Integrand size = 30, antiderivative size = 376 \[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=-\frac {2 b f^2 x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {3 b c f^2 x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 f^2 x^4 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {3}{8} f^2 x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))+\frac {1}{4} c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))+\frac {2 f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c}+\frac {5 f^2 \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))^2}{16 b c \sqrt {1-c^2 x^2}} \]
3/8*f^2*x*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)+1/4*c^2*f^2*x ^3*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)+2/3*f^2*(-c^2*x^2+1) *(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/c-2/3*b*f^2*x*(c*d*x+d )^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-3/16*b*c*f^2*x^2*(c*d*x+d)^(1/ 2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+2/9*b*c^2*f^2*x^3*(c*d*x+d)^(1/2)*( -c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-1/16*b*c^3*f^2*x^4*(c*d*x+d)^(1/2)*(-c* f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+5/16*f^2*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/ 2)*(-c*f*x+f)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)
Time = 2.32 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.78 \[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\frac {360 b f^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-720 a \sqrt {d} f^{5/2} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (256 b c x \left (-3+c^2 x^2\right )+48 a \sqrt {1-c^2 x^2} \left (16+9 c x-16 c^2 x^2+6 c^3 x^3\right )+144 b \cos (2 \arcsin (c x))-9 b \cos (4 \arcsin (c x))\right )-12 b f^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (-64 \left (1-c^2 x^2\right )^{3/2}-24 \sin (2 \arcsin (c x))+3 \sin (4 \arcsin (c x))\right )}{1152 c \sqrt {1-c^2 x^2}} \]
(360*b*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*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])/(Sqr t[d]*Sqrt[f]*(-1 + c^2*x^2))] + f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(256*b *c*x*(-3 + c^2*x^2) + 48*a*Sqrt[1 - c^2*x^2]*(16 + 9*c*x - 16*c^2*x^2 + 6* c^3*x^3) + 144*b*Cos[2*ArcSin[c*x]] - 9*b*Cos[4*ArcSin[c*x]]) - 12*b*f^2*S qrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-64*(1 - c^2*x^2)^(3/2) - 24*S in[2*ArcSin[c*x]] + 3*Sin[4*ArcSin[c*x]]))/(1152*c*Sqrt[1 - c^2*x^2])
Time = 0.72 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.49, 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.
| \(\displaystyle \int \sqrt {c d x+d} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \int f^2 (1-c x)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^2 \sqrt {c d x+d} \sqrt {f-c f x} \int (1-c x)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {f^2 \sqrt {c d x+d} \sqrt {f-c f x} \int \left (c^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) x^2-2 c \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) x+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^2 \sqrt {c d x+d} \sqrt {f-c f x} \left (\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c}+\frac {1}{4} c^2 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {5 (a+b \arcsin (c x))^2}{16 b c}-\frac {1}{16} b c^3 x^4+\frac {2}{9} b c^2 x^3-\frac {3}{16} b c x^2-\frac {2 b x}{3}\right )}{\sqrt {1-c^2 x^2}}\) |
(f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*((-2*b*x)/3 - (3*b*c*x^2)/16 + (2*b*c ^2*x^3)/9 - (b*c^3*x^4)/16 + (3*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/8 + (c^2*x^3*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/4 + (2*(1 - c^2*x^2)^(3 /2)*(a + b*ArcSin[c*x]))/(3*c) + (5*(a + b*ArcSin[c*x])^2)/(16*b*c)))/Sqrt [1 - c^2*x^2]
3.6.18.3.1 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))
\[\int \sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )d x\]
\[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { \sqrt {c d x + d} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
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*d*x + d)*sqrt(-c*f*x + f), x)
Timed out. \[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]
\[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { \sqrt {c d x + d} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
b*sqrt(d)*sqrt(f)*integrate((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(c*x + 1)* sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/24*(15*s qrt(-c^2*d*f*x^2 + d*f)*f^2*x + 15*d*f^3*arcsin(c*x)/(sqrt(d*f)*c) - 6*(-c ^2*d*f*x^2 + d*f)^(3/2)*f*x/d + 16*(-c^2*d*f*x^2 + d*f)^(3/2)*f/(c*d))*a
\[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int { \sqrt {c d x + d} {\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int \sqrt {d+c d x} (f-c f x)^{5/2} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x}\,{\left (f-c\,f\,x\right )}^{5/2} \,d x \]