Integrand size = 30, antiderivative size = 252 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=-\frac {b d^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {4 d^3 (1+c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {d^3 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {3 d^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{2 b c (d+c d x)^{3/2} (f-c f x)^{3/2}}+\frac {4 b d^3 \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (d+c d x)^{3/2} (f-c f x)^{3/2}} \]
-b*d^3*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+4*d^3*(c*x+1) *(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)+d^3*(-c ^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-3/2*d^3*( -c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))^2/b/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2) +4*b*d^3*(-c^2*x^2+1)^(3/2)*ln(-c*x+1)/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(514\) vs. \(2(252)=504\).
Time = 5.33 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.04 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\frac {d \left (\frac {2 a (-5+c x) \sqrt {d+c d x} \sqrt {f-c f x}}{-1+c x}+6 a \sqrt {d} \sqrt {f} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )-\frac {b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right ) \left ((-4+\arcsin (c x)) \arcsin (c x)-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )-\left (\arcsin (c x) (4+\arcsin (c x))-8 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}-\frac {2 b (1+c x) \sqrt {d+c d x} \sqrt {f-c f x} \left (\arcsin (c x)^2 \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+\left (c x-4 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-\arcsin (c x) \left (\left (2+\sqrt {1-c^2 x^2}\right ) \cos \left (\frac {1}{2} \arcsin (c x)\right )-\left (-2+\sqrt {1-c^2 x^2}\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right ) \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}\right )}{2 c f^2} \]
(d*((2*a*(-5 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(-1 + c*x) + 6*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))] - (b*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSi n[c*x]/2]*((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[ArcSin[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[Ar cSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2 ])^2) - (2*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[Cos[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)))/(2*c*f^2)
Time = 0.61 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5274, 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.
| \(\displaystyle \int \frac {(c d x+d)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d^3 (c x+1)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1)^3 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \int \left (-\frac {c x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {3 (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}+\frac {4 (c x+1) (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \left (\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c}+\frac {4 (c x+1) (a+b \arcsin (c x))}{c \sqrt {1-c^2 x^2}}-\frac {3 (a+b \arcsin (c x))^2}{2 b c}+\frac {4 b \log (1-c x)}{c}-b x\right )}{(c d x+d)^{3/2} (f-c f x)^{3/2}}\) |
(d^3*(1 - c^2*x^2)^(3/2)*(-(b*x) + (4*(1 + c*x)*(a + b*ArcSin[c*x]))/(c*Sq rt[1 - c^2*x^2]) + (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c - (3*(a + b*A rcSin[c*x])^2)/(2*b*c) + (4*b*Log[1 - c*x])/c))/((d + c*d*x)^(3/2)*(f - c* f*x)^(3/2))
3.6.29.3.1 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[(a + b*ArcSin[c*x] )^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]
\[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
integral((a*c*d*x + a*d + (b*c*d*x + b*d)*arcsin(c*x))*sqrt(c*d*x + d)*sqr t(-c*f*x + f)/(c^2*f^2*x^2 - 2*c*f^2*x + f^2), x)
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int \frac {\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- f \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
b*sqrt(d)*sqrt(f)*integrate((c*d*x + d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arcta n2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*f^2*x^2 - 2*c*f^2*x + f^2), x) - a*((-c^2*d*f*x^2 + d*f)^(3/2)/(c^3*f^3*x^2 - 2*c^2*f^3*x + c*f^3) + 6*sq rt(-c^2*d*f*x^2 + d*f)*d/(c^2*f^2*x - c*f^2) + 3*d^2*arcsin(c*x)/(c*f^2*sq rt(d/f)))
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (-c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))}{(f-c f x)^{3/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{3/2}}{{\left (f-c\,f\,x\right )}^{3/2}} \,d x \]