Integrand size = 30, antiderivative size = 163 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {2 b f^3 \left (1-c^2 x^2\right )^{5/2}}{3 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {f^3 (1-c x)^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {b f^3 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-2/3*b*f^3*(-c^2*x^2+1)^(5/2)/c/(c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-1 /3*f^3*(-c*x+1)^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x +f)^(5/2)-1/3*b*f^3*(-c^2*x^2+1)^(5/2)*ln(c*x+1)/c/(c*d*x+d)^(5/2)/(-c*f*x +f)^(5/2)
Time = 1.85 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {f \sqrt {d+c d x} \left ((-1+c x) \left (-a+a c x-b \sqrt {1-c^2 x^2}\right )+b (-1+c x)^2 \arcsin (c x)+b (1+c x) \sqrt {1-c^2 x^2} \log (-f (1+c x))\right )}{3 c d^3 (1+c x)^2 \sqrt {f-c f x}} \]
-1/3*(f*Sqrt[d + c*d*x]*((-1 + c*x)*(-a + a*c*x - b*Sqrt[1 - c^2*x^2]) + b *(-1 + c*x)^2*ArcSin[c*x] + b*(1 + c*x)*Sqrt[1 - c^2*x^2]*Log[-(f*(1 + c*x ))]))/(c*d^3*(1 + c*x)^2*Sqrt[f - c*f*x])
Time = 0.49 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.65, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {5178, 27, 5260, 27, 456, 49, 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.
| \(\displaystyle \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(c d x+d)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {f^3 (1-c x)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int -\frac {(1-c x)^3}{3 c \left (1-c^2 x^2\right )^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \frac {(1-c x)^3}{\left (1-c^2 x^2\right )^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 456 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \frac {1-c x}{(c x+1)^2}dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \int \left (\frac {2}{(c x+1)^2}+\frac {1}{-c x-1}\right )dx-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {1}{3} b \left (-\frac {2}{c (c x+1)}-\frac {\log (c x+1)}{c}\right )-\frac {(1-c x)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(f^3*(1 - c^2*x^2)^(5/2)*(-1/3*((1 - c*x)^3*(a + b*ArcSin[c*x]))/(c*(1 - c ^2*x^2)^(3/2)) + (b*(-2/(c*(1 + c*x)) - Log[1 + c*x]/c))/3))/((d + c*d*x)^ (5/2)*(f - c*f*x)^(5/2))
3.6.9.3.1 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_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
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_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 - c^2*x^2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IG tQ[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3] )
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}}{\left (c d x +d \right )^{\frac {5}{2}}}d x\]
Time = 0.32 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.19 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\left [\frac {{\left (b c^{3} d x^{3} + b c^{2} d x^{2} - b c d x - b d\right )} \sqrt {\frac {f}{d}} \log \left (\frac {c^{6} f x^{6} + 4 \, c^{5} f x^{5} + 5 \, c^{4} f x^{4} - 4 \, c^{2} f x^{2} - 4 \, c f x + {\left (c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {\frac {f}{d}} - 2 \, f}{c^{4} x^{4} + 2 \, c^{3} x^{3} - 2 \, c x - 1}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x - 2 \, a c x + {\left (b c^{2} x^{2} - 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d^{3} x^{3} + c^{3} d^{3} x^{2} - c^{2} d^{3} x - c d^{3}\right )}}, -\frac {{\left (b c^{3} d x^{3} + b c^{2} d x^{2} - b c d x - b d\right )} \sqrt {-\frac {f}{d}} \arctan \left (\frac {{\left (c^{2} x^{2} + 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-\frac {f}{d}}}{c^{4} f x^{4} + 2 \, c^{3} f x^{3} - c^{2} f x^{2} - 2 \, c f x}\right ) - {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x - 2 \, a c x + {\left (b c^{2} x^{2} - 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d^{3} x^{3} + c^{3} d^{3} x^{2} - c^{2} d^{3} x - c d^{3}\right )}}\right ] \]
[1/6*((b*c^3*d*x^3 + b*c^2*d*x^2 - b*c*d*x - b*d)*sqrt(f/d)*log((c^6*f*x^6 + 4*c^5*f*x^5 + 5*c^4*f*x^4 - 4*c^2*f*x^2 - 4*c*f*x + (c^4*x^4 + 4*c^3*x^ 3 + 6*c^2*x^2 + 4*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f) *sqrt(f/d) - 2*f)/(c^4*x^4 + 2*c^3*x^3 - 2*c*x - 1)) + 2*(a*c^2*x^2 - 2*sq rt(-c^2*x^2 + 1)*b*c*x - 2*a*c*x + (b*c^2*x^2 - 2*b*c*x + b)*arcsin(c*x) + a)*sqrt(c*d*x + d)*sqrt(-c*f*x + f))/(c^4*d^3*x^3 + c^3*d^3*x^2 - c^2*d^3 *x - c*d^3), -1/3*((b*c^3*d*x^3 + b*c^2*d*x^2 - b*c*d*x - b*d)*sqrt(-f/d)* arctan((c^2*x^2 + 2*c*x + 2)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f* x + f)*sqrt(-f/d)/(c^4*f*x^4 + 2*c^3*f*x^3 - c^2*f*x^2 - 2*c*f*x)) - (a*c^ 2*x^2 - 2*sqrt(-c^2*x^2 + 1)*b*c*x - 2*a*c*x + (b*c^2*x^2 - 2*b*c*x + b)*a rcsin(c*x) + a)*sqrt(c*d*x + d)*sqrt(-c*f*x + f))/(c^4*d^3*x^3 + c^3*d^3*x ^2 - c^2*d^3*x - c*d^3)]
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (d \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {2 \, \sqrt {f}}{c^{3} d^{\frac {5}{2}} x + c^{2} d^{\frac {5}{2}}} + \frac {\sqrt {f} \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} x + c d^{3}}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d^{3} x + c d^{3}}\right )} \]
-1/3*b*c*(2*sqrt(f)/(c^3*d^(5/2)*x + c^2*d^(5/2)) + sqrt(f)*log(c*x + 1)/( c^2*d^(5/2))) - 1/3*b*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d^3*x^2 + 2*c^2*d^3 *x + c*d^3) - sqrt(-c^2*d*f*x^2 + d*f)/(c^2*d^3*x + c*d^3))*arcsin(c*x) - 1/3*a*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) - sq rt(-c^2*d*f*x^2 + d*f)/(c^2*d^3*x + c*d^3))
\[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {f-c f x} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]