Integrand size = 33, antiderivative size = 225 \[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\frac {4 b^2 \sqrt {d+c d x} \sqrt {e-c e x}}{9 c^2}+\frac {2 b^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}{27 c^2}+\frac {2 b x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}-\frac {2 b c x^3 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{9 \sqrt {1-c^2 x^2}}-\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c^2} \]
4/9*b^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c^2+2/27*b^2*(-c^2*x^2+1)*(c*d*x+ d)^(1/2)*(-c*e*x+e)^(1/2)/c^2-1/3*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2*(c*d*x+ d)^(1/2)*(-c*e*x+e)^(1/2)/c^2+2/3*b*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)-2/9*b*c*x^3*(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 = 1.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.79 \[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (6 a b c x \sqrt {1-c^2 x^2} \left (-3+c^2 x^2\right )+9 a^2 \left (-1+c^2 x^2\right )^2-2 b^2 \left (7-8 c^2 x^2+c^4 x^4\right )+6 b \left (b c x \sqrt {1-c^2 x^2} \left (-3+c^2 x^2\right )+3 a \left (-1+c^2 x^2\right )^2\right ) \arcsin (c x)+9 b^2 \left (-1+c^2 x^2\right )^2 \arcsin (c x)^2\right )}{27 c^2 \left (-1+c^2 x^2\right )} \]
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(6*a*b*c*x*Sqrt[1 - c^2*x^2]*(-3 + c^2*x^ 2) + 9*a^2*(-1 + c^2*x^2)^2 - 2*b^2*(7 - 8*c^2*x^2 + c^4*x^4) + 6*b*(b*c*x *Sqrt[1 - c^2*x^2]*(-3 + c^2*x^2) + 3*a*(-1 + c^2*x^2)^2)*ArcSin[c*x] + 9* b^2*(-1 + c^2*x^2)^2*ArcSin[c*x]^2))/(27*c^2*(-1 + c^2*x^2))
Time = 0.69 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.68, 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, 353, 53, 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.
\(\displaystyle \int x \sqrt {c d x+d} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx\) |
\(\Big \downarrow \) 5238 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5182 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \int \left (1-c^2 x^2\right ) (a+b \arcsin (c x))dx}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5154 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-b c \int \frac {x \left (3-c^2 x^2\right )}{3 \sqrt {1-c^2 x^2}}dx-\frac {1}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{3} b c \int \frac {x \left (3-c^2 x^2\right )}{\sqrt {1-c^2 x^2}}dx-\frac {1}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{6} b c \int \frac {3-c^2 x^2}{\sqrt {1-c^2 x^2}}dx^2-\frac {1}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{6} b c \int \left (\sqrt {1-c^2 x^2}+\frac {2}{\sqrt {1-c^2 x^2}}\right )dx^2-\frac {1}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))\right )}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2 b \left (-\frac {1}{3} c^2 x^3 (a+b \arcsin (c x))+x (a+b \arcsin (c x))-\frac {1}{6} b c \left (-\frac {2 \left (1-c^2 x^2\right )^{3/2}}{3 c^2}-\frac {4 \sqrt {1-c^2 x^2}}{c^2}\right )\right )}{3 c}-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c^2}\right )}{\sqrt {1-c^2 x^2}}\) |
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-1/3*((1 - c^2*x^2)^(3/2)*(a + b*ArcSin[ c*x])^2)/c^2 + (2*b*(-1/6*(b*c*((-4*Sqrt[1 - c^2*x^2])/c^2 - (2*(1 - c^2*x ^2)^(3/2))/(3*c^2))) + x*(a + b*ArcSin[c*x]) - (c^2*x^3*(a + b*ArcSin[c*x] ))/3))/(3*c)))/Sqrt[1 - c^2*x^2]
3.6.77.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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]
\[\int x \sqrt {c d x +d}\, \sqrt {-c e x +e}\, \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
Time = 0.27 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.88 \[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\frac {{\left ({\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{4} x^{4} - 2 \, {\left (9 \, a^{2} - 8 \, b^{2}\right )} c^{2} x^{2} + 9 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arcsin \left (c x\right )^{2} + 9 \, a^{2} - 14 \, b^{2} + 18 \, {\left (a b c^{4} x^{4} - 2 \, a b c^{2} x^{2} + a b\right )} \arcsin \left (c x\right ) + 6 \, {\left (a b c^{3} x^{3} - 3 \, a b c x + {\left (b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}\right )} \sqrt {c d x + d} \sqrt {-c e x + e}}{27 \, {\left (c^{4} x^{2} - c^{2}\right )}} \]
1/27*((9*a^2 - 2*b^2)*c^4*x^4 - 2*(9*a^2 - 8*b^2)*c^2*x^2 + 9*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*arcsin(c*x)^2 + 9*a^2 - 14*b^2 + 18*(a*b*c^4*x^4 - 2*a*b*c^2*x^2 + a*b)*arcsin(c*x) + 6*(a*b*c^3*x^3 - 3*a*b*c*x + (b^2*c^3* x^3 - 3*b^2*c*x)*arcsin(c*x))*sqrt(-c^2*x^2 + 1))*sqrt(c*d*x + d)*sqrt(-c* e*x + e)/(c^4*x^2 - c^2)
\[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int x \sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]
Exception generated. \[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x \,d x } \]
Timed out. \[ \int x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x} \,d x \]