Integrand size = 32, antiderivative size = 486 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\frac {i e^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 b^2 e^3 \left (1-c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {e^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {2 b e^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {e^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}-\frac {4 b e^3 \left (1-c^2 x^2\right )^{5/2} (a+b \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}}+\frac {4 i b^2 e^3 \left (1-c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c (d+c d x)^{5/2} (e-c e x)^{5/2}} \]
1/3*I*e^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^2*e^3*(-c^2*x^2+1)^(5/2)*cot(1/4*Pi+1/2*arcsin(c*x))/c/(c* d*x+d)^(5/2)/(-c*e*x+e)^(5/2)+1/3*e^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x)) ^2*cot(1/4*Pi+1/2*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-2/3*b*e^ 3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x))*csc(1/4*Pi+1/2*arcsin(c*x))^2/c/(c* d*x+d)^(5/2)/(-c*e*x+e)^(5/2)-1/3*e^3*(-c^2*x^2+1)^(5/2)*(a+b*arcsin(c*x)) ^2*cot(1/4*Pi+1/2*arcsin(c*x))*csc(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*e^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 *e^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)
Time = 9.32 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.08 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (\frac {a^2 (-1+c x)^2}{(1+c x)^2}-\frac {a b \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right ) \left (\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 )-\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 )+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}-\frac {b^2 (-1+c x)^2 \left (-i \pi \arcsin (c x)+(1+i) \arcsin (c x)^2-4 \pi \log \left (1+e^{-i \arcsin (c x)}\right )-2 (\pi +2 \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 (\sin \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 {4 \arcsin (c x)^2 \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}-\frac {2 \arcsin (c x) (2+\arcsin (c x))}{\left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )^2}-\frac {2 \left (-4+\arcsin (c x)^2\right ) \sin \left (\frac {1}{2} \arcsin (c x)\right )}{\cos \left (\frac {1}{2} \arcsin (c x)\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 )^2}\right )}{3 c d^3 (-1+c x)} \]
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((a^2*(-1 + c*x)^2)/(1 + c*x)^2 - (a*b*(C os[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[C os[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 - (b^2*(-1 + c*x)^2*((-I)*Pi*ArcSin[c*x] + (1 + I)*ArcSin[c*x]^2 - 4*Pi*Log[1 + E^((-I)*ArcSin[c*x])] - 2*(Pi + 2*Ar cSin[c*x])*Log[1 - I*E^(I*ArcSin[c*x])] + 4*Pi*Log[Cos[ArcSin[c*x]/2]] + 2 *Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] + (4*I)*PolyLog[2, I*E^(I*ArcSin[c*x] )] + (4*ArcSin[c*x]^2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin [c*x]/2])^3 - (2*ArcSin[c*x]*(2 + ArcSin[c*x]))/(Cos[ArcSin[c*x]/2] + Sin[ ArcSin[c*x]/2])^2 - (2*(-4 + ArcSin[c*x]^2)*Sin[ArcSin[c*x]/2])/(Cos[ArcSi n[c*x]/2] + Sin[ArcSin[c*x]/2])))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])^2)))/(3*c*d^3*(-1 + c*x))
Time = 1.26 (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.
| \(\displaystyle \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(c d x+d)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {e^3 (1-c x)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {(a+b \arcsin (c x))^2}{(-c x-1) \sqrt {1-c^2 x^2}}+\frac {2 (a+b \arcsin (c x))^2}{(c x+1)^2 \sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {i (a+b \arcsin (c x))^2}{3 c}-\frac {4 b \log \left (1-i e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {\cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {2 b \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}-\frac {\cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}+\frac {4 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-\frac {4 b^2 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\) |
(e^3*(1 - c^2*x^2)^(5/2)*(((I/3)*(a + b*ArcSin[c*x])^2)/c - (4*b^2*Cot[Pi/ 4 + ArcSin[c*x]/2])/(3*c) + ((a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin[c*x]/ 2])/(3*c) - (2*b*(a + b*ArcSin[c*x])*Csc[Pi/4 + ArcSin[c*x]/2]^2)/(3*c) - ((a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin[c*x]/2]*Csc[Pi/4 + ArcSin[c*x]/2] ^2)/(3*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))/((d + c*d*x)^(5/2)*(e - c*e*x)^(5/2))
3.6.45.3.1 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]
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2} \sqrt {-c e x +e}}{\left (c d x +d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sqr t(-c*e*x + e)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)
\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int \frac {\sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {e-c\,e\,x}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \]