Integrand size = 32, antiderivative size = 697 \[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=-\frac {8 b^2 e (d+c d x)^{3/2} (e-c e x)^{3/2}}{225 c}-\frac {1}{32} b^2 e x (d+c d x)^{3/2} (e-c e x)^{3/2}-\frac {16 b^2 e (d+c d x)^{3/2} (e-c e x)^{3/2}}{75 c \left (1-c^2 x^2\right )}-\frac {15 b^2 e x (d+c d x)^{3/2} (e-c e x)^{3/2}}{64 \left (1-c^2 x^2\right )}-\frac {2 b^2 e (d+c d x)^{3/2} (e-c e x)^{3/2} \left (1-c^2 x^2\right )}{125 c}+\frac {9 b^2 e (d+c d x)^{3/2} (e-c e x)^{3/2} \arcsin (c x)}{64 c \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b e x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {3 b c e x^2 (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))}{8 \left (1-c^2 x^2\right )^{3/2}}+\frac {4 b c^2 e x^3 (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))}{15 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^4 e x^5 (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {b e (d+c d x)^{3/2} (e-c e x)^{3/2} \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{8 c}+\frac {1}{4} e x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2+\frac {3 e x (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^2}{8 \left (1-c^2 x^2\right )}+\frac {e (d+c d x)^{3/2} (e-c e x)^{3/2} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{5 c}+\frac {e (d+c d x)^{3/2} (e-c e x)^{3/2} (a+b \arcsin (c x))^3}{8 b c \left (1-c^2 x^2\right )^{3/2}} \]
-8/225*b^2*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)/c-1/32*b^2*e*x*(c*d*x+d)^(3/ 2)*(-c*e*x+e)^(3/2)-16/75*b^2*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)/c/(-c^2*x ^2+1)-15/64*b^2*e*x*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)/(-c^2*x^2+1)-2/125*b^ 2*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(-c^2*x^2+1)/c+9/64*b^2*e*(c*d*x+d)^( 3/2)*(-c*e*x+e)^(3/2)*arcsin(c*x)/c/(-c^2*x^2+1)^(3/2)-2/5*b*e*x*(c*d*x+d) ^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)^(3/2)-3/8*b*c*e*x^2 *(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)^(3/2)+4/1 5*b*c^2*e*x^3*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))/(-c^2*x^2 +1)^(3/2)-2/25*b*c^4*e*x^5*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c* x))/(-c^2*x^2+1)^(3/2)+1/4*e*x*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsi n(c*x))^2+3/8*e*x*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^2/(-c ^2*x^2+1)+1/5*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(-c^2*x^2+1)*(a+b*arcsin( c*x))^2/c+1/8*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^3/b/c/( -c^2*x^2+1)^(3/2)+1/8*b*e*(c*d*x+d)^(3/2)*(-c*e*x+e)^(3/2)*(a+b*arcsin(c*x ))*(-c^2*x^2+1)^(1/2)/c
Time = 4.54 (sec) , antiderivative size = 574, normalized size of antiderivative = 0.82 \[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\frac {d e^2 \left (36000 b^2 \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3-108000 a^2 \sqrt {d} \sqrt {e} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+1800 b \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^2 \left (10 b \cos (3 \arcsin (c x))+2 b \cos (5 \arcsin (c x))+5 \left (12 a+4 b \sqrt {1-c^2 x^2}+8 b \sin (2 \arcsin (c x))+b \sin (4 \arcsin (c x))\right )\right )+\sqrt {d+c d x} \sqrt {e-c e x} \left (72000 a b \cos (2 \arcsin (c x))-4000 b^2 \cos (3 \arcsin (c x))+4500 a b \cos (4 \arcsin (c x))-288 b^2 \cos (5 \arcsin (c x))-15 \left (4800 b^2 \sqrt {1-c^2 x^2}+512 a b c x \left (15-10 c^2 x^2+3 c^4 x^4\right )-480 a^2 \sqrt {1-c^2 x^2} \left (8+25 c x-16 c^2 x^2-10 c^3 x^3+8 c^4 x^4\right )+2400 b^2 \sin (2 \arcsin (c x))+75 b^2 \sin (4 \arcsin (c x))\right )\right )+60 b \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (1200 b \cos (2 \arcsin (c x))+75 b \cos (4 \arcsin (c x))+4 \left (-300 b c x+480 a \sqrt {1-c^2 x^2}-960 a c^2 x^2 \sqrt {1-c^2 x^2}+480 a c^4 x^4 \sqrt {1-c^2 x^2}+600 a \sin (2 \arcsin (c x))-50 b \sin (3 \arcsin (c x))+75 a \sin (4 \arcsin (c x))-6 b \sin (5 \arcsin (c x))\right )\right )\right )}{288000 c \sqrt {1-c^2 x^2}} \]
(d*e^2*(36000*b^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 108000*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))] + 1800*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^2*(10*b*Cos[3*ArcSin[c*x]] + 2*b*Cos[5*ArcSin[c*x]] + 5*(12*a + 4*b*Sqrt[1 - c^2*x^2] + 8*b*Sin[2*ArcSin[c*x]] + b*Sin[4*ArcSin [c*x]])) + Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(72000*a*b*Cos[2*ArcSin[c*x]] - 4000*b^2*Cos[3*ArcSin[c*x]] + 4500*a*b*Cos[4*ArcSin[c*x]] - 288*b^2*Cos[5 *ArcSin[c*x]] - 15*(4800*b^2*Sqrt[1 - c^2*x^2] + 512*a*b*c*x*(15 - 10*c^2* x^2 + 3*c^4*x^4) - 480*a^2*Sqrt[1 - c^2*x^2]*(8 + 25*c*x - 16*c^2*x^2 - 10 *c^3*x^3 + 8*c^4*x^4) + 2400*b^2*Sin[2*ArcSin[c*x]] + 75*b^2*Sin[4*ArcSin[ c*x]])) + 60*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]*(1200*b*Cos[2*A rcSin[c*x]] + 75*b*Cos[4*ArcSin[c*x]] + 4*(-300*b*c*x + 480*a*Sqrt[1 - c^2 *x^2] - 960*a*c^2*x^2*Sqrt[1 - c^2*x^2] + 480*a*c^4*x^4*Sqrt[1 - c^2*x^2] + 600*a*Sin[2*ArcSin[c*x]] - 50*b*Sin[3*ArcSin[c*x]] + 75*a*Sin[4*ArcSin[c *x]] - 6*b*Sin[5*ArcSin[c*x]]))))/(288000*c*Sqrt[1 - c^2*x^2])
Time = 0.94 (sec) , antiderivative size = 374, 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.
| \(\displaystyle \int (c d x+d)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {(c d x+d)^{3/2} (e-c e x)^{3/2} \int e (1-c x) \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e (c d x+d)^{3/2} (e-c e x)^{3/2} \int (1-c x) \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {e (c d x+d)^{3/2} (e-c e x)^{3/2} \int \left (\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2-c x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2\right )dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e (c d x+d)^{3/2} (e-c e x)^{3/2} \left (-\frac {2}{25} b c^4 x^5 (a+b \arcsin (c x))+\frac {4}{15} b c^2 x^3 (a+b \arcsin (c x))+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2+\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2+\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{5 c}+\frac {b \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{8 c}-\frac {3}{8} b c x^2 (a+b \arcsin (c x))-\frac {2}{5} b x (a+b \arcsin (c x))+\frac {(a+b \arcsin (c x))^3}{8 b c}+\frac {9 b^2 \arcsin (c x)}{64 c}-\frac {1}{32} b^2 x \left (1-c^2 x^2\right )^{3/2}-\frac {15}{64} b^2 x \sqrt {1-c^2 x^2}-\frac {2 b^2 \left (1-c^2 x^2\right )^{5/2}}{125 c}-\frac {8 b^2 \left (1-c^2 x^2\right )^{3/2}}{225 c}-\frac {16 b^2 \sqrt {1-c^2 x^2}}{75 c}\right )}{\left (1-c^2 x^2\right )^{3/2}}\) |
(e*(d + c*d*x)^(3/2)*(e - c*e*x)^(3/2)*((-16*b^2*Sqrt[1 - c^2*x^2])/(75*c) - (15*b^2*x*Sqrt[1 - c^2*x^2])/64 - (8*b^2*(1 - c^2*x^2)^(3/2))/(225*c) - (b^2*x*(1 - c^2*x^2)^(3/2))/32 - (2*b^2*(1 - c^2*x^2)^(5/2))/(125*c) + (9 *b^2*ArcSin[c*x])/(64*c) - (2*b*x*(a + b*ArcSin[c*x]))/5 - (3*b*c*x^2*(a + b*ArcSin[c*x]))/8 + (4*b*c^2*x^3*(a + b*ArcSin[c*x]))/15 - (2*b*c^4*x^5*( a + b*ArcSin[c*x]))/25 + (b*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(8*c) + ( 3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/8 + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x])^2)/4 + ((1 - c^2*x^2)^(5/2)*(a + b*ArcSin[c*x])^2)/(5*c) + (a + b*ArcSin[c*x])^3/(8*b*c)))/(1 - c^2*x^2)^(3/2)
3.6.53.3.1 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))
\[\int \left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
\[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
integral((a^2*c^3*d*e^2*x^3 - a^2*c^2*d*e^2*x^2 - a^2*c*d*e^2*x + a^2*d*e^ 2 + (b^2*c^3*d*e^2*x^3 - b^2*c^2*d*e^2*x^2 - b^2*c*d*e^2*x + b^2*d*e^2)*ar csin(c*x)^2 + 2*(a*b*c^3*d*e^2*x^3 - a*b*c^2*d*e^2*x^2 - a*b*c*d*e^2*x + a *b*d*e^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e), x)
Timed out. \[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Timed out} \]
Exception generated. \[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (d+c d x)^{3/2} (e-c e x)^{5/2} (a+b \arcsin (c x))^2 \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{3/2}\,{\left (e-c\,e\,x\right )}^{5/2} \,d x \]