Integrand size = 30, antiderivative size = 98 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\frac {d (1+c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {b d \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]
d*(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 )+b*d*(-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 = 0.92 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {f-c f x} \left (-a \sqrt {1-c^2 x^2}-b \sqrt {1-c^2 x^2} \arcsin (c x)+b (-1+c x) \log (f-c f x)\right )}{c d f^2 (-1+c x) \sqrt {1-c^2 x^2}} \]
(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-(a*Sqrt[1 - c^2*x^2]) - b*Sqrt[1 - c^2* x^2]*ArcSin[c*x] + b*(-1 + c*x)*Log[f - c*f*x]))/(c*d*f^2*(-1 + c*x)*Sqrt[ 1 - c^2*x^2])
Time = 0.42 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5178, 27, 5260, 27, 451, 16}
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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{\sqrt {c d x+d} (f-c f x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d (c x+1) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 5260 |
\(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \left (\frac {(c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-b c \int \frac {c x+1}{c \left (1-c^2 x^2\right )}dx\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \left (\frac {(c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-b \int \frac {c x+1}{1-c^2 x^2}dx\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 451 |
\(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \left (\frac {(c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-b \int \frac {1}{1-c x}dx\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {d \left (1-c^2 x^2\right )^{3/2} \left (\frac {(c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}+\frac {b \log (1-c x)}{c}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
(d*(1 - c^2*x^2)^(3/2)*(((1 + c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c^2*x^ 2]) + (b*Log[1 - c*x])/c))/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))
3.6.31.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c^2/a In t[1/(c - d*x), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
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] )
\[\int \frac {a +b \arcsin \left (c x \right )}{\sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {3}{2}}}d x\]
Time = 0.31 (sec) , antiderivative size = 354, normalized size of antiderivative = 3.61 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\left [\frac {{\left (b c x - b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} - 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} + 4 \, c d f x - {\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) - 2 \, \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{2 \, {\left (c^{2} d f^{2} x - c d f^{2}\right )}}, \frac {{\left (b c x - b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} - 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} + 2 \, c d f x}\right ) - \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d f^{2} x - c d f^{2}}\right ] \]
[1/2*((b*c*x - b)*sqrt(d*f)*log((c^6*d*f*x^6 - 4*c^5*d*f*x^5 + 5*c^4*d*f*x ^4 - 4*c^2*d*f*x^2 + 4*c*d*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(d*f) - 2*d*f)/(c ^4*x^4 - 2*c^3*x^3 + 2*c*x - 1)) - 2*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*a rcsin(c*x) + a))/(c^2*d*f^2*x - c*d*f^2), ((b*c*x - b)*sqrt(-d*f)*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)*s qrt(-d*f)/(c^4*d*f*x^4 - 2*c^3*d*f*x^3 - c^2*d*f*x^2 + 2*c*d*f*x)) - sqrt( c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a))/(c^2*d*f^2*x - c*d*f^2)]
\[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\sqrt {d \left (c x + 1\right )} \left (- f \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=-\frac {\sqrt {-c^{2} d f x^{2} + d f} b \arcsin \left (c x\right )}{c^{2} d f^{2} x - c d f^{2}} - \frac {\sqrt {-c^{2} d f x^{2} + d f} a}{c^{2} d f^{2} x - c d f^{2}} + \frac {b \log \left (c x - 1\right )}{c \sqrt {d} f^{\frac {3}{2}}} \]
-sqrt(-c^2*d*f*x^2 + d*f)*b*arcsin(c*x)/(c^2*d*f^2*x - c*d*f^2) - sqrt(-c^ 2*d*f*x^2 + d*f)*a/(c^2*d*f^2*x - c*d*f^2) + b*log(c*x - 1)/(c*sqrt(d)*f^( 3/2))
\[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{\sqrt {c d x + d} {\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\sqrt {d+c\,d\,x}\,{\left (f-c\,f\,x\right )}^{3/2}} \,d x \]