Integrand size = 30, antiderivative size = 324 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=-\frac {4 b d^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {b d^4 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^4 (1+c x)^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {2 d^4 (1+c x) \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d^4 \left (1-c^2 x^2\right )^{5/2} \arcsin (c x) (a+b \arcsin (c x))}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {8 b d^4 \left (1-c^2 x^2\right )^{5/2} \log (1-c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-4/3*b*d^4*(-c^2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)- 1/2*b*d^4*(-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^4*(c*x+1)^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c *f*x+f)^(5/2)-2*d^4*(c*x+1)*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^( 5/2)/(-c*f*x+f)^(5/2)+d^4*(-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)-8/3*b*d^4*(-c^2*x^2+1)^(5/2)*ln(-c*x+1 )/c/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)
Time = 6.75 (sec) , antiderivative size = 601, normalized size of antiderivative = 1.85 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\frac {d \left (\frac {16 a (-1+2 c x) \sqrt {d+c d x} \sqrt {f-c f x}}{(-1+c x)^2}-12 a \sqrt {d} \sqrt {f} \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 )+\frac {2 b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left (-4+3 \arcsin (c x)-6 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-\cos \left (\frac {3}{2} \arcsin (c x)\right ) \left (\arcsin (c x)-2 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+2 \left (2+\left (2+\sqrt {1-c^2 x^2}\right ) \arcsin (c x)+2 \left (2+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^4 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left (-8-6 \arcsin (c x)+9 \arcsin (c x)^2-84 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+\cos \left (\frac {3}{2} \arcsin (c x)\right ) \left (-\arcsin (c x) (14+3 \arcsin (c x))+28 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )+2 \left (4+2 \left (2+7 \sqrt {1-c^2 x^2}\right ) \arcsin (c x)-3 \left (2+\sqrt {1-c^2 x^2}\right ) \arcsin (c x)^2+28 \left (2+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^4 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}\right )}{12 c f^3} \]
(d*((16*a*(-1 + 2*c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(-1 + c*x)^2 - 12* a*Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sq rt[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] - 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]]) + 2*(2 + (2 + Sqrt[1 - c^2*x^2])*ArcSin[c*x] + 2*(2 + Sq rt[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])) + (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*ArcS in[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[ArcSin[c*x]/2])^4*(Co s[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]))))/(12*c*f^3)
Time = 0.56 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.53, 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.
| \(\displaystyle \int \frac {(c d x+d)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^4 (c x+1)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (\frac {2 (c x+1)^3}{3 c \left (1-c^2 x^2\right )^2}-\frac {2 (c x+1)}{c \left (1-c^2 x^2\right )}+\frac {\arcsin (c x)}{c \sqrt {1-c^2 x^2}}\right )dx+\frac {2 (c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {2 (c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}+\frac {\arcsin (c x) (a+b \arcsin (c x))}{c}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{5/2} \left (\frac {2 (c x+1)^3 (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-\frac {2 (c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}+\frac {\arcsin (c x) (a+b \arcsin (c x))}{c}-b c \left (\frac {\arcsin (c x)^2}{2 c^2}+\frac {4}{3 c^2 (1-c x)}+\frac {8 \log (1-c x)}{3 c^2}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(d^4*(1 - c^2*x^2)^(5/2)*((2*(1 + c*x)^3*(a + b*ArcSin[c*x]))/(3*c*(1 - c^ 2*x^2)^(3/2)) - (2*(1 + c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c^2*x^2]) + (ArcSin[c*x]*(a + b*ArcSin[c*x]))/c - b*c*(4/(3*c^2*(1 - c*x)) + ArcSin[c* x]^2/(2*c^2) + (8*Log[1 - c*x])/(3*c^2))))/((d + c*d*x)^(5/2)*(f - c*f*x)^ (5/2))
3.6.35.3.1 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] )
\[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(a*c*d*x + a*d + (b*c*d*x + b*d)*arcsin(c*x))*sqrt(c*d*x + d)*sq rt(-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)
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int \frac {\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- f \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
-b*sqrt(d)*sqrt(f)*integrate((c*d*x + d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arct an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^3*f^3*x^3 - 3*c^2*f^3*x^2 + 3*c* f^3*x - f^3), x) - 1/3*a*((-c^2*d*f*x^2 + d*f)^(3/2)/(c^4*f^4*x^3 - 3*c^3* f^4*x^2 + 3*c^2*f^4*x - c*f^4) - 2*sqrt(-c^2*d*f*x^2 + d*f)*d/(c^3*f^3*x^2 - 2*c^2*f^3*x + c*f^3) - 7*sqrt(-c^2*d*f*x^2 + d*f)*d/(c^2*f^3*x - c*f^3) - 3*d^2*arcsin(c*x)/(c*f^3*sqrt(d/f)))
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}}{{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]