Integrand size = 30, antiderivative size = 376 \[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\frac {2 b d^2 x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {3 b c d^2 x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}-\frac {2 b c^2 d^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 d^2 x^4 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {3}{8} d^2 x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))+\frac {1}{4} c^2 d^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))-\frac {2 d^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c}+\frac {5 d^2 \sqrt {d+c d x} \sqrt {f-c f x} (a+b \arcsin (c x))^2}{16 b c \sqrt {1-c^2 x^2}} \]
3/8*d^2*x*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)+1/4*c^2*d^2*x ^3*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)-2/3*d^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*d^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*d^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*d^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*d^2*x^4*(c*d*x+d)^(1/2)*(-c* f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+5/16*d^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 = 2.69 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.78 \[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\frac {360 b d^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x)^2-720 a d^{5/2} \sqrt {f} \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+d^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (-256 b c x \left (-3+c^2 x^2\right )+48 a \sqrt {1-c^2 x^2} \left (-16+9 c x+16 c^2 x^2+6 c^3 x^3\right )+144 b \cos (2 \arcsin (c x))-9 b \cos (4 \arcsin (c x))\right )+12 b d^2 \sqrt {d+c d x} \sqrt {f-c f x} \arcsin (c x) \left (-64 \left (1-c^2 x^2\right )^{3/2}+24 \sin (2 \arcsin (c x))-3 \sin (4 \arcsin (c x))\right )}{1152 c \sqrt {1-c^2 x^2}} \]
(360*b*d^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*a*d^(5/2)*S qrt[f]*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))] + d^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*d^2 *Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-64*(1 - c^2*x^2)^(3/2) + 24 *Sin[2*ArcSin[c*x]] - 3*Sin[4*ArcSin[c*x]]))/(1152*c*Sqrt[1 - c^2*x^2])
Time = 0.74 (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.
| \(\displaystyle \int (c d x+d)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\sqrt {c d x+d} \sqrt {f-c f x} \int d^2 (c x+1)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \sqrt {c d x+d} \sqrt {f-c f x} \int (c x+1)^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 5262 |
| \(\displaystyle \frac {d^2 \sqrt {c d x+d} \sqrt {f-c f x} \int \left (c^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) x^2+2 c \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) x+\sqrt {1-c^2 x^2} (a+b \arcsin (c x))\right )dx}{\sqrt {1-c^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \sqrt {c d x+d} \sqrt {f-c f x} \left (\frac {3}{8} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))-\frac {2 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))}{3 c}+\frac {1}{4} c^2 x^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {5 (a+b \arcsin (c x))^2}{16 b c}-\frac {1}{16} b c^3 x^4-\frac {2}{9} b c^2 x^3-\frac {3}{16} b c x^2+\frac {2 b x}{3}\right )}{\sqrt {1-c^2 x^2}}\) |
(d^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]
3.6.4.3.1 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))
\[\int \left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}d x\]
\[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
integral((a*c^2*d^2*x^2 + 2*a*c*d^2*x + a*d^2 + (b*c^2*d^2*x^2 + 2*b*c*d^2 *x + b*d^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*f*x + f), x)
Timed out. \[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\text {Timed out} \]
\[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
b*sqrt(d)*sqrt(f)*integrate((c^2*d^2*x^2 + 2*c*d^2*x + d^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)*d^2*x + 15*d^3*f*arcsin(c*x)/(sqrt(d*f)*c) - 6*(-c ^2*d*f*x^2 + d*f)^(3/2)*d*x/f - 16*(-c^2*d*f*x^2 + d*f)^(3/2)*d/(c*f))*a
\[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int { {\left (c d x + d\right )}^{\frac {5}{2}} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )} \,d x } \]
Timed out. \[ \int (d+c d x)^{5/2} \sqrt {f-c f x} (a+b \arcsin (c x)) \, dx=\int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}\,\sqrt {f-c\,f\,x} \,d x \]