Integrand size = 30, antiderivative size = 99 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=-\frac {f (1-c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {b f \left (1-c^2 x^2\right )^{3/2} \log (1+c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]
-f*(-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*f*(-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 = 0.82 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\frac {\sqrt {d+c d x} \left (a (-1+c x)+b (-1+c x) \arcsin (c x)+b \sqrt {1-c^2 x^2} \log (-f (1+c x))\right )}{c d^2 (1+c x) \sqrt {f-c f x}} \]
(Sqrt[d + c*d*x]*(a*(-1 + c*x) + b*(-1 + c*x)*ArcSin[c*x] + b*Sqrt[1 - c^2 *x^2]*Log[-(f*(1 + c*x))]))/(c*d^2*(1 + c*x)*Sqrt[f - c*f*x])
Time = 0.42 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83, 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, 25, 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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{(c d x+d)^{3/2} \sqrt {f-c f x}} \, dx\) |
\(\Big \downarrow \) 5178 |
\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {f (1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 5260 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (-b c \int -\frac {1-c x}{c \left (1-c^2 x^2\right )}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b c \int \frac {1-c x}{c \left (1-c^2 x^2\right )}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b \int \frac {1-c x}{1-c^2 x^2}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 451 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (b \int \frac {1}{c x+1}dx-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {f \left (1-c^2 x^2\right )^{3/2} \left (\frac {b \log (c x+1)}{c}-\frac {(1-c x) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
(f*(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))
3.6.26.3.1 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] )
\[\int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {3}{2}} \sqrt {-c f x +f}}d x\]
Time = 0.31 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.52 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\left [\frac {{\left (b c x + b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} + 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} - 4 \, c d f x - {\left (c^{4} x^{4} + 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} + 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} + 2 \, c^{3} x^{3} - 2 \, c x - 1}\right ) - 2 \, \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{2 \, {\left (c^{2} d^{2} f x + c d^{2} f\right )}}, \frac {{\left (b c x + b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} + 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} + 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} - 2 \, c d f x}\right ) - \sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d^{2} f x + c d^{2} f}\right ] \]
[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^2*f*x + c*d^2*f), ((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^2*f*x + c*d^2*f)]
\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \sqrt {- f \left (c x - 1\right )}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=-\frac {\sqrt {-c^{2} d f x^{2} + d f} b \arcsin \left (c x\right )}{c^{2} d^{2} f x + c d^{2} f} - \frac {\sqrt {-c^{2} d f x^{2} + d f} a}{c^{2} d^{2} f x + c d^{2} f} + \frac {b \log \left (c x + 1\right )}{c d^{\frac {3}{2}} \sqrt {f}} \]
-sqrt(-c^2*d*f*x^2 + d*f)*b*arcsin(c*x)/(c^2*d^2*f*x + c*d^2*f) - sqrt(-c^ 2*d*f*x^2 + d*f)*a/(c^2*d^2*f*x + c*d^2*f) + b*log(c*x + 1)/(c*d^(3/2)*sqr t(f))
\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {3}{2}} \sqrt {-c f x + f}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{3/2} \sqrt {f-c f x}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{3/2}\,\sqrt {f-c\,f\,x}} \,d x \]