Integrand size = 32, antiderivative size = 714 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {2 a b e^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b^2 e^3 \left (1-c^2 x^2\right )^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b^2 e^3 x \left (1-c^2 x^2\right )^{3/2} \arcsin (c x)}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 e^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 e^3 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i e^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {e^3 \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {e^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x))^3}{b c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {16 i b e^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 b e^3 \left (1-c^2 x^2\right )^{3/2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {8 i b^2 e^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {8 i b^2 e^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {4 i b^2 e^3 \left (1-c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]
2*a*b*e^3*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+2*b^2*e^3* (-c^2*x^2+1)^2/c/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+2*b^2*e^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*e^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*e^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*e^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)-e^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)-e^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*e ^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*e^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*e^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*e^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* e^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 = 12.92 (sec) , antiderivative size = 1086, normalized size of antiderivative = 1.52 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx =\text {Too large to display} \]
(-3*a^2*e*(5 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2]*(Cos [ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 9*a^2*Sqrt[d]*e^(3/2)*(1 + c*x)*Sq rt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt [e]*(-1 + c^2*x^2))]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 3*a*b*e*( 1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(Cos[ArcSin[c*x]/2]*(ArcSin[c*x]* (4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + ((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2 ]])*Sin[ArcSin[c*x]/2]) - b^2*e*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]* ((6 + 6*I)*ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2] + I*Sin[ArcSin[c*x]/2]) + Arc Sin[c*x]^3*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - (6*I)*ArcSin[c*x]*( Pi - (4*I)*Log[1 - I*E^(I*ArcSin[c*x])])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[ c*x]/2]) - 12*Pi*(2*Log[1 + E^((-I)*ArcSin[c*x])] + Log[1 - I*E^(I*ArcSin[ c*x])] - 2*Log[Cos[ArcSin[c*x]/2]] - Log[Sin[(Pi + 2*ArcSin[c*x])/4]])*(Co s[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + (24*I)*PolyLog[2, I*E^(I*ArcSin[c *x])]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])) - 6*a*b*e*(1 + c*x)*Sqrt[ d + c*d*x]*Sqrt[e - c*e*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[Ar cSin[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])) - b^2*e *(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(2*ArcSin[c*x]^3*(Cos[ArcSin...
Time = 1.19 (sec) , antiderivative size = 327, 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.
| \(\displaystyle \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(c d x+d)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {e^3 (1-c x)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x)^3 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 5274 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{3/2} \int \left (\frac {c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {3 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {4 (1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {16 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {4 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {4 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^3}{b c}-\frac {4 i (a+b \arcsin (c x))^2}{c}+\frac {8 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+2 a b x+\frac {8 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}-\frac {8 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}+2 b^2 x \arcsin (c x)+\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\) |
(e^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] - (S qrt[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*A rcSin[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))
3.6.50.3.1 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]
\[\int \frac {\left (-c e x +e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
integral(-(a^2*c*e*x - a^2*e + (b^2*c*e*x - b^2*e)*arcsin(c*x)^2 + 2*(a*b* c*e*x - a*b*e)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-c*e*x + e)/(c^2*d^2*x^2 + 2*c*d^2*x + d^2), x)
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (e-c\,e\,x\right )}^{3/2}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \]