + \frac {e^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((a*d**2*x + 2*a*d*e*x**3/3 + a*e**2*x**5/5 + b*d**2*x*asin(c*x) + 2*b*d*e*x**3*asin(c*x)/3 + b*e**2*x**5*asin(c*x)/5 + b*d**2*sqrt(-c**2*x **2 + 1)/c + 2*b*d*e*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + b*e**2*x**4*sqrt(-c **2*x**2 + 1)/(25*c) + 4*b*d*e*sqrt(-c**2*x**2 + 1)/(9*c**3) + 4*b*e**2*x* *2*sqrt(-c**2*x**2 + 1)/(75*c**3) + 8*b*e**2*sqrt(-c**2*x**2 + 1)/(75*c**5 ), Ne(c, 0)), (a*(d**2*x + 2*d*e*x**3/3 + e**2*x**5/5), True))
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a e^{2} x^{5} + \frac {2}{3} \, a d e x^{3} + \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d e + \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b e^{2} + a d^{2} x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d^{2}}{c} \]
1/5*a*e^2*x^5 + 2/3*a*d*e*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*d*e + 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^2 + a*d^2*x + (c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*b*d^2/c
Time = 0.29 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.77 \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a e^{2} x^{5} + \frac {2}{3} \, a d e x^{3} + b d^{2} x \arcsin \left (c x\right ) + a d^{2} x + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b d e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {2 \, b d e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b e^{2} x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d^{2}}{c} + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b e^{2} x \arcsin \left (c x\right )}{5 \, c^{4}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d e}{9 \, c^{3}} + \frac {b e^{2} x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b d e}{3 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b e^{2}}{25 \, c^{5}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e^{2}}{15 \, c^{5}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e^{2}}{5 \, c^{5}} \]
1/5*a*e^2*x^5 + 2/3*a*d*e*x^3 + b*d^2*x*arcsin(c*x) + a*d^2*x + 2/3*(c^2*x ^2 - 1)*b*d*e*x*arcsin(c*x)/c^2 + 2/3*b*d*e*x*arcsin(c*x)/c^2 + 1/5*(c^2*x ^2 - 1)^2*b*e^2*x*arcsin(c*x)/c^4 + sqrt(-c^2*x^2 + 1)*b*d^2/c + 2/5*(c^2* x^2 - 1)*b*e^2*x*arcsin(c*x)/c^4 - 2/9*(-c^2*x^2 + 1)^(3/2)*b*d*e/c^3 + 1/ 5*b*e^2*x*arcsin(c*x)/c^4 + 2/3*sqrt(-c^2*x^2 + 1)*b*d*e/c^3 + 1/25*(c^2*x ^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*e^2/c^5 - 2/15*(-c^2*x^2 + 1)^(3/2)*b*e^2/c ^5 + 1/5*sqrt(-c^2*x^2 + 1)*b*e^2/c^5
Timed out. \[ \int \left (d+e x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2 \,d x \]