Integrand size = 32, antiderivative size = 455 \[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=-\frac {4 b^2 e \sqrt {d+c d x} \sqrt {e-c e x}}{9 c}-\frac {1}{4} b^2 e x \sqrt {d+c d x} \sqrt {e-c e x}-\frac {2 b^2 e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}{27 c}+\frac {b^2 e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)}{4 c \sqrt {1-c^2 x^2}}-\frac {2 b e x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}-\frac {b c e x^2 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{2 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 e x^3 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{9 \sqrt {1-c^2 x^2}}+\frac {1}{2} e x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2+\frac {e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c}+\frac {e \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^3}{6 b c \sqrt {1-c^2 x^2}} \]
-4/9*b^2*e*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c-1/4*b^2*e*x*(c*d*x+d)^(1/2)* (-c*e*x+e)^(1/2)-2/27*b^2*e*(-c^2*x^2+1)*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/ c+1/2*e*x*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)+1/3*e*(-c^2 *x^2+1)*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c+1/4*b^2*e*a rcsin(c*x)*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c/(-c^2*x^2+1)^(1/2)-2/3*b*e*x *(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)-1/2 *b*c*e*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)+2/9*b*c^2*e*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)+1/6*e*(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 = 2.90 (sec) , antiderivative size = 440, normalized size of antiderivative = 0.97 \[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {36 b^2 e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-108 a^2 \sqrt {d} e^{3/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+18 b e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 \left (6 a+3 b \sqrt {1-c^2 x^2}+b \cos (3 \arcsin (c x))+3 b \sin (2 \arcsin (c x))\right )+e \sqrt {d+c d x} \sqrt {e-c e x} \left (54 a b \cos (2 \arcsin (c x))-4 b^2 \cos (3 \arcsin (c x))-3 \left (4 \left (9 b^2 \sqrt {1-c^2 x^2}-4 a b c x \left (-3+c^2 x^2\right )+3 a^2 \sqrt {1-c^2 x^2} \left (-2-3 c x+2 c^2 x^2\right )\right )+9 b^2 \sin (2 \arcsin (c x))\right )\right )-6 b e \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (-9 b \cos (2 \arcsin (c x))+2 \left (9 b c x-12 a \sqrt {1-c^2 x^2}+12 a c^2 x^2 \sqrt {1-c^2 x^2}-9 a \sin (2 \arcsin (c x))+b \sin (3 \arcsin (c x))\right )\right )}{216 c \sqrt {1-c^2 x^2}} \]
(36*b^2*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 108*a^2*Sqrt[d]* e^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sq rt[d]*Sqrt[e]*(-1 + c^2*x^2))] + 18*b*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Ar cSin[c*x]^2*(6*a + 3*b*Sqrt[1 - c^2*x^2] + b*Cos[3*ArcSin[c*x]] + 3*b*Sin[ 2*ArcSin[c*x]]) + e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(54*a*b*Cos[2*ArcSin[c *x]] - 4*b^2*Cos[3*ArcSin[c*x]] - 3*(4*(9*b^2*Sqrt[1 - c^2*x^2] - 4*a*b*c* x*(-3 + c^2*x^2) + 3*a^2*Sqrt[1 - c^2*x^2]*(-2 - 3*c*x + 2*c^2*x^2)) + 9*b ^2*Sin[2*ArcSin[c*x]])) - 6*b*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x ]*(-9*b*Cos[2*ArcSin[c*x]] + 2*(9*b*c*x - 12*a*Sqrt[1 - c^2*x^2] + 12*a*c^ 2*x^2*Sqrt[1 - c^2*x^2] - 9*a*Sin[2*ArcSin[c*x]] + b*Sin[3*ArcSin[c*x]]))) /(216*c*Sqrt[1 - c^2*x^2])
Time = 0.75 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5178, 27, 5262, 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.
| \(\displaystyle \int \sqrt {c d x+d} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int e (1-c x) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \sqrt {c d x+d} \sqrt {e-c e x} \int (1-c x) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {e \sqrt {c d x+d} \sqrt {e-c e x} \int \left (\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-c x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e \sqrt {c d x+d} \sqrt {e-c e x} \left (\frac {2}{9} b c^2 x^3 (a+b \arcsin (c x))+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c}-\frac {1}{2} b c x^2 (a+b \arcsin (c x))-\frac {2}{3} b x (a+b \arcsin (c x))+\frac {(a+b \arcsin (c x))^3}{6 b c}+\frac {b^2 \arcsin (c x)}{4 c}-\frac {1}{4} b^2 x \sqrt {1-c^2 x^2}-\frac {2 b^2 \left (1-c^2 x^2\right )^{3/2}}{27 c}-\frac {4 b^2 \sqrt {1-c^2 x^2}}{9 c}\right )}{\sqrt {1-c^2 x^2}}\) |
(e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-4*b^2*Sqrt[1 - c^2*x^2])/(9*c) - (b^ 2*x*Sqrt[1 - c^2*x^2])/4 - (2*b^2*(1 - c^2*x^2)^(3/2))/(27*c) + (b^2*ArcSi n[c*x])/(4*c) - (2*b*x*(a + b*ArcSin[c*x]))/3 - (b*c*x^2*(a + b*ArcSin[c*x ]))/2 + (2*b*c^2*x^3*(a + b*ArcSin[c*x]))/9 + (x*Sqrt[1 - c^2*x^2]*(a + b* ArcSin[c*x])^2)/2 + ((1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/(3*c) + (a + b*ArcSin[c*x])^3/(6*b*c)))/Sqrt[1 - c^2*x^2]
3.6.48.3.1 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_.) + 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[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
\[\int \sqrt {c d x +d}\, \left (-c e x +e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
\[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
integral(-(a^2*c*e*x - a^2*e + (b^2*c*e*x - b^2*e)*arcsin(c*x)^2 + 2*(a*b* c*e*x - a*b*e)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e), x)
\[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int \sqrt {d \left (c x + 1\right )} \left (- e \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}\, dx \]
Exception generated. \[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int \sqrt {d+c d x} (e-c e x)^{3/2} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,{\left (e-c\,e\,x\right )}^{3/2} \,d x \]