Integrand size = 32, antiderivative size = 613 \[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=-\frac {8 b^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}}{9 c}-\frac {15}{64} b^2 e^2 x \sqrt {d+c d x} \sqrt {e-c e x}-\frac {1}{32} b^2 c^2 e^2 x^3 \sqrt {d+c d x} \sqrt {e-c e x}-\frac {4 b^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}{27 c}+\frac {15 b^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)}{64 c \sqrt {1-c^2 x^2}}-\frac {4 b e^2 x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}-\frac {3 b c e^2 x^2 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{8 \sqrt {1-c^2 x^2}}+\frac {4 b c^2 e^2 x^3 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 e^2 x^4 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))}{8 \sqrt {1-c^2 x^2}}+\frac {3}{8} e^2 x \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2+\frac {1}{4} c^2 e^2 x^3 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2+\frac {2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{3 c}+\frac {5 e^2 \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^3}{24 b c \sqrt {1-c^2 x^2}} \]
-8/9*b^2*e^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c-15/64*b^2*e^2*x*(c*d*x+d)^ (1/2)*(-c*e*x+e)^(1/2)-1/32*b^2*c^2*e^2*x^3*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/ 2)-4/27*b^2*e^2*(-c^2*x^2+1)*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c+3/8*e^2*x* (a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)+1/4*c^2*e^2*x^3*(a+b* arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)+2/3*e^2*(-c^2*x^2+1)*(a+b* arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c+15/64*b^2*e^2*arcsin(c*x )*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/c/(-c^2*x^2+1)^(1/2)-4/3*b*e^2*x*(a+b*a rcsin(c*x))*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)-3/8*b*c*e^ 2*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 )+4/9*b*c^2*e^2*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)-1/8*b*c^3*e^2*x^4*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*e* x+e)^(1/2)/(-c^2*x^2+1)^(1/2)+5/24*e^2*(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 = 3.43 (sec) , antiderivative size = 555, normalized size of antiderivative = 0.91 \[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\frac {1440 b^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-4320 a^2 \sqrt {d} e^{5/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )-12 b e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (576 b c x-768 a \sqrt {1-c^2 x^2}+768 a c^2 x^2 \sqrt {1-c^2 x^2}-144 b \cos (2 \arcsin (c x))+9 b \cos (4 \arcsin (c x))-288 a \sin (2 \arcsin (c x))+64 b \sin (3 \arcsin (c x))+36 a \sin (4 \arcsin (c x))\right )+72 b e^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 \left (60 a+48 b \sqrt {1-c^2 x^2}+16 b \cos (3 \arcsin (c x))+24 b \sin (2 \arcsin (c x))-3 b \sin (4 \arcsin (c x))\right )+e^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (1728 a b \cos (2 \arcsin (c x))-256 b^2 \cos (3 \arcsin (c x))+3 \left (-3072 a b c x+1024 a b c^3 x^3+1536 a^2 \sqrt {1-c^2 x^2}-2304 b^2 \sqrt {1-c^2 x^2}+864 a^2 c x \sqrt {1-c^2 x^2}-1536 a^2 c^2 x^2 \sqrt {1-c^2 x^2}+576 a^2 c^3 x^3 \sqrt {1-c^2 x^2}-36 a b \cos (4 \arcsin (c x))-288 b^2 \sin (2 \arcsin (c x))+9 b^2 \sin (4 \arcsin (c x))\right )\right )}{6912 c \sqrt {1-c^2 x^2}} \]
(1440*b^2*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 4320*a^2*Sqr t[d]*e^(5/2)*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))] - 12*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c* e*x]*ArcSin[c*x]*(576*b*c*x - 768*a*Sqrt[1 - c^2*x^2] + 768*a*c^2*x^2*Sqrt [1 - c^2*x^2] - 144*b*Cos[2*ArcSin[c*x]] + 9*b*Cos[4*ArcSin[c*x]] - 288*a* Sin[2*ArcSin[c*x]] + 64*b*Sin[3*ArcSin[c*x]] + 36*a*Sin[4*ArcSin[c*x]]) + 72*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^2*(60*a + 48*b*Sqrt[1 - c^2*x^2] + 16*b*Cos[3*ArcSin[c*x]] + 24*b*Sin[2*ArcSin[c*x]] - 3*b*Sin[ 4*ArcSin[c*x]]) + e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(1728*a*b*Cos[2*ArcS in[c*x]] - 256*b^2*Cos[3*ArcSin[c*x]] + 3*(-3072*a*b*c*x + 1024*a*b*c^3*x^ 3 + 1536*a^2*Sqrt[1 - c^2*x^2] - 2304*b^2*Sqrt[1 - c^2*x^2] + 864*a^2*c*x* Sqrt[1 - c^2*x^2] - 1536*a^2*c^2*x^2*Sqrt[1 - c^2*x^2] + 576*a^2*c^3*x^3*S qrt[1 - c^2*x^2] - 36*a*b*Cos[4*ArcSin[c*x]] - 288*b^2*Sin[2*ArcSin[c*x]] + 9*b^2*Sin[4*ArcSin[c*x]])))/(6912*c*Sqrt[1 - c^2*x^2])
Time = 1.14 (sec) , antiderivative size = 334, normalized size of antiderivative = 0.54, 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.
| \(\displaystyle \int \sqrt {c d x+d} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int e^2 (1-c x)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^2 \sqrt {c d x+d} \sqrt {e-c e x} \int (1-c x)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {e^2 \sqrt {c d x+d} \sqrt {e-c e x} \int \left (c^2 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-2 c x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^2 \sqrt {c d x+d} \sqrt {e-c e x} \left (-\frac {1}{8} b c^3 x^4 (a+b \arcsin (c x))+\frac {4}{9} b c^2 x^3 (a+b \arcsin (c x))+\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{3 c}+\frac {1}{4} c^2 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {3}{8} b c x^2 (a+b \arcsin (c x))-\frac {4}{3} b x (a+b \arcsin (c x))+\frac {5 (a+b \arcsin (c x))^3}{24 b c}+\frac {15 b^2 \arcsin (c x)}{64 c}-\frac {15}{64} b^2 x \sqrt {1-c^2 x^2}-\frac {4 b^2 \left (1-c^2 x^2\right )^{3/2}}{27 c}-\frac {8 b^2 \sqrt {1-c^2 x^2}}{9 c}-\frac {1}{32} b^2 c^2 x^3 \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}\) |
(e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-8*b^2*Sqrt[1 - c^2*x^2])/(9*c) - ( 15*b^2*x*Sqrt[1 - c^2*x^2])/64 - (b^2*c^2*x^3*Sqrt[1 - c^2*x^2])/32 - (4*b ^2*(1 - c^2*x^2)^(3/2))/(27*c) + (15*b^2*ArcSin[c*x])/(64*c) - (4*b*x*(a + b*ArcSin[c*x]))/3 - (3*b*c*x^2*(a + b*ArcSin[c*x]))/8 + (4*b*c^2*x^3*(a + b*ArcSin[c*x]))/9 - (b*c^3*x^4*(a + b*ArcSin[c*x]))/8 + (3*x*Sqrt[1 - c^2 *x^2]*(a + b*ArcSin[c*x])^2)/8 + (c^2*x^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[ c*x])^2)/4 + (2*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/(3*c) + (5*(a + b*ArcSin[c*x])^3)/(24*b*c)))/Sqrt[1 - c^2*x^2]
3.6.54.3.1 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))
\[\int \sqrt {c d x +d}\, \left (-c e x +e \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
\[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
integral((a^2*c^2*e^2*x^2 - 2*a^2*c*e^2*x + a^2*e^2 + (b^2*c^2*e^2*x^2 - 2 *b^2*c*e^2*x + b^2*e^2)*arcsin(c*x)^2 + 2*(a*b*c^2*e^2*x^2 - 2*a*b*c*e^2*x + a*b*e^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e), x)
Timed out. \[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Timed out} \]
Exception generated. \[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { \sqrt {c d x + d} {\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int \sqrt {d+c d x} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,{\left (e-c\,e\,x\right )}^{5/2} \,d x \]