Integrand size = 30, antiderivative size = 255 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=-\frac {b f \left (1-c^2 x^2\right )^{5/2}}{6 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {f (1-c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 f x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f \left (1-c^2 x^2\right )^{5/2} \text {arctanh}(c x)}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b f \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-1/6*b*f*(-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*(-c*x+1)*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5 /2)+2/3*f*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5 /2)+1/6*b*f*(-c^2*x^2+1)^(5/2)*arctanh(c*x)/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^( 5/2)+1/3*b*f*(-c^2*x^2+1)^(5/2)*ln(-c^2*x^2+1)/c/(c*d*x+d)^(5/2)/(-c*f*x+f )^(5/2)
Time = 1.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.71 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\frac {\sqrt {d+c d x} \left (-4 a+8 a c x+8 a c^2 x^2-2 b \sqrt {1-c^2 x^2}+4 b \left (-1+2 c x+2 c^2 x^2\right ) \arcsin (c x)+5 b (1+c x) \sqrt {1-c^2 x^2} \log (-f (1+c x))+3 b \sqrt {1-c^2 x^2} \log (f-c f x)+3 b c x \sqrt {1-c^2 x^2} \log (f-c f x)\right )}{12 c d^3 f (1+c x)^2 \sqrt {f-c f x}} \]
(Sqrt[d + c*d*x]*(-4*a + 8*a*c*x + 8*a*c^2*x^2 - 2*b*Sqrt[1 - c^2*x^2] + 4 *b*(-1 + 2*c*x + 2*c^2*x^2)*ArcSin[c*x] + 5*b*(1 + c*x)*Sqrt[1 - c^2*x^2]* Log[-(f*(1 + c*x))] + 3*b*Sqrt[1 - c^2*x^2]*Log[f - c*f*x] + 3*b*c*x*Sqrt[ 1 - c^2*x^2]*Log[f - c*f*x]))/(12*c*d^3*f*(1 + c*x)^2*Sqrt[f - c*f*x])
Time = 0.49 (sec) , antiderivative size = 159, 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, 5260, 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.
| \(\displaystyle \int \frac {a+b \arcsin (c x)}{(c d x+d)^{5/2} (f-c f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {f (1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {f \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (\frac {2 x}{3 \left (1-c^2 x^2\right )}-\frac {1-c x}{3 c \left (1-c^2 x^2\right )^2}\right )dx+\frac {2 x (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}-\frac {(1-c x) (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f \left (1-c^2 x^2\right )^{5/2} \left (\frac {2 x (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}-\frac {(1-c x) (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-b c \left (-\frac {\text {arctanh}(c x)}{6 c^2}+\frac {1-c x}{6 c^2 \left (1-c^2 x^2\right )}-\frac {\log \left (1-c^2 x^2\right )}{3 c^2}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(f*(1 - c^2*x^2)^(5/2)*(-1/3*((1 - c*x)*(a + b*ArcSin[c*x]))/(c*(1 - c^2*x ^2)^(3/2)) + (2*x*(a + b*ArcSin[c*x]))/(3*Sqrt[1 - c^2*x^2]) - b*c*((1 - c *x)/(6*c^2*(1 - c^2*x^2)) - ArcTanh[c*x]/(6*c^2) - Log[1 - c^2*x^2]/(3*c^2 ))))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))
3.6.33.3.1 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_.))*((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] )
\[\int \frac {a +b \arcsin \left (c x \right )}{\left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/(c^5*d^3*f^2 *x^5 + c^4*d^3*f^2*x^4 - 2*c^3*d^3*f^2*x^3 - 2*c^2*d^3*f^2*x^2 + c*d^3*f^2 *x + d^3*f^2), x)
Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=-\frac {1}{12} \, b c {\left (\frac {2 \, \sqrt {d} \sqrt {f}}{c^{3} d^{3} f^{2} x + c^{2} d^{3} f^{2}} - \frac {5 \, \log \left (c x + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {3}{2}}} - \frac {3 \, \log \left (c x - 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {3}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d^{2} f x + \sqrt {-c^{2} d f x^{2} + d f} c d^{2} f} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {1}{\sqrt {-c^{2} d f x^{2} + d f} c^{2} d^{2} f x + \sqrt {-c^{2} d f x^{2} + d f} c d^{2} f} - \frac {2 \, x}{\sqrt {-c^{2} d f x^{2} + d f} d^{2} f}\right )} \]
-1/12*b*c*(2*sqrt(d)*sqrt(f)/(c^3*d^3*f^2*x + c^2*d^3*f^2) - 5*log(c*x + 1 )/(c^2*d^(5/2)*f^(3/2)) - 3*log(c*x - 1)/(c^2*d^(5/2)*f^(3/2))) - 1/3*b*(1 /(sqrt(-c^2*d*f*x^2 + d*f)*c^2*d^2*f*x + sqrt(-c^2*d*f*x^2 + d*f)*c*d^2*f) - 2*x/(sqrt(-c^2*d*f*x^2 + d*f)*d^2*f))*arcsin(c*x) - 1/3*a*(1/(sqrt(-c^2 *d*f*x^2 + d*f)*c^2*d^2*f*x + sqrt(-c^2*d*f*x^2 + d*f)*c*d^2*f) - 2*x/(sqr t(-c^2*d*f*x^2 + d*f)*d^2*f))
\[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{{\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{(d+c d x)^{5/2} (f-c f x)^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+c\,d\,x\right )}^{5/2}\,{\left (f-c\,f\,x\right )}^{3/2}} \,d x \]