Integrand size = 30, antiderivative size = 164 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=-\frac {2 b d^3 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d^3 (1+c x)^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {b d^3 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-2/3*b*d^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*d^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*d^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 = 1.48 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {f-c f x} \left ((1+c x) \left (-b+b c x+a \sqrt {1-c^2 x^2}\right )+b (1+c x) \sqrt {1-c^2 x^2} \arcsin (c x)-b (-1+c x)^2 \log (f-c f x)\right )}{3 c f^3 (-1+c x)^2 \sqrt {1-c^2 x^2}} \]
(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*((1 + c*x)*(-b + b*c*x + a*Sqrt[1 - c^2*x ^2]) + b*(1 + c*x)*Sqrt[1 - c^2*x^2]*ArcSin[c*x] - b*(-1 + c*x)^2*Log[f - c*f*x]))/(3*c*f^3*(-1 + c*x)^2*Sqrt[1 - c^2*x^2])
Time = 0.47 (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.
| \(\displaystyle \int \frac {\sqrt {c d x+d} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^3 (c x+1)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-b c \int \frac {(c x+1)^3}{3 c \left (1-c^2 x^2\right )^2}dx\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {1}{3} b \int \frac {(c x+1)^3}{\left (1-c^2 x^2\right )^2}dx\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 456 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {1}{3} b \int \frac {c x+1}{(1-c x)^2}dx\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {1}{3} b \int \left (\frac {1}{c x-1}+\frac {2}{(c x-1)^2}\right )dx\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {(c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {1}{3} b \left (\frac {2}{c (1-c x)}+\frac {\log (1-c x)}{c}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(d^3*(1 - c^2*x^2)^(5/2)*(((1 + c*x)^3*(a + b*ArcSin[c*x]))/(3*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))
3.6.36.3.1 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] )
\[\int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}d x\]
Time = 0.35 (sec) , antiderivative size = 520, normalized size of antiderivative = 3.17 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\left [\frac {{\left (b c^{3} f x^{3} - b c^{2} f x^{2} - b c f x + b f\right )} \sqrt {\frac {d}{f}} \log \left (\frac {c^{6} d x^{6} - 4 \, c^{5} d x^{5} + 5 \, c^{4} d x^{4} - 4 \, c^{2} d x^{2} + 4 \, c d x + {\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {\frac {d}{f}} - 2 \, d}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 2 \, a c x + {\left (b c^{2} x^{2} + 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} f^{3} x^{3} - c^{3} f^{3} x^{2} - c^{2} f^{3} x + c f^{3}\right )}}, -\frac {{\left (b c^{3} f x^{3} - b c^{2} f x^{2} - b c f x + b f\right )} \sqrt {-\frac {d}{f}} \arctan \left (\frac {{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-\frac {d}{f}}}{c^{4} d x^{4} - 2 \, c^{3} d x^{3} - c^{2} d x^{2} + 2 \, c d x}\right ) - {\left (a c^{2} x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 2 \, a c x + {\left (b c^{2} x^{2} + 2 \, b c x + b\right )} \arcsin \left (c x\right ) + a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} f^{3} x^{3} - c^{3} f^{3} x^{2} - c^{2} f^{3} x + c f^{3}\right )}}\right ] \]
[1/6*((b*c^3*f*x^3 - b*c^2*f*x^2 - b*c*f*x + b*f)*sqrt(d/f)*log((c^6*d*x^6 - 4*c^5*d*x^5 + 5*c^4*d*x^4 - 4*c^2*d*x^2 + 4*c*d*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)/(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*f^3*x^3 - c^3*f^3*x^2 - c^2*f^3 *x + c*f^3), -1/3*((b*c^3*f*x^3 - b*c^2*f*x^2 - b*c*f*x + b*f)*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)*sqrt(-d/f)/(c^4*d*x^4 - 2*c^3*d*x^3 - c^2*d*x^2 + 2*c*d*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*f^3*x^3 - c^3*f^3*x ^2 - c^2*f^3*x + c*f^3)]
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- f \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\frac {1}{3} \, b c {\left (\frac {2 \, \sqrt {d}}{c^{3} f^{\frac {5}{2}} x - c^{2} f^{\frac {5}{2}}} - \frac {\sqrt {d} \log \left (c x - 1\right )}{c^{2} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} - 2 \, c^{2} f^{3} x + c f^{3}} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} f^{3} x - c f^{3}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} - 2 \, c^{2} f^{3} x + c f^{3}} + \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} f^{3} x - c f^{3}}\right )} \]
1/3*b*c*(2*sqrt(d)/(c^3*f^(5/2)*x - c^2*f^(5/2)) - sqrt(d)*log(c*x - 1)/(c ^2*f^(5/2))) + 1/3*b*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*f^3*x^2 - 2*c^2*f^3* x + c*f^3) + sqrt(-c^2*d*f*x^2 + d*f)/(c^2*f^3*x - c*f^3))*arcsin(c*x) + 1 /3*a*(2*sqrt(-c^2*d*f*x^2 + d*f)/(c^3*f^3*x^2 - 2*c^2*f^3*x + c*f^3) + sqr t(-c^2*d*f*x^2 + d*f)/(c^2*f^3*x - c*f^3))
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x}}{{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]