Integrand size = 30, antiderivative size = 419 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\frac {b d^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b d^5 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^5 (1+c x)^4 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {10 d^5 (1+c x)^2 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 d^5 \left (1-c^2 x^2\right )^3 (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 d^5 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x) (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {28 b d^5 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
b*d^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-8/3*b*d^5*(-c^ 2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-5/2*b*d^5*(-c^2 *x^2+1)^(5/2)*arcsin(c*x)^2/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+2/3*d^5*(c* x+1)^4*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-1 0/3*d^5*(c*x+1)^2*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f *x+f)^(5/2)-5*d^5*(-c^2*x^2+1)^3*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f *x+f)^(5/2)+5*d^5*(-c^2*x^2+1)^(5/2)*arcsin(c*x)*(a+b*arcsin(c*x))/c/(c*d* x+d)^(5/2)/(-c*f*x+f)^(5/2)-28/3*b*d^5*(-c^2*x^2+1)^(5/2)*ln(-c*x+1)/c/(c* d*x+d)^(5/2)/(-c*f*x+f)^(5/2)
Leaf count is larger than twice the leaf count of optimal. \(850\) vs. \(2(419)=838\).
Time = 7.25 (sec) , antiderivative size = 850, normalized size of antiderivative = 2.03 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx =\text {Too large to display} \]
(d^2*((-4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(23 - 34*c*x + 3*c^2*x^2))/(-1 + c*x)^2 - 60*a*Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f* x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + (2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f* x]*(Cos[ArcSin[c*x]/2]*(-4 + 3*ArcSin[c*x] - 6*Log[Cos[ArcSin[c*x]/2] - Si n[ArcSin[c*x]/2]]) - Cos[(3*ArcSin[c*x])/2]*(ArcSin[c*x] - 2*Log[Cos[ArcSi n[c*x]/2] - Sin[ArcSin[c*x]/2]]) + 2*(2 + (2 + Sqrt[1 - c^2*x^2])*ArcSin[c *x] + 2*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2 ]])*Sin[ArcSin[c*x]/2]))/((Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^4*(Cos [ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f *x]*(Cos[ArcSin[c*x]/2]*(-8 - 6*ArcSin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos [ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) + Cos[(3*ArcSin[c*x])/2]*(-(ArcSin[ c*x]*(14 + 3*ArcSin[c*x])) + 28*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2 ]]) + 2*(4 + 2*(2 + 7*Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 3*(2 + Sqrt[1 - c^2 *x^2])*ArcSin[c*x]^2 + 28*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/((Cos[ArcSin[c*x]/2] - Sin[ArcS in[c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (b*Sqrt[d + c*d *x]*Sqrt[f - c*f*x]*(2*(-7 + 6*c*x + 3*Cos[2*ArcSin[c*x]] + 52*(-1 + c*x)* Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] - Sin[Ar cSin[c*x]/2]) + 18*ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]) ^3 + ArcSin[c*x]*(-24*Cos[ArcSin[c*x]/2] - 35*Cos[(3*ArcSin[c*x])/2] + ...
Time = 0.62 (sec) , antiderivative size = 209, 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, 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.
| \(\displaystyle \int \frac {(c d x+d)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^5 (c x+1)^5 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^5 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (\frac {2 (c x+1)^4}{3 c \left (1-c^2 x^2\right )^2}-\frac {20 (c x+1)}{3 c \left (1-c^2 x^2\right )}+\frac {5 \arcsin (c x)}{c \sqrt {1-c^2 x^2}}-\frac {5}{3 c}\right )dx+\frac {2 (c x+1)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {20 (c x+1) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}-\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^5 \left (1-c^2 x^2\right )^{5/2} \left (\frac {2 (c x+1)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {20 (c x+1) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}-\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}-b c \left (\frac {5 \arcsin (c x)^2}{2 c^2}+\frac {8}{3 c^2 (1-c x)}+\frac {28 \log (1-c x)}{3 c^2}-\frac {x}{c}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(d^5*(1 - c^2*x^2)^(5/2)*((2*(1 + c*x)^4*(a + b*ArcSin[c*x]))/(3*c*(1 - c^ 2*x^2)^(3/2)) - (20*(1 + c*x)*(a + b*ArcSin[c*x]))/(3*c*Sqrt[1 - c^2*x^2]) - (5*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*c) + (5*ArcSin[c*x]*(a + b *ArcSin[c*x]))/c - b*c*(-(x/c) + 8/(3*c^2*(1 - c*x)) + (5*ArcSin[c*x]^2)/( 2*c^2) + (28*Log[1 - c*x])/(3*c^2))))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2) )
3.6.34.3.1 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] )
\[\int \frac {\left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
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)/(c^3*f^3*x^3 - 3*c^2*f^3*x^2 + 3*c*f^3*x - f^3), x)
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
-1/3*(3*(-c^2*d*f*x^2 + d*f)^(5/2)/(c^5*f^5*x^4 - 4*c^4*f^5*x^3 + 6*c^3*f^ 5*x^2 - 4*c^2*f^5*x + c*f^5) + 5*(-c^2*d*f*x^2 + d*f)^(3/2)*d/(c^4*f^4*x^3 - 3*c^3*f^4*x^2 + 3*c^2*f^4*x - c*f^4) - 10*sqrt(-c^2*d*f*x^2 + d*f)*d^2/ (c^3*f^3*x^2 - 2*c^2*f^3*x + c*f^3) - 35*sqrt(-c^2*d*f*x^2 + d*f)*d^2/(c^2 *f^3*x - c*f^3) - 15*d^3*arcsin(c*x)/(c*f^3*sqrt(d/f)))*a + b*sqrt(d)*inte grate((c^2*d^2*x^2 + 2*c*d^2*x + d^2)*sqrt(c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(-c*x + 1)), x)/ sqrt(f)
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}}{{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]