Integrand size = 30, antiderivative size = 463 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=-\frac {3 b d^4 x \left (1-c^2 x^2\right )^{3/2}}{2 (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {b c d^4 x^2 \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {5 b d^4 (1+c x)^2 \left (1-c^2 x^2\right )^{3/2}}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {15 b d^4 \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)^2}{4 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {2 d^4 (1+c x)^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {15 d^4 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {5 d^4 (1+c x) \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {15 d^4 \left (1-c^2 x^2\right )^{3/2} \arcsin (c x) (a+b \arcsin (c x))}{2 c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {8 b d^4 \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]
-3/2*b*d^4*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+b*c*d^4*x ^2*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-5/4*b*d^4*(c*x+1)^2 *(-c^2*x^2+1)^(3/2)/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+15/4*b*d^4*(-c^2*x^ 2+1)^(3/2)*arcsin(c*x)^2/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+2*d^4*(c*x+1)^ 3*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+15/2*d ^4*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+5/2 *d^4*(c*x+1)*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f) ^(3/2)-15/2*d^4*(-c^2*x^2+1)^(3/2)*arcsin(c*x)*(a+b*arcsin(c*x))/c/(c*d*x+ d)^(3/2)/(-c*f*x+f)^(3/2)+8*b*d^4*(-c^2*x^2+1)^(3/2)*ln(-c*x+1)/c/(c*d*x+d )^(3/2)/(-c*f*x+f)^(3/2)
Time = 6.11 (sec) , antiderivative size = 768, normalized size of antiderivative = 1.66 \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\frac {d^2 \left (\frac {8 a \sqrt {d+c d x} \sqrt {f-c f x} \left (-24+7 c x+c^2 x^2\right )}{-1+c x}+120 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 {8 b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left ((-4+\arcsin (c x)) \arcsin (c x)-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-\left (\arcsin (c x) (4+\arcsin (c x))-8 \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 )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\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 )^2}-\frac {32 b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (\arcsin (c x)^2 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+\left (c x-4 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\arcsin (c x) \left (\left (2+\sqrt {1-c^2 x^2}\right ) \cos \left (\frac {1}{2} \arcsin (c x)\right )-\left (-2+\sqrt {1-c^2 x^2}\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\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 )^2}+\frac {b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (-20 \arcsin (c x)^2 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 \left (-16 c x+\cos (2 \arcsin (c x))+32 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 \arcsin (c x) \left (24 \cos \left (\frac {1}{2} \arcsin (c x)\right )+7 \cos \left (\frac {3}{2} \arcsin (c x)\right )+\cos \left (\frac {5}{2} \arcsin (c x)\right )+24 \sin \left (\frac {1}{2} \arcsin (c x)\right )-7 \sin \left (\frac {3}{2} \arcsin (c x)\right )+\sin \left (\frac {5}{2} \arcsin (c x)\right )\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\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 )^2}\right )}{16 c f^2} \]
(d^2*((8*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-24 + 7*c*x + c^2*x^2))/(-1 + c*x) + 120*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))] - (8*b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2]*((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[A rcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) - (ArcSin[c*x]*(4 + ArcSin[c*x]) - 8* Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2) - (32*b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x] *(ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]) + (c*x - 4*Log[C os[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[ c*x]/2]) - ArcSin[c*x]*((2 + Sqrt[1 - c^2*x^2])*Cos[ArcSin[c*x]/2] - (-2 + Sqrt[1 - c^2*x^2])*Sin[ArcSin[c*x]/2])))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c *x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2) + (b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-20*ArcSin[c*x]^2*(Cos[Arc Sin[c*x]/2] - Sin[ArcSin[c*x]/2]) + 2*(-16*c*x + Cos[2*ArcSin[c*x]] + 32*L og[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] - Sin[Arc Sin[c*x]/2]) + 2*ArcSin[c*x]*(24*Cos[ArcSin[c*x]/2] + 7*Cos[(3*ArcSin[c*x] )/2] + Cos[(5*ArcSin[c*x])/2] + 24*Sin[ArcSin[c*x]/2] - 7*Sin[(3*ArcSin[c* x])/2] + Sin[(5*ArcSin[c*x])/2])))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2)))/(1...
Time = 0.59 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.41, 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.
| \(\displaystyle \int \frac {(c d x+d)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d^4 (c x+1)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1)^4 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \left (-b c \int \left (\frac {x}{2}-\frac {15 \arcsin (c x)}{2 c \sqrt {1-c^2 x^2}}+\frac {4}{c}+\frac {8 (c x+1)}{c \left (1-c^2 x^2\right )}\right )dx+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {8 (c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-\frac {15 \arcsin (c x) (a+b \arcsin (c x))}{2 c}\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 \left (1-c^2 x^2\right )^{3/2} \left (\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))+\frac {4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {8 (c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-\frac {15 \arcsin (c x) (a+b \arcsin (c x))}{2 c}-b c \left (-\frac {15 \arcsin (c x)^2}{4 c^2}-\frac {8 \log (1-c x)}{c^2}+\frac {4 x}{c}+\frac {x^2}{4}\right )\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
(d^4*(1 - c^2*x^2)^(3/2)*((8*(1 + c*x)*(a + b*ArcSin[c*x]))/(c*Sqrt[1 - c^ 2*x^2]) + (4*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + (x*Sqrt[1 - c^2*x^ 2]*(a + b*ArcSin[c*x]))/2 - (15*ArcSin[c*x]*(a + b*ArcSin[c*x]))/(2*c) - b *c*((4*x)/c + x^2/4 - (15*ArcSin[c*x]^2)/(4*c^2) - (8*Log[1 - c*x])/c^2))) /((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))
3.6.28.3.1 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] )
\[\int \frac {\left (c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
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^2*f^2*x^2 - 2 *c*f^2*x + f^2), x)
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
-1/2*(c^2*d^3*x^3/(sqrt(-c^2*d*f*x^2 + d*f)*f) + 8*c*d^3*x^2/(sqrt(-c^2*d* f*x^2 + d*f)*f) - 17*d^3*x/(sqrt(-c^2*d*f*x^2 + d*f)*f) + 15*d^3*arcsin(c* x)/(sqrt(d*f)*c*f) - 24*d^3/(sqrt(-c^2*d*f*x^2 + d*f)*c*f))*a - b*sqrt(d)* integrate((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*f*x - f)*sqrt(-c*x + 1)), x)/sqrt(f)
\[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {5}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{5/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}}{{\left (f-c\,f\,x\right )}^{3/2}} \,d x \]