Integrand size = 30, antiderivative size = 414 \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {b d x (d+c d x)^{3/2} (f-c f x)^{3/2}}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {5 b c d x^2 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^2 d x^3 (d+c d x)^{3/2} (f-c f x)^{3/2}}{15 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 d x^4 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^4 d x^5 (d+c d x)^{3/2} (f-c f x)^{3/2}}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} d x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))+\frac {3 d x (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))}{8 \left (1-c^2 x^2\right )}-\frac {d (d+c d x)^{3/2} (f-c f x)^{3/2} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{5 c}+\frac {3 d (d+c d x)^{3/2} (f-c f x)^{3/2} (a+b \arcsin (c x))^2}{16 b c \left (1-c^2 x^2\right )^{3/2}} \]
1/5*b*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-5/16*b*c*d*x ^2*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-2/15*b*c^2*d*x^3*(c *d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/16*b*c^3*d*x^4*(c*d*x+ d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/25*b*c^4*d*x^5*(c*d*x+d)^(3 /2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/4*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f) ^(3/2)*(a+b*arcsin(c*x))+3/8*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arc sin(c*x))/(-c^2*x^2+1)-1/5*d*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(-c^2*x^2+1) *(a+b*arcsin(c*x))/c+3/16*d*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c *x))^2/b/c/(-c^2*x^2+1)^(3/2)
Time = 2.74 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.74 \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\frac {d^2 f \left (1800 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-3600 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\sqrt {d+c d x} \sqrt {f-c f x} \left (128 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right )-240 a \sqrt {1-c^2 x^2} \left (8-25 c x-16 c^2 x^2+10 c^3 x^3+8 c^4 x^4\right )+1200 b \cos (2 \arcsin (c x))+75 b \cos (4 \arcsin (c x))\right )-60 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (32 \left (1-c^2 x^2\right )^{5/2}-40 \sin (2 \arcsin (c x))-5 \sin (4 \arcsin (c x))\right )\right )}{9600 c \sqrt {1-c^2 x^2}} \]
(d^2*f*(1800*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 3600*a*Sqrt [d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]) /(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(128* b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) - 240*a*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) + 1200*b*Cos[2*ArcSin[c*x]] + 75*b* Cos[4*ArcSin[c*x]]) - 60*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(32 *(1 - c^2*x^2)^(5/2) - 40*Sin[2*ArcSin[c*x]] - 5*Sin[4*ArcSin[c*x]])))/(96 00*c*Sqrt[1 - c^2*x^2])
Time = 0.61 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.46, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
| \(\displaystyle \int (c d x+d)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {(c d x+d)^{3/2} (f-c f x)^{3/2} \int d (c x+1) \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d (c d x+d)^{3/2} (f-c f x)^{3/2} \int (c x+1) \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d (c d x+d)^{3/2} (f-c f x)^{3/2} \int \left (c x (a+b \arcsin (c x)) \left (1-c^2 x^2\right )^{3/2}+(a+b \arcsin (c x)) \left (1-c^2 x^2\right )^{3/2}\right )dx}{\left (1-c^2 x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d (c d x+d)^{3/2} (f-c f x)^{3/2} \left (\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))+\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {\left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))}{5 c}+\frac {3 (a+b \arcsin (c x))^2}{16 b c}+\frac {1}{25} b c^4 x^5+\frac {1}{16} b c^3 x^4-\frac {2}{15} b c^2 x^3-\frac {5}{16} b c x^2+\frac {b x}{5}\right )}{\left (1-c^2 x^2\right )^{3/2}}\) |
(d*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*((b*x)/5 - (5*b*c*x^2)/16 - (2*b*c^ 2*x^3)/15 + (b*c^3*x^4)/16 + (b*c^4*x^5)/25 + (3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/8 + (x*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/4 - ((1 - c^2*x^2)^(5/2)*(a + b*ArcSin[c*x]))/(5*c) + (3*(a + b*ArcSin[c*x])^2)/(16* b*c)))/(1 - c^2*x^2)^(3/2)
3.6.10.3.1 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))
\[\int \left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )d x\]
\[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
integral(-(a*c^3*d^2*f*x^3 + a*c^2*d^2*f*x^2 - a*c*d^2*f*x - a*d^2*f + (b* c^3*d^2*f*x^3 + b*c^2*d^2*f*x^2 - b*c*d^2*f*x - b*d^2*f)*arcsin(c*x))*sqrt (c*d*x + d)*sqrt(-c*f*x + f), x)
Timed out. \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]
\[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
b*sqrt(d)*sqrt(f)*integrate(-(c^3*d^2*f*x^3 + c^2*d^2*f*x^2 - c*d^2*f*x - d^2*f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/40*(15*sqrt(-c^2*d*f*x^2 + d*f)*d^2*f*x + 15*d^3*f^2*arcsin(c *x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*d*x - 8*(-c^2*d*f*x^2 + d*f)^(5/2)/(c*f))*a
\[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int (d+c d x)^{5/2} (f-c f x)^{3/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}\,{\left (f-c\,f\,x\right )}^{3/2} \,d x \]