Integrand size = 32, antiderivative size = 713 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=-\frac {2 a b d^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 b^2 d^3 \left (1-c^2 x^2\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 b^2 d^3 x \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)}{(d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 d^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 d^3 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i d^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d^3 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {d^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^3}{b c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {16 i b d^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 b d^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 i b^2 d^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 i b^2 d^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i b^2 d^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]
-2*a*b*d^3*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-2*b^2*d^3 *(-c^2*x^2+1)^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-2*b^2*d^3*x*(-c^2*x^2+1 )^(3/2)*arcsin(c*x)/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+4*d^3*(-c^2*x^2+1)*(a +b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+4*d^3*x*(-c^2*x^2+1)* (a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-4*I*d^3*(-c^2*x^2+1)^ (3/2)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+d^3*(-c^2*x^2 +1)^2*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-d^3*(-c^2*x^2 +1)^(3/2)*(a+b*arcsin(c*x))^3/b/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+16*I*b* d^3*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/ c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+8*b*d^3*(-c^2*x^2+1)^(3/2)*(a+b*arcsin( c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2 )-8*I*b^2*d^3*(-c^2*x^2+1)^(3/2)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/ c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+8*I*b^2*d^3*(-c^2*x^2+1)^(3/2)*polylog( 2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)-4*I*b^2 *d^3*(-c^2*x^2+1)^(3/2)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/(c*d*x+ d)^(3/2)/(-c*e*x+e)^(3/2)
Time = 15.84 (sec) , antiderivative size = 1255, normalized size of antiderivative = 1.76 \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx =\text {Too large to display} \]
(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((a^2*d)/e^2 - (4*a^2*d)/(e^2*(-1 + c*x))))/c + (3*a^2*d^(3/2)*ArcTan[(c*x*Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[e]*(-1 + c*x)*(1 + c*x))])/(c*e^(3/2)) - (a*b*d*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]*(Cos[Arc Sin[c*x]/2]*((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[ArcSin[c*x]/2] - S in[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]))/(c*e^2*Sqrt[(-d - c*d*x) *(e - c*e*x)]*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*a*b*d*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2]*(-( c*x) + 2*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*ArcSin[c*x] - ArcSin[c*x]^2 + 4*L og[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]]) + (c*x + 2*ArcSin[c*x] - Sqrt [1 - c^2*x^2]*ArcSin[c*x] + ArcSin[c*x]^2 - 4*Log[Cos[ArcSin[c*x]/2] - Sin [ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(c*e^2*Sqrt[(-d - c*d*x)*(e - c*e*x )]*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^2*d*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[ e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2))]*((-18*I)*Pi*ArcSin[c*x] - (6 - 6*I)* ArcSin[c*x]^2 + ArcSin[c*x]^3 - 24*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 12*( Pi - 2*ArcSin[c*x])*Log[1 + I*E^(I*ArcSin[c*x])] + 24*Pi*Log[Cos[ArcSin[c* x]/2]] - 12*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + (24*I)*PolyLog[2, (-...
Time = 1.14 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.46, 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.
| \(\displaystyle \int \frac {(c d x+d)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {d^3 (c x+1)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \int \frac {(c x+1)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \int \left (-\frac {c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {3 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {4 (c x+1) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (1-c^2 x^2\right )^{3/2} \left (\frac {16 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {4 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {4 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^3}{b c}-\frac {4 i (a+b \arcsin (c x))^2}{c}+\frac {8 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-2 a b x-\frac {8 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}+\frac {8 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}-2 b^2 x \arcsin (c x)-\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
(d^3*(1 - c^2*x^2)^(3/2)*(-2*a*b*x - (2*b^2*Sqrt[1 - c^2*x^2])/c - 2*b^2*x *ArcSin[c*x] - ((4*I)*(a + b*ArcSin[c*x])^2)/c + (4*(a + b*ArcSin[c*x])^2) /(c*Sqrt[1 - c^2*x^2]) + (4*x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] + ( Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/c - (a + b*ArcSin[c*x])^3/(b*c) + ((16*I)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (8*b*(a + b* ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/c - ((8*I)*b^2*PolyLog[2, (-I )*E^(I*ArcSin[c*x])])/c + ((8*I)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c - ((4*I)*b^2*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/c))/((d + c*d*x)^(3/2)*(e - c*e*x)^(3/2))
3.6.65.3.1 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]
\[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
integral((a^2*c*d*x + a^2*d + (b^2*c*d*x + b^2*d)*arcsin(c*x)^2 + 2*(a*b*c *d*x + a*b*d)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^2*e^2*x^2 - 2*c*e^2*x + e^2), x)
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int \frac {\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int { \frac {{\left (c d x + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+c d x)^{3/2} (a+b \arcsin (c x))^2}{(e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\right )}^{3/2}}{{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]