Integrand size = 30, antiderivative size = 420 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=-\frac {b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b f^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {5 b f^5 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 f^5 (1-c x)^4 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^3 (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {5 f^5 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x) (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {28 b f^5 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-b*f^5*x*(-c^2*x^2+1)^(5/2)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)-8/3*b*f^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*f^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*f^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)+ 10/3*f^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*f^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*f^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*f^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. \(847\) vs. \(2(420)=840\).
Time = 9.97 (sec) , antiderivative size = 847, normalized size of antiderivative = 2.02 \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx =\text {Too large to display} \]
(f^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] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2]*(-8 + 6*Arc Sin[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]*((14 - 3*ArcSin[c*x])*ArcSin[c*x] + 28*Log[Co s[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]))/((1 - c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) + (2*b*Sqrt[ d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ (3*ArcSin[c*x])/2]*(ArcSin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c* x]/2]]) - Cos[ArcSin[c*x]/2]*(4 + 3*ArcSin[c*x] + 6*Log[Cos[ArcSin[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]))/((1 - c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/ 2])*(2*(4 + 6*c*x + 6*c^2*x^2 + 52*(1 + c*x)*Log[Cos[ArcSin[c*x]/2] + Sin[ ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 18*ArcSin[c*x ]^2*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3 + ArcSin[c*x]*(-24*Cos[...
Time = 0.61 (sec) , antiderivative size = 208, 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.
| \(\displaystyle \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(c d x+d)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {f^5 (1-c x)^5 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^5 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^5 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {f^5 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (-\frac {2 (1-c x)^4}{3 c \left (1-c^2 x^2\right )^2}+\frac {20 (1-c x)}{3 c \left (1-c^2 x^2\right )}+\frac {5 \arcsin (c x)}{c \sqrt {1-c^2 x^2}}+\frac {5}{3 c}\right )dx-\frac {2 (1-c x)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {20 (1-c x) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}+\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^5 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {2 (1-c x)^4 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}+\frac {20 (1-c x) (a+b \arcsin (c x))}{3 c \sqrt {1-c^2 x^2}}+\frac {5 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{3 c}+\frac {5 \arcsin (c x) (a+b \arcsin (c x))}{c}-b c \left (\frac {5 \arcsin (c x)^2}{2 c^2}+\frac {8}{3 c^2 (c x+1)}+\frac {28 \log (c x+1)}{3 c^2}+\frac {x}{c}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(f^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))
3.6.21.3.1 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] )
\[\int \frac {\left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (c d x +d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
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)/(c^3*d^3*x^3 + 3 *c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
Timed out. \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
1/3*(3*(-c^2*d*f*x^2 + d*f)^(5/2)/(c^5*d^5*x^4 + 4*c^4*d^5*x^3 + 6*c^3*d^5 *x^2 + 4*c^2*d^5*x + c*d^5) - 5*(-c^2*d*f*x^2 + d*f)^(3/2)*f/(c^4*d^4*x^3 + 3*c^3*d^4*x^2 + 3*c^2*d^4*x + c*d^4) - 10*sqrt(-c^2*d*f*x^2 + d*f)*f^2/( c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) + 35*sqrt(-c^2*d*f*x^2 + d*f)*f^2/(c^2* d^3*x + c*d^3) + 15*f^3*arcsin(c*x)/(c*d^3*sqrt(f/d)))*a + b*sqrt(f)*integ rate((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))/((c^2*d^2*x^2 + 2*c*d^2*x + d^2)*sqrt(c*x + 1)), x)/s qrt(d)
\[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int { \frac {{\left (-c f x + f\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(f-c f x)^{5/2} (a+b \arcsin (c x))}{(d+c d x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\right )}^{5/2}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]