Integrand size = 30, antiderivative size = 345 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=-\frac {11 b f^3 x \sqrt {1-c^2 x^2}}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 b c f^3 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {b c^2 f^3 x^3 \sqrt {1-c^2 x^2}}{9 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {11 f^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {3 f^3 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {c f^3 x^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {5 f^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]
11/3*f^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2) -3/2*f^3*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2) +1/3*c*f^3*x^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))/(c*d*x+d)^(1/2)/(-c*f*x+f)^( 1/2)-11/3*b*f^3*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+3/4* b*c*f^3*x^2*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-1/9*b*c^2* f^3*x^3*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)+5/4*f^3*(a+b*a rcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)
Time = 6.20 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.79 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\frac {90 b f^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-180 a \sqrt {d} f^{5/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )-6 b f^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (9 (-5+2 c x) \sqrt {1-c^2 x^2}+\cos (3 \arcsin (c x))\right )+f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (-270 b c x+12 a \sqrt {1-c^2 x^2} \left (22-9 c x+2 c^2 x^2\right )-27 b \cos (2 \arcsin (c x))+2 b \sin (3 \arcsin (c x))\right )}{72 c d \sqrt {1-c^2 x^2}} \]
(90*b*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 180*a*Sqrt[d]*f^ (5/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt [d]*Sqrt[f]*(-1 + c^2*x^2))] - 6*b*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Arc Sin[c*x]*(9*(-5 + 2*c*x)*Sqrt[1 - c^2*x^2] + Cos[3*ArcSin[c*x]]) + f^2*Sqr t[d + c*d*x]*Sqrt[f - c*f*x]*(-270*b*c*x + 12*a*Sqrt[1 - c^2*x^2]*(22 - 9* c*x + 2*c^2*x^2) - 27*b*Cos[2*ArcSin[c*x]] + 2*b*Sin[3*ArcSin[c*x]]))/(72* c*d*Sqrt[1 - c^2*x^2])
Time = 0.76 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.50, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5262, 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.
| \(\displaystyle \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {c d x+d}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {f^3 (1-c x)^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^3 \sqrt {1-c^2 x^2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {f^3 \sqrt {1-c^2 x^2} \int \left (-\frac {c^3 (a+b \arcsin (c x)) x^3}{\sqrt {1-c^2 x^2}}+\frac {3 c^2 (a+b \arcsin (c x)) x^2}{\sqrt {1-c^2 x^2}}-\frac {3 c (a+b \arcsin (c x)) x}{\sqrt {1-c^2 x^2}}+\frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}\right )dx}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^3 \sqrt {1-c^2 x^2} \left (\frac {1}{3} c x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {3}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {11 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 (a+b \arcsin (c x))^2}{4 b c}-\frac {1}{9} b c^2 x^3+\frac {3}{4} b c x^2-\frac {11 b x}{3}\right )}{\sqrt {c d x+d} \sqrt {f-c f x}}\) |
(f^3*Sqrt[1 - c^2*x^2]*((-11*b*x)/3 + (3*b*c*x^2)/4 - (b*c^2*x^3)/9 + (11* Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*c) - (3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/2 + (c*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/3 + (5* (a + b*ArcSin[c*x])^2)/(4*b*c)))/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x])
3.6.19.3.1 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_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ ) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & & EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ [n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
\[\int \frac {\left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {c d x +d}}d x\]
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
integral((a*c^2*f^2*x^2 - 2*a*c*f^2*x + a*f^2 + (b*c^2*f^2*x^2 - 2*b*c*f^2 *x + b*f^2)*arcsin(c*x))*sqrt(-c*f*x + f)/sqrt(c*d*x + d), x)
Timed out. \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\text {Timed out} \]
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
1/6*(2*sqrt(-c^2*d*f*x^2 + d*f)*c*f^2*x^2/d - 9*sqrt(-c^2*d*f*x^2 + d*f)*f ^2*x/d + 15*f^3*arcsin(c*x)/(sqrt(d*f)*c) + 22*sqrt(-c^2*d*f*x^2 + d*f)*f^ 2/(c*d))*a + b*sqrt(f)*integrate((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/sqrt(c*x + 1), x)/sqrt(d)
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{\sqrt {c d x + d}} \,d x } \]
Timed out. \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{\sqrt {d+c d x}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{5/2}}{\sqrt {d+c\,d\,x}} \,d x \]