Integrand size = 30, antiderivative size = 242 \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=-\frac {2 b f^2 x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b c f^2 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {2 f^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {f^2 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 f^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]
2*f^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-1/ 2*f^2*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-2* b*f^2*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/4*b*c*f^2*x^ 2*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+3/4*f^2*(a+b*arcsin( c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
Time = 5.21 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.98 \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\frac {-4 b f (-4+c x) \sqrt {d+c d x} \sqrt {f-c f x} \sqrt {1-c^2 x^2} \arcsin (c x)+6 b f \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-12 a \sqrt {d} f^{3/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )-f \sqrt {d+c d x} \sqrt {f-c f x} \left (16 b c x+4 a (-4+c x) \sqrt {1-c^2 x^2}+b \cos (2 \arcsin (c x))\right )}{8 c d \sqrt {1-c^2 x^2}} \]
(-4*b*f*(-4 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[1 - c^2*x^2]*ArcSi n[c*x] + 6*b*f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 12*a*Sqrt[d ]*f^(3/2)*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))] - f*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(16*b *c*x + 4*a*(-4 + c*x)*Sqrt[1 - c^2*x^2] + b*Cos[2*ArcSin[c*x]]))/(8*c*d*Sq rt[1 - c^2*x^2])
Time = 0.64 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.52, 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.
| \(\displaystyle \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {c d x+d}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {f^2 (1-c x)^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^2 \sqrt {1-c^2 x^2} \int \frac {(1-c x)^2 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {f^2 \sqrt {1-c^2 x^2} \int \left (\frac {c^2 (a+b \arcsin (c x)) x^2}{\sqrt {1-c^2 x^2}}-\frac {2 c (a+b \arcsin (c x)) x}{\sqrt {1-c^2 x^2}}+\frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}\right )dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \sqrt {1-c^2 x^2} \left (-\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {3 (a+b \arcsin (c x))^2}{4 b c}+\frac {1}{4} b c x^2-2 b x\right )}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
(f^2*Sqrt[1 - c^2*x^2]*(-2*b*x + (b*c*x^2)/4 + (2*Sqrt[1 - c^2*x^2]*(a + b *ArcSin[c*x]))/c - (x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 + (3*(a + b *ArcSin[c*x])^2)/(4*b*c)))/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x])
3.6.13.3.1 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))
\[\int \frac {\left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {c d x +d}}d x\]
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
integral(-(a*c*f*x - a*f + (b*c*f*x - b*f)*arcsin(c*x))*sqrt(-c*f*x + f)/s qrt(c*d*x + d), x)
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int \frac {\left (- f \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {d \left (c x + 1\right )}}\, dx \]
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
-1/2*(sqrt(-c^2*d*f*x^2 + d*f)*f*x/d - 3*f^2*arcsin(c*x)/(sqrt(d*f)*c) - 4 *sqrt(-c^2*d*f*x^2 + d*f)*f/(c*d))*a - b*sqrt(f)*integrate((c*f*x - f)*sqr t(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/sqrt(c*x + 1), x)/s qrt(d)
\[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
Timed out. \[ \int \frac {(f-c f x)^{3/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{3/2}}{\sqrt {d+c\,d\,x}} \,d x \]