Integrand size = 30, antiderivative size = 162 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=-\frac {2 f^2 (1-c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {f^2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{2 b c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {2 b f^2 \left (1-c^2 x^2\right )^{3/2} \log (1+c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]
-2*f^2*(-c*x+1)*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f )^(3/2)-1/2*f^2*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))^2/b/c/(c*d*x+d)^(3/2) /(-c*f*x+f)^(3/2)+2*b*f^2*(-c^2*x^2+1)^(3/2)*ln(c*x+1)/c/(c*d*x+d)^(3/2)/( -c*f*x+f)^(3/2)
Time = 2.45 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.53 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=-\frac {\frac {4 a \sqrt {d+c d x} \sqrt {f-c f x}}{1+c x}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left (\arcsin (c x) (4+\arcsin (c x))-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+\left ((-4+\arcsin (c x)) \arcsin (c x)-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}}{2 c d^2} \]
-1/2*((4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(1 + c*x) - 2*a*Sqrt[d]*Sqrt[f ]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2* x^2))] + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2]*(ArcSin[c* x]*(4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + ( (-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x ]/2]])*Sin[ArcSin[c*x]/2]))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] + Sin[A rcSin[c*x]/2])))/(c*d^2)
Time = 0.56 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5274, 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.
| \(\displaystyle \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(c d x+d)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {f^2 (1-c x)^2 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^2 \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x)^2 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {f^2 \left (1-c^2 x^2\right )^{3/2} \int \left (\frac {2 (1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}-\frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {2 (1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^2}{2 b c}+\frac {2 b \log (c x+1)}{c}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
(f^2*(1 - c^2*x^2)^(3/2)*((-2*(1 - c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c ^2*x^2]) - (a + b*ArcSin[c*x])^2/(2*b*c) + (2*b*Log[1 + c*x])/c))/((d + c* d*x)^(3/2)*(f - c*f*x)^(3/2))
3.6.8.3.1 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[(a + b*ArcSin[c*x] )^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}}{\left (c d x +d \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/(c^2*d^2*x^2 + 2*c*d^2*x + d^2), x)
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=\int \frac {\sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
-a*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^2*d^2*x + c*d^2) + f*arcsin(c*x)/(c*d^2* sqrt(f/d))) + b*sqrt(f)*integrate(sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1 )*sqrt(-c*x + 1))/((c*d*x + d)*sqrt(c*x + 1)), x)/sqrt(d)
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{3/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \]