\(\int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx\) [79]

Optimal result

Integrand size = 32, antiderivative size = 685 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {2 b^2 e^2 \left (1-c^2 x^2\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b e^2 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 e^2 (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 e^2 x (a+b \arcsin (c x))^2}{d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 i e^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {e^2 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {e^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{b c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {16 i b e^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {8 b e^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {8 i b^2 e^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {8 i b^2 e^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 i b^2 e^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

2*b^2*e^2*(-c^2*x^2+1)/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*b*e^2*x*(-c^ 
2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-4*e^2* 
(a+b*arcsin(c*x))^2/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+4*e^2*x*(a+b*arcs 
in(c*x))^2/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-4*I*e^2*(-c^2*x^2+1)^(1/2)*( 
a+b*arcsin(c*x))^2/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-e^2*(-c^2*x^2+1)*( 
a+b*arcsin(c*x))^2/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-e^2*(-c^2*x^2+1)^( 
1/2)*(a+b*arcsin(c*x))^3/b/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-16*I*b*e^2 
*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c/d 
/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+8*b*e^2*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c 
*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/ 
2)+8*I*b^2*e^2*(-c^2*x^2+1)^(1/2)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2))) 
/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-8*I*b^2*e^2*(-c^2*x^2+1)^(1/2)*polyl 
og(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-4* 
I*b^2*e^2*(-c^2*x^2+1)^(1/2)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c/d/ 
(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)
 

Mathematica [A] (verified)

Time = 10.90 (sec) , antiderivative size = 1086, normalized size of antiderivative = 1.59 \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[((e - c*e*x)^(3/2)*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(3/2),x]
 

Output:

(-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...
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.48, 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

\(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {16 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {4 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {4 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {(a+b \arcsin (c x))^3}{b c}-\frac {4 i (a+b \arcsin (c x))^2}{c}+\frac {8 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+2 a b x+\frac {8 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}-\frac {8 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {4 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}+2 b^2 x \arcsin (c x)+\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

Input:

Int[((e - c*e*x)^(3/2)*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(3/2),x]
 

Output:

(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))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5178

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]
 

rule 5274

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]
 

Maple [A] (verified)

Time = 4.05 (sec) , antiderivative size = 564, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, \left (a +b \arcsin \left (c x \right )\right )^{3} e}{\left (c x +1\right ) \left (c x -1\right ) d^{2} c b}-\frac {\sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, c x -1\right ) \left (2 i b^{2} \arcsin \left (c x \right )+\arcsin \left (c x \right )^{2} b^{2}+2 i a b +2 \arcsin \left (c x \right ) a b +a^{2}-2 b^{2}\right ) e}{2 \left (c x +1\right ) \left (c x -1\right ) d^{2} c}-\frac {\sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, c x +c^{2} x^{2}-1\right ) \left (-2 i b^{2} \arcsin \left (c x \right )+\arcsin \left (c x \right )^{2} b^{2}-2 i a b +2 \arcsin \left (c x \right ) a b +a^{2}-2 b^{2}\right ) e}{2 \left (c x +1\right ) \left (c x -1\right ) d^{2} c}-\frac {4 e \left (\arcsin \left (c x \right )^{2} b^{2}+2 \arcsin \left (c x \right ) a b +a^{2}\right ) \left (i \sqrt {-c^{2} x^{2}+1}+c x -1\right ) \sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}}{d^{2} c \left (c^{2} x^{2}-1\right )}+\frac {8 i \sqrt {-c^{2} x^{2}+1}\, \sqrt {d \left (c x +1\right )}\, \sqrt {-e \left (c x -1\right )}\, b \left (2 i \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) b +\arcsin \left (c x \right )^{2} b +i \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right ) a -2 i a \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 \arctan \left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) a +2 \operatorname {polylog}\left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right ) b \right ) e}{\left (c x +1\right ) \left (c x -1\right ) d^{2} c}\)	\(564\)

Input:

int((-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

(-c^2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x+1)/(c*x-1)/d^ 
2/c*(a+b*arcsin(c*x))^3*e/b-1/2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(c^2* 
x^2-I*c*x*(-c^2*x^2+1)^(1/2)-1)*(2*I*b^2*arcsin(c*x)+arcsin(c*x)^2*b^2+2*I 
*a*b+2*arcsin(c*x)*a*b+a^2-2*b^2)*e/(c*x+1)/(c*x-1)/d^2/c-1/2*(-e*(c*x-1)) 
^(1/2)*(d*(c*x+1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*c*x+c^2*x^2-1)*(-2*I*b^2*ar 
csin(c*x)+arcsin(c*x)^2*b^2-2*I*b*a+2*arcsin(c*x)*a*b+a^2-2*b^2)*e/(c*x+1) 
/(c*x-1)/d^2/c-4*e*(arcsin(c*x)^2*b^2+2*arcsin(c*x)*a*b+a^2)*(I*(-c^2*x^2+ 
1)^(1/2)+c*x-1)*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/d^2/c/(c^2*x^2-1)+8*I 
*(-c^2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x+1)/(c*x-1)/d 
^2/c*b*(2*I*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*b+arcsin(c*x)^2 
*b+I*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*a-2*I*a*ln(I*c*x+(-c^2*x^2+1)^(1/2 
))+2*arctan(I*c*x+(-c^2*x^2+1)^(1/2))*a+2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^ 
(1/2)))*b)*e
 

Fricas [F]

\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((-c*e*x+e)**(3/2)*(a+b*asin(c*x))**2/(c*d*x+d)**(3/2),x)
 

Output:

Integral((-e*(c*x - 1))**(3/2)*(a + b*asin(c*x))**2/(d*(c*x + 1))**(3/2), 
x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="maxima")
 

Output:

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
 

Giac [F]

\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int { \frac {{\left (-c e x + e\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-c*e*x+e)^(3/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(3/2),x, algorith 
m="giac")
 

Output:

integrate((-c*e*x + e)^(3/2)*(b*arcsin(c*x) + a)^2/(c*d*x + d)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (e-c\,e\,x\right )}^{3/2}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \] Input:

int(((a + b*asin(c*x))^2*(e - c*e*x)^(3/2))/(d + c*d*x)^(3/2),x)
 

Output:

int(((a + b*asin(c*x))^2*(e - c*e*x)^(3/2))/(d + c*d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(e-c e x)^{3/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {\sqrt {e}\, e \left (6 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a^{2}-\sqrt {-c x +1}\, a^{2} c x -5 \sqrt {-c x +1}\, a^{2}-2 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b \,c^{2}+2 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b c -\sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2} x}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c^{2}+\sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2}}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c \right )}{\sqrt {d}\, \sqrt {c x +1}\, c d} \] Input:

int((-c*e*x+e)^(3/2)*(a+b*asin(c*x))^2/(c*d*x+d)^(3/2),x)
 

Output:

(sqrt(e)*e*(6*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2 - sqrt( - 
c*x + 1)*a**2*c*x - 5*sqrt( - c*x + 1)*a**2 - 2*sqrt(c*x + 1)*int((sqrt( - 
 c*x + 1)*asin(c*x)*x)/(sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c**2 + 2 
*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x + 1)*c*x + sqrt( 
c*x + 1)),x)*a*b*c - sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2*x)/( 
sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c**2 + sqrt(c*x + 1)*int((sqrt( 
 - c*x + 1)*asin(c*x)**2)/(sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c))/ 
(sqrt(d)*sqrt(c*x + 1)*c*d)