Integrand size = 32, antiderivative size = 222 \[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=-\frac {1}{4} b^2 x \sqrt {d+c d x} \sqrt {e-c e x}+\frac {b^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)}{4 c \sqrt {1-c^2 x^2}}-\frac {b c x^2 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2+\frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^3}{6 b c \sqrt {1-c^2 x^2}} \]
-1/4*b^2*x*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)+1/2*x*(c*d*x+d)^(1/2)*(-c*e*x+ e)^(1/2)*(a+b*arcsin(c*x))^2+1/4*b^2*arcsin(c*x)*(c*d*x+d)^(1/2)*(-c*e*x+e )^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/2*b*c*x^2*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2) *(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)+1/6*(a+b*arcsin(c*x))^3*(c*d*x+d)^(1/ 2)*(-c*e*x+e)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)
Time = 0.44 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.30 \[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\frac {4 b^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-12 a^2 \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+6 b \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) (b \cos (2 \arcsin (c x))+2 a \sin (2 \arcsin (c x)))+6 b \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 (2 a+b \sin (2 \arcsin (c x)))+3 \sqrt {d+c d x} \sqrt {e-c e x} \left (4 a^2 c x \sqrt {1-c^2 x^2}+2 a b \cos (2 \arcsin (c x))-b^2 \sin (2 \arcsin (c x))\right )}{24 c \sqrt {1-c^2 x^2}} \]
(4*b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 12*a^2*Sqrt[d]*Sqrt [e]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d ]*Sqrt[e]*(-1 + c^2*x^2))] + 6*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c* x]*(b*Cos[2*ArcSin[c*x]] + 2*a*Sin[2*ArcSin[c*x]]) + 6*b*Sqrt[d + c*d*x]*S qrt[e - c*e*x]*ArcSin[c*x]^2*(2*a + b*Sin[2*ArcSin[c*x]]) + 3*Sqrt[d + c*d *x]*Sqrt[e - c*e*x]*(4*a^2*c*x*Sqrt[1 - c^2*x^2] + 2*a*b*Cos[2*ArcSin[c*x] ] - b^2*Sin[2*ArcSin[c*x]]))/(24*c*Sqrt[1 - c^2*x^2])
Time = 0.69 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5178, 5156, 5138, 262, 223, 5152}
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.
\(\displaystyle \int \sqrt {c d x+d} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5156 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {1}{2} \int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx-b c \int x (a+b \arcsin (c x))dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (-b c \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \int \frac {x^2}{\sqrt {1-c^2 x^2}}dx\right )+\frac {1}{2} \int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (-b c \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )+\frac {1}{2} \int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {1}{2} \int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-b c \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-b c \left (\frac {1}{2} x^2 (a+b \arcsin (c x))-\frac {1}{2} b c \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )\right )+\frac {(a+b \arcsin (c x))^3}{6 b c}\right )}{\sqrt {1-c^2 x^2}}\) |
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) ^2)/2 + (a + b*ArcSin[c*x])^3/(6*b*c) - b*c*((x^2*(a + b*ArcSin[c*x]))/2 - (b*c*(-1/2*(x*Sqrt[1 - c^2*x^2])/c^2 + ArcSin[c*x]/(2*c^3)))/2)))/Sqrt[1 - c^2*x^2]
3.6.78.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/2), x] + (Simp[(1/2 )*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int[(a + b*ArcSin[c*x])^n/Sqrt[ 1 - c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2 ]] Int[x*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x ] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 2)^q) Int[(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, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
\[\int \sqrt {c d x +d}\, \sqrt {-c e x +e}\, \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
\[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
\[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int \sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]
Exception generated. \[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x} \,d x \]