Integrand size = 30, antiderivative size = 273 \[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\frac {b d x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {b c d x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}-\frac {b c^2 d x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}+\frac {1}{2} d x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))-\frac {d \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c}+\frac {d \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))^2}{4 b c \sqrt {1-c^2 x^2}} \]
1/2*d*x*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)-1/3*d*(-c^2*x^2 +1)*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/c+1/3*b*d*x*(c*d*x+ d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-1/4*b*c*d*x^2*(c*d*x+d)^(1/2) *(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-1/9*b*c^2*d*x^3*(c*d*x+d)^(1/2)*(-c*f *x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+1/4*d*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*( -c*f*x+f)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)
Time = 1.88 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.95 \[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\frac {18 b d \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-36 a d^{3/2} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+d \sqrt {d+c d x} \sqrt {f-c f x} \left (-8 b c x \left (-3+c^2 x^2\right )+12 a \sqrt {1-c^2 x^2} \left (-2+3 c x+2 c^2 x^2\right )+9 b \cos (2 \arcsin (c x))\right )+6 b d \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (-4 \left (1-c^2 x^2\right )^{3/2}+3 \sin (2 \arcsin (c x))\right )}{72 c \sqrt {1-c^2 x^2}} \]
(18*b*d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 36*a*d^(3/2)*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))] + d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-8*b*c*x*(- 3 + c^2*x^2) + 12*a*Sqrt[1 - c^2*x^2]*(-2 + 3*c*x + 2*c^2*x^2) + 9*b*Cos[2 *ArcSin[c*x]]) + 6*b*d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-4*(1 - c^2*x^2)^(3/2) + 3*Sin[2*ArcSin[c*x]]))/(72*c*Sqrt[1 - c^2*x^2])
Time = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.51, 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.
| \(\displaystyle \int (c d x+d)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \int d (c x+1) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \sqrt {c d x+d} \sqrt {f-c f x} \int (c x+1) \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d \sqrt {c d x+d} \sqrt {f-c f x} \int \left (c x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \sqrt {c d x+d} \sqrt {f-c f x} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {\left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c}+\frac {(a+b \arcsin (c x))^2}{4 b c}-\frac {1}{9} b c^2 x^3-\frac {1}{4} b c x^2+\frac {b x}{3}\right )}{\sqrt {1-c^2 x^2}}\) |
(d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*((b*x)/3 - (b*c*x^2)/4 - (b*c^2*x^3)/9 + (x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 - ((1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(3*c) + (a + b*ArcSin[c*x])^2/(4*b*c)))/Sqrt[1 - c^2*x^2]
3.6.5.3.1 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))
\[\int \left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}d x\]
\[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
\[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int \left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )\, dx \]
\[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
b*sqrt(d)*sqrt(f)*integrate((c*d*x + d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arcta n2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/6*(3*sqrt(-c^2*d*f*x^2 + d*f )*d*x + 3*d^2*f*arcsin(c*x)/(sqrt(d*f)*c) - 2*(-c^2*d*f*x^2 + d*f)^(3/2)/( c*f))*a
\[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {3}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}\,\sqrt {f-c\,f\,x} \,d x \]