\(\int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx\) [92]

Optimal result

Integrand size = 32, antiderivative size = 742 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=-\frac {b^2}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 x}{3 d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b (a+b \arcsin (c x))}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2}}+\frac {b x (a+b \arcsin (c x))}{3 d^2 e \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2}}+\frac {2 x (a+b \arcsin (c x))^2}{3 d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {(a+b \arcsin (c x))^2}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}+\frac {x (a+b \arcsin (c x))^2}{3 d^2 e \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}-\frac {2 i \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c d^2 e \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 9.23 (sec) , antiderivative size = 739, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(a + b*ArcSin[c*x])^2/((d + c*d*x)^(5/2)*(e - c*e*x)^(3/2)),x]
 

Output:

(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*(-1/4*a^2/(d^3*e^2*(-1 + c*x)) - 
a^2/(6*d^3*e^2*(1 + c*x)^2) - (5*a^2)/(12*d^3*e^2*(1 + c*x))))/c + (a*b*Sq 
rt[d + c*d*x]*Sqrt[e - c*e*x]*(2*ArcSin[c*x]*(-2*c*x + Cos[2*ArcSin[c*x]]) 
 - Sqrt[1 - c^2*x^2]*(-1 + 3*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] 
+ 5*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + c*x*(3*Log[Cos[ArcSin[c 
*x]/2] - Sin[ArcSin[c*x]/2]] + 5*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/ 
2]]))))/(3*c*d^2*e*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[-(d*e*(1 - c^2*x^2) 
)]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2) + (b^2*Sqrt[d + c*d*x]*Sqr 
t[e - c*e*x]*Sqrt[1 - c^2*x^2]*((-7*I)*Pi*ArcSin[c*x] + (1 + 4*I)*ArcSin[c 
*x]^2 - 16*Pi*Log[1 + E^((-I)*ArcSin[c*x])] - 5*(Pi + 2*ArcSin[c*x])*Log[1 
 - I*E^(I*ArcSin[c*x])] + 3*(Pi - 2*ArcSin[c*x])*Log[1 + I*E^(I*ArcSin[c*x 
])] + 16*Pi*Log[Cos[ArcSin[c*x]/2]] - 3*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4 
]] + 5*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] + (6*I)*PolyLog[2, (-I)*E^(I*Ar 
cSin[c*x])] + (10*I)*PolyLog[2, I*E^(I*ArcSin[c*x])] - (3*ArcSin[c*x]^2*Si 
n[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]) - (2*ArcSin[c* 
x]^2*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^3 + (Ar 
cSin[c*x]*(2 + ArcSin[c*x]))/(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^2 - 
 ((4 + 5*ArcSin[c*x]^2)*Sin[ArcSin[c*x]/2])/(Cos[ArcSin[c*x]/2] + Sin[ArcS 
in[c*x]/2])))/(6*c*d^2*e*Sqrt[(-d - c*d*x)*(e - c*e*x)]*Sqrt[-(d*e*(1 - c^ 
2*x^2))])
 

Rubi [A] (verified)

Time = 1.29 (sec) , antiderivative size = 386, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5178, 27, 5262, 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.

\(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{(c d x+d)^{5/2} (e-c e x)^{3/2}} \, dx\)

\(\Big \downarrow \) 5178

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {e (1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e \left (1-c^2 x^2\right )^{5/2} \int \frac {(1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 5262

\(\displaystyle \frac {e \left (1-c^2 x^2\right )^{5/2} \int \left (\frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}-\frac {c x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{5/2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*ArcSin[c*x])^2/((d + c*d*x)^(5/2)*(e - c*e*x)^(3/2)),x]
 

Output:

(e*(1 - c^2*x^2)^(5/2)*(-1/3*b^2/(c*Sqrt[1 - c^2*x^2]) + (b^2*x)/(3*Sqrt[1 
 - c^2*x^2]) - (b*(a + b*ArcSin[c*x]))/(3*c*(1 - c^2*x^2)) + (b*x*(a + b*A 
rcSin[c*x]))/(3*(1 - c^2*x^2)) - (((2*I)/3)*(a + b*ArcSin[c*x])^2)/c - (a 
+ b*ArcSin[c*x])^2/(3*c*(1 - c^2*x^2)^(3/2)) + (x*(a + b*ArcSin[c*x])^2)/( 
3*(1 - c^2*x^2)^(3/2)) + (2*x*(a + b*ArcSin[c*x])^2)/(3*Sqrt[1 - c^2*x^2]) 
 - (((2*I)/3)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (4*b*(a 
 + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(3*c) + ((I/3)*b^2*PolyL 
og[2, (-I)*E^(I*ArcSin[c*x])])/c - ((I/3)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x 
])])/c - (((2*I)/3)*b^2*PolyLog[2, -E^((2*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 5262

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_ 
) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + 
 b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] & 
& EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ 
[n, 0] && (m == 1 || p > 0 || (n == 1 && p > -1) || (m == 2 && p < -2))
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3172 vs. \(2 (685 ) = 1370\).

Time = 3.44 (sec) , antiderivative size = 3173, normalized size of antiderivative = 4.28

method	result	size
default	\(\text {Expression too large to display}\)	\(3173\)
parts	\(\text {Expression too large to display}\)	\(3173\)

Input:

int((a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

4/3*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2- 
9*c*x-5)/d^3/e^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x-8/3*b^2*(-e*(c*x-1))^(1/ 
2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2*c^2*arc 
sin(c*x)^2*x^3-6*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3 
*x^3+c^2*x^2-9*c*x-5)/d^3/e^2*c*arcsin(c*x)^2*x^2+5/3*b^2*(-e*(c*x-1))^(1/ 
2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2/c*arcsi 
n(c*x)*(-c^2*x^2+1)^(1/2)+10/3*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4 
*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2*c*(-c^2*x^2+1)*x^2+4/3*b^2*(-e 
*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^ 
3/e^2*c^2*(-c^2*x^2+1)*x^3-2/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/ 
(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2*arcsin(c*x)*x-3*I*b^2*(-e*(c 
*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e 
^2*(-c^2*x^2+1)^(1/2)*x-10/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4 
*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2*c*arcsin(c*x)*x^2-4/3*I*b^2*(- 
e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d 
^3/e^2*c*(-c^2*x^2+1)^(1/2)*x^2-10/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^ 
(1/2)/(4*c^4*x^4+9*c^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2/c*arcsin(c*x)^2*(-c^2* 
x^2+1)^(1/2)-2/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(4*c^4*x^4+9*c 
^3*x^3+c^2*x^2-9*c*x-5)/d^3/e^2/c*arcsin(c*x)*(-c^2*x^2+1)+4/3*I*b^2*(-e*( 
c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^3/e^2/c/(c^2*x^2-1...
 

Fricas [F]

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

integrate((a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(3/2),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^5*d^3*e^2*x^5 + c^4*d^3*e^2*x^4 - 2*c^3*d^3*e^2*x^3 - 2*c 
^2*d^3*e^2*x^2 + c*d^3*e^2*x + d^3*e^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+b*asin(c*x))**2/(c*d*x+d)**(5/2)/(-c*e*x+e)**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

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

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

integrate((a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(3/2),x, algorith 
m="giac")
 

Output:

integrate((b*arcsin(c*x) + a)^2/((c*d*x + d)^(5/2)*(-c*e*x + e)^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{5/2}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \] Input:

int((a + b*asin(c*x))^2/((d + c*d*x)^(5/2)*(e - c*e*x)^(3/2)),x)
 

Output:

int((a + b*asin(c*x))^2/((d + c*d*x)^(5/2)*(e - c*e*x)^(3/2)), x)
 

Reduce [F]

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

int((a+b*asin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

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