Integrand size = 27, antiderivative size = 279 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {16 b^2 d \sqrt {d-c^2 d x^2}}{75 c^2}+\frac {8 b^2 d \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{225 c^2}+\frac {2 b^2 d \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2}}{125 c^2}+\frac {2 b d x \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{5 c \sqrt {1-c^2 x^2}}-\frac {4 b c d x^3 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{15 \sqrt {1-c^2 x^2}}+\frac {2 b c^3 d x^5 \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{25 \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d} \]
-1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arcsin(c*x))^2/c^2/d+16/75*b^2*d*(-c^2*d*x^ 2+d)^(1/2)/c^2+8/225*b^2*d*(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/2)/c^2+2/125*b^2 *d*(-c^2*x^2+1)^2*(-c^2*d*x^2+d)^(1/2)/c^2+2/5*b*d*x*(a+b*arcsin(c*x))*(-c ^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-4/15*b*c*d*x^3*(a+b*arcsin(c*x))*(- c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+2/25*b*c^3*d*x^5*(a+b*arcsin(c*x))*( -c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)
Time = 0.14 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.57 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}+\frac {2 b d \sqrt {d-c^2 d x^2} \left (15 a c x \left (15-10 c^2 x^2+3 c^4 x^4\right )+b \sqrt {1-c^2 x^2} \left (149-38 c^2 x^2+9 c^4 x^4\right )+15 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right ) \arcsin (c x)\right )}{1125 c^2 \sqrt {1-c^2 x^2}} \]
-1/5*((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x])^2)/(c^2*d) + (2*b*d*Sqrt[d - c^2*d*x^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]))/(1125*c^2*Sqrt[1 - c^2*x^2])
Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.68, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5182, 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.
| \(\displaystyle \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5182 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \int \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))dx}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
| \(\Big \downarrow \) 5154 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \left (-b c \int \frac {x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{15 \sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \left (-\frac {1}{15} b c \int \frac {x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{\sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \left (-\frac {1}{30} b c \int \frac {3 c^4 x^4-10 c^2 x^2+15}{\sqrt {1-c^2 x^2}}dx^2+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \left (-\frac {1}{30} b c \int \left (3 \left (1-c^2 x^2\right )^{3/2}+4 \sqrt {1-c^2 x^2}+\frac {8}{\sqrt {1-c^2 x^2}}\right )dx^2+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b d \sqrt {d-c^2 d x^2} \left (\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))-\frac {1}{30} b c \left (-\frac {6 \left (1-c^2 x^2\right )^{5/2}}{5 c^2}-\frac {8 \left (1-c^2 x^2\right )^{3/2}}{3 c^2}-\frac {16 \sqrt {1-c^2 x^2}}{c^2}\right )\right )}{5 c \sqrt {1-c^2 x^2}}-\frac {\left (d-c^2 d x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2 d}\) |
-1/5*((d - c^2*d*x^2)^(5/2)*(a + b*ArcSin[c*x])^2)/(c^2*d) + (2*b*d*Sqrt[d - c^2*d*x^2]*(-1/30*(b*c*((-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))) + x*(a + b*ArcSin[c*x]) - (2*c^2*x^3*(a + b*ArcSin[c*x]))/3 + (c^4*x^5*(a + b*ArcSin[c*x]))/5))/( 5*c*Sqrt[1 - c^2*x^2])
3.3.20.3.1 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] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]
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]
Result contains complex when optimal does not.
Time = 0.49 (sec) , antiderivative size = 1151, normalized size of antiderivative = 4.13
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1151\) |
| parts | \(\text {Expression too large to display}\) | \(1151\) |
-1/5*a^2*(-c^2*d*x^2+d)^(5/2)/c^2/d+b^2*(-1/4000*(-d*(c^2*x^2-1))^(1/2)*(1 6*c^6*x^6-28*c^4*x^4-16*I*(-c^2*x^2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2 *x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(10*I*arcsin(c*x)+25*a rcsin(c*x)^2-2)*d/c^2/(c^2*x^2-1)+1/288*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4- 5*c^2*x^2-4*I*c^3*x^3*(-c^2*x^2+1)^(1/2)+3*I*(-c^2*x^2+1)^(1/2)*x*c+1)*(6* I*arcsin(c*x)+9*arcsin(c*x)^2-2)*d/c^2/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^( 1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)^2-2+2*I*arcsin(c*x) )*d/c^2/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+ c^2*x^2-1)*(arcsin(c*x)^2-2-2*I*arcsin(c*x))*d/c^2/(c^2*x^2-1)+1/288*(-d*( c^2*x^2-1))^(1/2)*(4*I*c^3*x^3*(-c^2*x^2+1)^(1/2)+4*c^4*x^4-3*I*(-c^2*x^2+ 1)^(1/2)*x*c-5*c^2*x^2+1)*(-6*I*arcsin(c*x)+9*arcsin(c*x)^2-2)*d/c^2/(c^2* x^2-1)-1/4000*(-d*(c^2*x^2-1))^(1/2)*(16*I*c^5*x^5*(-c^2*x^2+1)^(1/2)+16*c ^6*x^6-20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-28*c^4*x^4+5*I*(-c^2*x^2+1)^(1/2)*x *c+13*c^2*x^2-1)*(-10*I*arcsin(c*x)+25*arcsin(c*x)^2-2)*d/c^2/(c^2*x^2-1)) +2*a*b*(-1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^6*x^6-28*c^4*x^4-16*I*(-c^2*x^ 2+1)^(1/2)*x^5*c^5+13*c^2*x^2+20*I*(-c^2*x^2+1)^(1/2)*x^3*c^3-5*I*(-c^2*x^ 2+1)^(1/2)*x*c-1)*(I+5*arcsin(c*x))*d/c^2/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1) )^(1/2)*(c^2*x^2-I*(-c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)+I)*d/c^2/(c^2*x^ 2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c+c^2*x^2-1)*(arc sin(c*x)-I)*d/c^2/(c^2*x^2-1)+1/96*(-d*(c^2*x^2-1))^(1/2)*(4*I*c^3*x^3*...
Time = 0.27 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.06 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {30 \, {\left (3 \, a b c^{5} d x^{5} - 10 \, a b c^{3} d x^{3} + 15 \, a b c d x + {\left (3 \, b^{2} c^{5} d x^{5} - 10 \, b^{2} c^{3} d x^{3} + 15 \, b^{2} c d x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + {\left (9 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{6} d x^{6} - {\left (675 \, a^{2} - 94 \, b^{2}\right )} c^{4} d x^{4} + {\left (675 \, a^{2} - 374 \, b^{2}\right )} c^{2} d x^{2} + 225 \, {\left (b^{2} c^{6} d x^{6} - 3 \, b^{2} c^{4} d x^{4} + 3 \, b^{2} c^{2} d x^{2} - b^{2} d\right )} \arcsin \left (c x\right )^{2} - {\left (225 \, a^{2} - 298 \, b^{2}\right )} d + 450 \, {\left (a b c^{6} d x^{6} - 3 \, a b c^{4} d x^{4} + 3 \, a b c^{2} d x^{2} - a b d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}}{1125 \, {\left (c^{4} x^{2} - c^{2}\right )}} \]
-1/1125*(30*(3*a*b*c^5*d*x^5 - 10*a*b*c^3*d*x^3 + 15*a*b*c*d*x + (3*b^2*c^ 5*d*x^5 - 10*b^2*c^3*d*x^3 + 15*b^2*c*d*x)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + (9*(25*a^2 - 2*b^2)*c^6*d*x^6 - (675*a^2 - 94*b^2) *c^4*d*x^4 + (675*a^2 - 374*b^2)*c^2*d*x^2 + 225*(b^2*c^6*d*x^6 - 3*b^2*c^ 4*d*x^4 + 3*b^2*c^2*d*x^2 - b^2*d)*arcsin(c*x)^2 - (225*a^2 - 298*b^2)*d + 450*(a*b*c^6*d*x^6 - 3*a*b*c^4*d*x^4 + 3*a*b*c^2*d*x^2 - a*b*d)*arcsin(c* x))*sqrt(-c^2*d*x^2 + d))/(c^4*x^2 - c^2)
\[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int x \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]
Time = 0.30 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.85 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b^{2} \arcsin \left (c x\right )^{2}}{5 \, c^{2} d} + \frac {2}{1125} \, b^{2} {\left (\frac {9 \, \sqrt {-c^{2} x^{2} + 1} c^{2} d^{\frac {5}{2}} x^{4} - 38 \, \sqrt {-c^{2} x^{2} + 1} d^{\frac {5}{2}} x^{2} + \frac {149 \, \sqrt {-c^{2} x^{2} + 1} d^{\frac {5}{2}}}{c^{2}}}{d} + \frac {15 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} - 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} \arcsin \left (c x\right )}{c d}\right )} - \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a b \arcsin \left (c x\right )}{5 \, c^{2} d} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a^{2}}{5 \, c^{2} d} + \frac {2 \, {\left (3 \, c^{4} d^{\frac {5}{2}} x^{5} - 10 \, c^{2} d^{\frac {5}{2}} x^{3} + 15 \, d^{\frac {5}{2}} x\right )} a b}{75 \, c d} \]
-1/5*(-c^2*d*x^2 + d)^(5/2)*b^2*arcsin(c*x)^2/(c^2*d) + 2/1125*b^2*((9*sqr t(-c^2*x^2 + 1)*c^2*d^(5/2)*x^4 - 38*sqrt(-c^2*x^2 + 1)*d^(5/2)*x^2 + 149* sqrt(-c^2*x^2 + 1)*d^(5/2)/c^2)/d + 15*(3*c^4*d^(5/2)*x^5 - 10*c^2*d^(5/2) *x^3 + 15*d^(5/2)*x)*arcsin(c*x)/(c*d)) - 2/5*(-c^2*d*x^2 + d)^(5/2)*a*b*a rcsin(c*x)/(c^2*d) - 1/5*(-c^2*d*x^2 + d)^(5/2)*a^2/(c^2*d) + 2/75*(3*c^4* d^(5/2)*x^5 - 10*c^2*d^(5/2)*x^3 + 15*d^(5/2)*x)*a*b/(c*d)
Exception generated. \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: TypeError} \]
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
Timed out. \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]