Integrand size = 30, antiderivative size = 141 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=-\frac {b f x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]
f*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-b*f*x* (-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+1/2*f*(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 = 1.62 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\frac {\frac {2 \sqrt {d+c d x} \sqrt {f-c f x} \left (-b c x+a \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}+2 b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2}{\sqrt {1-c^2 x^2}}-2 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )}{2 c d} \]
((2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-(b*c*x) + a*Sqrt[1 - c^2*x^2]))/Sqrt [1 - c^2*x^2] + 2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x] + (b*Sqrt[ d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 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))])/(2*c*d)
Time = 0.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62, 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.
\(\displaystyle \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {c d x+d}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {f (1-c x) (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f \sqrt {1-c^2 x^2} \int \frac {(1-c x) (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
\(\Big \downarrow \) 5262 |
\(\displaystyle \frac {f \sqrt {1-c^2 x^2} \int \left (\frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}-\frac {c x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right )dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f \sqrt {1-c^2 x^2} \left (\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {(a+b \arcsin (c x))^2}{2 b c}-b x\right )}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
(f*Sqrt[1 - c^2*x^2]*(-(b*x) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + (a + b*ArcSin[c*x])^2/(2*b*c)))/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x])
3.6.7.3.1 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))
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}}{\sqrt {c d x +d}}d x\]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int \frac {\sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {d \left (c x + 1\right )}}\, dx \]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
a*(f*arcsin(c*x)/(c*d*sqrt(f/d)) + sqrt(-c^2*d*f*x^2 + d*f)/(c*d)) + b*sqr t(f)*integrate(sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/s qrt(c*x + 1), x)/sqrt(d)
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{\sqrt {d+c\,d\,x}} \,d x \]