\(\int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx\) [74]

Optimal result

Integrand size = 32, antiderivative size = 493 \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\frac {i e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 b^2 e \sqrt {1-c^2 x^2} \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b e \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {e \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 b e \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 i b^2 e \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

1/3*I*e*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))^2/c/d^2/(c*d*x+d)^(1/2)/(-c*e 
*x+e)^(1/2)-4/3*b^2*e*(-c^2*x^2+1)^(1/2)*cot(1/4*Pi+1/2*arcsin(c*x))/c/d^2 
/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/3*e*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x 
))^2*cot(1/4*Pi+1/2*arcsin(c*x))/c/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2/ 
3*b*e*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))*csc(1/4*Pi+1/2*arcsin(c*x))^2/c 
/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/3*e*(-c^2*x^2+1)^(1/2)*(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/d^2/( 
c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-4/3*b*e*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x 
))*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/ 
2)+4/3*I*b^2*e*(-c^2*x^2+1)^(1/2)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/ 
c/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 7.87 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.07 \[ \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)} \] Input:

Integrate[(Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(5/2),x]
 

Output:

(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))
 

Rubi [A] (verified)

Time = 1.36 (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

\(\displaystyle \frac {e^3 \left (1-c^2 x^2\right )^{5/2} \left (\frac {i (a+b \arcsin (c x))^2}{3 c}-\frac {4 b \log \left (1-i e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {\cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {2 b \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}-\frac {\cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}+\frac {4 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-\frac {4 b^2 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

Input:

Int[(Sqrt[e - c*e*x]*(a + b*ArcSin[c*x])^2)/(d + c*d*x)^(5/2),x]
 

Output:

(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))
 

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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1137 vs. \(2 (429 ) = 858\).

Time = 3.93 (sec) , antiderivative size = 1138, normalized size of antiderivative = 2.31

method	result	size
default	\(\text {Expression too large to display}\)	\(1138\)
parts	\(\text {Expression too large to display}\)	\(1138\)

Input:

int((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

a^2*(-1/c/d*(-c*e*x+e)^(1/2)/(c*d*x+d)^(3/2)-e*(-1/3/c/d/e/(c*d*x+d)^(3/2) 
*(-c*e*x+e)^(1/2)-1/3/c/e/d^2/(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)))+1/3*b^2*( 
4+2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))+4*c*x+4*I*c*x*(-c^2*x^2+1)^(1/ 
2)+2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+arcsin(c*x)^2+4*c^3*x^3-2*arcsin(c*x)* 
(-c^2*x^2+1)^(1/2)*c*x+4*c^2*x^2+2*I*arcsin(c*x)+3*arcsin(c*x)^2*x^2*c^2+4 
*I*(-c^2*x^2+1)^(1/2)*x^2*c^2+6*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*x^ 
2*c^2+2*I*arcsin(c*x)*x^2*c^2+2*I*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*arcsi 
n(c*x)+6*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*x*c-2*I*(-c^2*x^2+1)^(1/2 
)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I*polylog(2,I*(I*c*x+(-c^2*x^2 
+1)^(1/2)))*(-c^2*x^2+1)^(1/2)*x^2*c^2+4*I*arcsin(c*x)*x*c+2*polylog(2,I*( 
I*c*x+(-c^2*x^2+1)^(1/2)))*x^3*c^3+2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1 
)^(1/2)))*(-c^2*x^2+1)^(1/2)+6*I*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1 
/2)))*c^2*x^2+6*I*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*x*c-2*arc 
sin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)*x^2*c^2+2*I 
*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))*x^3*c^3)*(-I*c*x*(-c^2*x^2 
+1)^(1/2)+c^2*x^2-I*(-c^2*x^2+1)^(1/2)-2*c*x+1)*(-e*(c*x-1))^(1/2)*(d*(c*x 
+1))^(1/2)/(3*c^5*x^5+3*c^4*x^4-2*c^3*x^3-2*c^2*x^2-c*x-1)/c/d^3+2/3*a*b*( 
I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x^3*c^3+3*I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I 
)*c^2*x^2-(-c^2*x^2+1)^(1/2)*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x^2*c^2+I*c^2* 
x^2+3*c^2*x^2*arcsin(c*x)+3*I*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)*x*c+2*I*c*...
 

Fricas [F]

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

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2),x, algorith 
m="fricas")
 

Output:

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)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2),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]

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

integrate((-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2),x, algorith 
m="giac")
 

Output:

integrate(sqrt(-c*e*x + e)*(b*arcsin(c*x) + a)^2/(c*d*x + d)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e-c e x} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {e-c\,e\,x}}{{\left (d+c\,d\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(e)*(sqrt( - c*x + 1)*a**2*c*x - sqrt( - c*x + 1)*a**2 + 6*sqrt(c*x + 
 1)*int((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x 
 + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c**2*x + 6*sqrt(c*x + 1)*int((sqrt( - c* 
x + 1)*asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c* 
x + 1)),x)*a*b*c + 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2)/(sq 
rt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c**2* 
x + 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2)/(sqrt(c*x + 1)*c** 
2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c))/(3*sqrt(d)*sqrt( 
c*x + 1)*c*d**2*(c*x + 1))