Integrand size = 32, antiderivative size = 486 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=-\frac {i d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 b d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1-i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 i b^2 d^3 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 b d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 b^2 d^3 \left (1-c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {d^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
-1/3*I*d^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e* x+e)^(5/2)-4/3*b*d^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*ln(1-I/(I*c*x+(- c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-4/3*I*b^2*d^3*(-c^2* x^2+1)^(5/2)*polylog(2,I/(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(5/2)/(-c *e*x+e)^(5/2)-2/3*b*d^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*sec(1/4*Pi+1/ 2*arcsin(c*x))^2/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+4/3*b^2*d^3*(-c^2*x^2+ 1)^(5/2)*tan(1/4*Pi+1/2*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/ 3*d^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))^2*tan(1/4*Pi+1/2*arcsin(c*x))/c /(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*d^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c *x))^2*sec(1/4*Pi+1/2*arcsin(c*x))^2*tan(1/4*Pi+1/2*arcsin(c*x))/c/(c*d*x+ d)^(5/2)/(-c*e*x+e)^(5/2)
Time = 10.90 (sec) , antiderivative size = 687, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\frac {\sqrt {-e (-1+c x)} \sqrt {d (1+c x)} \left (\frac {2 a^2}{3 e^3 (-1+c x)^2}+\frac {a^2}{3 e^3 (-1+c x)}\right )}{c}+\frac {a b \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {-d e \left (1-c^2 x^2\right )} \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+2 \arcsin (c x)+\sqrt {1-c^2 x^2} \arcsin (c x)+4 \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+2 \sqrt {1-c^2 x^2} \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 )}{3 c e^3 \sqrt {(-d-c d x) (e-c e x)} \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^2 (1+c x) \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {-d e \left (1-c^2 x^2\right )} \left (-3 i \pi \arcsin (c x)+\frac {4 \arcsin (c x)}{-1+c x}-(1-i) \arcsin (c x)^2-\frac {2 \arcsin (c x)^2}{-1+c x}-4 \pi \log \left (1+e^{-i \arcsin (c x)}\right )+2 \pi \log \left (1+i e^{i \arcsin (c x)}\right )-4 \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )+4 \pi \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )-2 \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+4 i \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+\frac {2 \left (4+\arcsin (c x)^2+c x \left (-4+\arcsin (c x)^2\right )\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^3}\right )}{3 c e^3 \sqrt {(-d-c d x) (e-c e x)} \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 )^2} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((2*a^2)/(3*e^3*(-1 + c*x)^2) + a ^2/(3*e^3*(-1 + c*x))))/c + (a*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d* e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2]*(-4 + 3*ArcSin[c*x] - 6*Log[Cos[ArcS in[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*ArcSin[c*x] + S qrt[1 - c^2*x^2]*ArcSin[c*x] + 4*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/ 2]] + 2*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]])*Si n[ArcSin[c*x]/2]))/(3*c*e^3*Sqrt[(-d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x ]/2] - Sin[ArcSin[c*x]/2])^4*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) + (b^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]* ((-3*I)*Pi*ArcSin[c*x] + (4*ArcSin[c*x])/(-1 + c*x) - (1 - I)*ArcSin[c*x]^ 2 - (2*ArcSin[c*x]^2)/(-1 + c*x) - 4*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 2* Pi*Log[1 + I*E^(I*ArcSin[c*x])] - 4*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x] )] + 4*Pi*Log[Cos[ArcSin[c*x]/2]] - 2*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + (4*I)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (2*(4 + ArcSin[c*x]^2 + c*x* (-4 + ArcSin[c*x]^2))*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin [c*x]/2])^3))/(3*c*e^3*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[1 - c^2*x^2]*(C os[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2)
Time = 1.20 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.53, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
| \(\displaystyle \int \frac {\sqrt {c d x+d} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^3 (c x+1)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {(a+b \arcsin (c x))^2}{(c x-1) \sqrt {1-c^2 x^2}}+\frac {2 (a+b \arcsin (c x))^2}{(c x-1)^2 \sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{5/2} \left (-\frac {i (a+b \arcsin (c x))^2}{3 c}-\frac {4 b \log \left (1-i e^{-i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}-\frac {\tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {2 b \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}+\frac {\tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \sec ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,i e^{-i \arcsin (c x)}\right )}{3 c}+\frac {4 b^2 \tan \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(d^3*(1 - c^2*x^2)^(5/2)*(((-1/3*I)*(a + b*ArcSin[c*x])^2)/c - (4*b*(a + b *ArcSin[c*x])*Log[1 - I/E^(I*ArcSin[c*x])])/(3*c) - (((4*I)/3)*b^2*PolyLog [2, I/E^(I*ArcSin[c*x])])/c - (2*b*(a + b*ArcSin[c*x])*Sec[Pi/4 + ArcSin[c *x]/2]^2)/(3*c) + (4*b^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c) - ((a + b*ArcSin [c*x])^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c) + ((a + b*ArcSin[c*x])^2*Sec[Pi/ 4 + ArcSin[c*x]/2]^2*Tan[Pi/4 + ArcSin[c*x]/2])/(3*c)))/((d + c*d*x)^(5/2) *(e - c*e*x)^(5/2))
3.6.72.3.1 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]
\[\int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sq rt(-c*e*x + e)/(c^3*e^3*x^3 - 3*c^2*e^3*x^2 + 3*c*e^3*x - e^3), x)
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- e \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int { \frac {\sqrt {c d x + d} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+c d x} (a+b \arcsin (c x))^2}{(e-c e x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}}{{\left (e-c\,e\,x\right )}^{5/2}} \,d x \]