Integrand size = 33, antiderivative size = 338 \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {16 b^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}{75 c^2}+\frac {8 b^2 d e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}{225 c^2}+\frac {2 b^2 d e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )^2}{125 c^2}+\frac {2 b d e x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{5 c \sqrt {1-c^2 x^2}}-\frac {4 b c d e x^3 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{15 \sqrt {1-c^2 x^2}}+\frac {2 b c^3 d e x^5 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{25 \sqrt {1-c^2 x^2}}-\frac {d e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{5 c^2} \]
16/75*b^2*d*e*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c^2+8/225*b^2*d*e*(-c^2*x^2 +1)*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c^2+2/125*b^2*d*e*(-c^2*x^2+1)^2*(c*d *x+d)^(1/2)*(-c*e*x+e)^(1/2)/c^2-1/5*d*e*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))^ 2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c^2+2/5*b*d*e*x*(a+b*arcsin(c*x))*(c*d* x+d)^(1/2)*(-c*e*x+e)^(1/2)/c/(-c^2*x^2+1)^(1/2)-4/15*b*c*d*e*x^3*(a+b*arc sin(c*x))*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)+2/25*b*c^3*d *e*x^5*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/ 2)
Time = 2.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.61 \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {d e \sqrt {d+c d x} \sqrt {e-c e x} \left (225 a^2 \left (-1+c^2 x^2\right )^3+30 a b c x \sqrt {1-c^2 x^2} \left (15-10 c^2 x^2+3 c^4 x^4\right )+2 b^2 \left (149-187 c^2 x^2+47 c^4 x^4-9 c^6 x^6\right )+30 b \left (15 a \left (-1+c^2 x^2\right )^3+b c x \sqrt {1-c^2 x^2} \left (15-10 c^2 x^2+3 c^4 x^4\right )\right ) \arcsin (c x)+225 b^2 \left (-1+c^2 x^2\right )^3 \arcsin (c x)^2\right )}{1125 c^2 \left (-1+c^2 x^2\right )} \]
-1/1125*(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(225*a^2*(-1 + c^2*x^2)^3 + 3 0*a*b*c*x*Sqrt[1 - c^2*x^2]*(15 - 10*c^2*x^2 + 3*c^4*x^4) + 2*b^2*(149 - 1 87*c^2*x^2 + 47*c^4*x^4 - 9*c^6*x^6) + 30*b*(15*a*(-1 + c^2*x^2)^3 + b*c*x *Sqrt[1 - c^2*x^2]*(15 - 10*c^2*x^2 + 3*c^4*x^4))*ArcSin[c*x] + 225*b^2*(- 1 + c^2*x^2)^3*ArcSin[c*x]^2))/(c^2*(-1 + c^2*x^2))
Time = 0.83 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.57, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {5238, 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.
| \(\displaystyle \int x (c d x+d)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5238 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \int x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5182 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \int \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))dx}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5154 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-b c \int \frac {x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{15 \sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{15} b c \int \frac {x \left (3 c^4 x^4-10 c^2 x^2+15\right )}{\sqrt {1-c^2 x^2}}dx+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{30} b c \int \frac {3 c^4 x^4-10 c^2 x^2+15}{\sqrt {1-c^2 x^2}}dx^2+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 1140 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{30} b c \int \left (3 \left (1-c^2 x^2\right )^{3/2}+4 \sqrt {1-c^2 x^2}+\frac {8}{\sqrt {1-c^2 x^2}}\right )dx^2+\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (\frac {1}{5} c^4 x^5 (a+b \arcsin (c x))-\frac {2}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))-\frac {1}{30} b c \left (-\frac {6 \left (1-c^2 x^2\right )^{5/2}}{5 c^2}-\frac {8 \left (1-c^2 x^2\right )^{3/2}}{3 c^2}-\frac {16 \sqrt {1-c^2 x^2}}{c^2}\right )\right )}{5 c}-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-1/5*((1 - c^2*x^2)^(5/2)*(a + b*Arc Sin[c*x])^2)/c^2 + (2*b*(-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*Ar cSin[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]
3.6.82.3.1 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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[((-d^2)*(g/e))^In tPart[q]*(d + e*x)^FracPart[q]*((f + g*x)^FracPart[q]/(1 - c^2*x^2)^FracPar t[q]) Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n , x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] & & EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
\[\int x \left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
Time = 0.30 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.89 \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {{\left (9 \, {\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{6} d e x^{6} - {\left (675 \, a^{2} - 94 \, b^{2}\right )} c^{4} d e x^{4} + {\left (675 \, a^{2} - 374 \, b^{2}\right )} c^{2} d e x^{2} - {\left (225 \, a^{2} - 298 \, b^{2}\right )} d e + 225 \, {\left (b^{2} c^{6} d e x^{6} - 3 \, b^{2} c^{4} d e x^{4} + 3 \, b^{2} c^{2} d e x^{2} - b^{2} d e\right )} \arcsin \left (c x\right )^{2} + 450 \, {\left (a b c^{6} d e x^{6} - 3 \, a b c^{4} d e x^{4} + 3 \, a b c^{2} d e x^{2} - a b d e\right )} \arcsin \left (c x\right ) + 30 \, {\left (3 \, a b c^{5} d e x^{5} - 10 \, a b c^{3} d e x^{3} + 15 \, a b c d e x + {\left (3 \, b^{2} c^{5} d e x^{5} - 10 \, b^{2} c^{3} d e x^{3} + 15 \, b^{2} c d e x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{1125 \, {\left (c^{4} x^{2} - c^{2}\right )}} \]
-1/1125*(9*(25*a^2 - 2*b^2)*c^6*d*e*x^6 - (675*a^2 - 94*b^2)*c^4*d*e*x^4 + (675*a^2 - 374*b^2)*c^2*d*e*x^2 - (225*a^2 - 298*b^2)*d*e + 225*(b^2*c^6* d*e*x^6 - 3*b^2*c^4*d*e*x^4 + 3*b^2*c^2*d*e*x^2 - b^2*d*e)*arcsin(c*x)^2 + 450*(a*b*c^6*d*e*x^6 - 3*a*b*c^4*d*e*x^4 + 3*a*b*c^2*d*e*x^2 - a*b*d*e)*a rcsin(c*x) + 30*(3*a*b*c^5*d*e*x^5 - 10*a*b*c^3*d*e*x^3 + 15*a*b*c*d*e*x + (3*b^2*c^5*d*e*x^5 - 10*b^2*c^3*d*e*x^3 + 15*b^2*c*d*e*x)*arcsin(c*x))*sq rt(-c^2*x^2 + 1))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^4*x^2 - c^2)
Timed out. \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Timed out} \]
Exception generated. \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x \,d x } \]
Timed out. \[ \int x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{3/2}\,{\left (e-c\,e\,x\right )}^{3/2} \,d x \]