\(\int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx\) [86]

Optimal result

Integrand size = 32, antiderivative size = 717 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=-\frac {2 b^2 e^3 \left (1-c^2 x^2\right )}{c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b e^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {28 i e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {5 e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {16 b^2 e^3 \sqrt {1-c^2 x^2} \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {28 e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {8 b e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} \arcsin (c x)\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {112 b e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1-i e^{i \arcsin (c x)}\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {112 i b^2 e^3 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c d^2 \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

-2*b^2*e^3*(-c^2*x^2+1)/c/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2*b*e^3*x*( 
-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2 
8/3*I*e^3*(-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)+e^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c/d^2/(c*d*x+d)^(1/2)/( 
-c*e*x+e)^(1/2)+5/3*e^3*(-c^2*x^2+1)^(1/2)*(a+b*arcsin(c*x))^3/b/c/d^2/(c* 
d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-16/3*b^2*e^3*(-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)+28/3*e^3*(-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)-8/3*b*e^3*(-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)-4/3*e^3*(- 
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*P 
i+1/2*arcsin(c*x))^2/c/d^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-112/3*b*e^3*(- 
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)+112/3*I*b^2*e^3*(-c^2*x^2+1)^(1/2)*poly 
log(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 [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2338\) vs. \(2(717)=1434\).

Time = 22.06 (sec) , antiderivative size = 2338, normalized size of antiderivative = 3.26 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)]*((a^2*e^2)/d^3 - (8*a^2*e^2)/(3*d 
^3*(1 + c*x)^2) + (28*a^2*e^2)/(3*d^3*(1 + c*x))))/c - (5*a^2*e^(5/2)*ArcT 
an[(c*x*Sqrt[-(e*(-1 + c*x))]*Sqrt[d*(1 + c*x)])/(Sqrt[d]*Sqrt[e]*(-1 + c* 
x)*(1 + c*x))])/(c*d^(5/2)) - (a*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqr 
t[-(d*e*(1 - c^2*x^2))]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[Arc 
Sin[c*x]/2]*(-8 + 6*ArcSin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos[ArcSin[c*x] 
/2] + Sin[ArcSin[c*x]/2]]) + Cos[(3*ArcSin[c*x])/2]*((14 - 3*ArcSin[c*x])* 
ArcSin[c*x] + 28*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + 2*(-4 + 4 
*ArcSin[c*x] + 6*ArcSin[c*x]^2 + Sqrt[1 - c^2*x^2]*(ArcSin[c*x]*(14 + 3*Ar 
cSin[c*x]) - 28*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) - 56*Log[Cos 
[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(3*c*d^3*(-1 + 
 c*x)*Sqrt[(-d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x] 
/2])^4) - (a*b*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[-(d*e*(1 - c^2*x^2 
))]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(Cos[(3*ArcSin[c*x])/2]*(Arc 
Sin[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*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*ArcSin[c*x] - 4*Log[Cos[ArcSi 
n[c*x]/2] + Sin[ArcSin[c*x]/2]] - 2*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/ 
2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(3*c*d^3*(-1 + c*x)*Sqrt[(- 
d - c*d*x)*(e - c*e*x)]*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) - ...
 

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 344, 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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5274

\(\displaystyle \frac {e^5 \left (1-c^2 x^2\right )^{5/2} \int \left (-\frac {c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {12 (a+b \arcsin (c x))^2}{(c x+1) \sqrt {1-c^2 x^2}}+\frac {8 (a+b \arcsin (c x))^2}{(c x+1)^2 \sqrt {1-c^2 x^2}}+\frac {5 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}\right )dx}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^5 \left (1-c^2 x^2\right )^{5/2} \left (\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {5 (a+b \arcsin (c x))^3}{3 b c}+\frac {28 i (a+b \arcsin (c x))^2}{3 c}-\frac {112 b \log \left (1-i e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{3 c}+\frac {28 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-\frac {8 b \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))}{3 c}-\frac {4 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) \csc ^2\left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right ) (a+b \arcsin (c x))^2}{3 c}-2 a b x+\frac {112 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{3 c}-2 b^2 x \arcsin (c x)-\frac {16 b^2 \cot \left (\frac {1}{2} \arcsin (c x)+\frac {\pi }{4}\right )}{3 c}-\frac {2 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

Input:

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

Output:

(e^5*(1 - c^2*x^2)^(5/2)*(-2*a*b*x - (2*b^2*Sqrt[1 - c^2*x^2])/c - 2*b^2*x 
*ArcSin[c*x] + (((28*I)/3)*(a + b*ArcSin[c*x])^2)/c + (Sqrt[1 - c^2*x^2]*( 
a + b*ArcSin[c*x])^2)/c + (5*(a + b*ArcSin[c*x])^3)/(3*b*c) - (16*b^2*Cot[ 
Pi/4 + ArcSin[c*x]/2])/(3*c) + (28*(a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin 
[c*x]/2])/(3*c) - (8*b*(a + b*ArcSin[c*x])*Csc[Pi/4 + ArcSin[c*x]/2]^2)/(3 
*c) - (4*(a + b*ArcSin[c*x])^2*Cot[Pi/4 + ArcSin[c*x]/2]*Csc[Pi/4 + ArcSin 
[c*x]/2]^2)/(3*c) - (112*b*(a + b*ArcSin[c*x])*Log[1 - I*E^(I*ArcSin[c*x]) 
])/(3*c) + (((112*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 [A] (verified)

Time = 4.10 (sec) , antiderivative size = 905, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {5 \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^{2}}{3 \left (c x +1\right ) d^{3} \left (c x -1\right ) 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}}{2 \left (c x +1\right ) d^{3} \left (c x -1\right ) 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}}{2 \left (c x +1\right ) d^{3} \left (c x -1\right ) c}+\frac {4 e^{2} \left (-14 i a b +63 \arcsin \left (c x \right )^{2} b^{2} c^{2} x^{2}-14 i b^{2} \arcsin \left (c x \right )+126 a b \,c^{2} x^{2} \arcsin \left (c x \right )-28 i \arcsin \left (c x \right ) b^{2} c x -14 i a b \,c^{2} x^{2}+96 c \arcsin \left (c x \right )^{2} b^{2} x -14 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) b^{2} c x +63 a^{2} c^{2} x^{2}-32 x^{2} c^{2} b^{2}-4 i \sqrt {-c^{2} x^{2}+1}\, b^{2} c x +192 c \arcsin \left (c x \right ) a b x -14 \sqrt {-c^{2} x^{2}+1}\, a b c x -28 i a b c x -14 i \arcsin \left (c x \right ) b^{2} c^{2} x^{2}+37 \arcsin \left (c x \right )^{2} b^{2}-10 \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) b^{2}+96 c x \,a^{2}-56 x c \,b^{2}-4 i \sqrt {-c^{2} x^{2}+1}\, b^{2}+74 \arcsin \left (c x \right ) a b -10 \sqrt {-c^{2} x^{2}+1}\, a b +37 a^{2}-24 b^{2}\right ) \left (7 i \sqrt {-c^{2} x^{2}+1}\, c x +7 c^{2} x^{2}+7 i \sqrt {-c^{2} x^{2}+1}-2 c x -5\right ) \sqrt {-e \left (c x -1\right )}\, \sqrt {d \left (c x +1\right )}}{3 \left (63 c^{5} x^{5}+159 c^{4} x^{4}+70 c^{3} x^{3}-122 c^{2} x^{2}-133 c x -37\right ) c \,d^{3}}-\frac {56 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^{2}}{3 \left (c x +1\right ) d^{3} \left (c x -1\right ) c}\)	\(905\)

Input:

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

Output:

-5/3*(-c^2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x+1)/d^3/( 
c*x-1)/c*(a+b*arcsin(c*x))^3*e^2/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^2/(c*x+1)/d^3/(c*x-1)/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*arcsin(c*x)+arcsin(c*x)^2*b^2-2*I*b*a+2*arcsin(c*x)*a*b+a^2-2*b^2)* 
e^2/(c*x+1)/d^3/(c*x-1)/c+4/3*e^2*(-14*I*a*b+63*arcsin(c*x)^2*b^2*c^2*x^2- 
14*I*b^2*arcsin(c*x)+126*a*b*c^2*x^2*arcsin(c*x)-28*I*arcsin(c*x)*b^2*c*x- 
14*I*a*b*c^2*x^2+96*c*arcsin(c*x)^2*b^2*x-14*(-c^2*x^2+1)^(1/2)*arcsin(c*x 
)*b^2*c*x+63*a^2*c^2*x^2-32*x^2*c^2*b^2-4*I*(-c^2*x^2+1)^(1/2)*b^2*c*x+192 
*c*arcsin(c*x)*a*b*x-14*(-c^2*x^2+1)^(1/2)*a*b*c*x-28*I*a*b*c*x-14*I*arcsi 
n(c*x)*b^2*c^2*x^2+37*arcsin(c*x)^2*b^2-10*(-c^2*x^2+1)^(1/2)*arcsin(c*x)* 
b^2+96*c*x*a^2-56*x*c*b^2-4*I*(-c^2*x^2+1)^(1/2)*b^2+74*arcsin(c*x)*a*b-10 
*(-c^2*x^2+1)^(1/2)*a*b+37*a^2-24*b^2)*(7*I*(-c^2*x^2+1)^(1/2)*x*c+7*c^2*x 
^2+7*I*(-c^2*x^2+1)^(1/2)-2*c*x-5)*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/(6 
3*c^5*x^5+159*c^4*x^4+70*c^3*x^3-122*c^2*x^2-133*c*x-37)/c/d^3-56/3*I*(-c^ 
2*x^2+1)^(1/2)*(d*(c*x+1))^(1/2)*(-e*(c*x-1))^(1/2)/(c*x+1)/d^3/(c*x-1)/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...
 

Fricas [F]

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

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

Output:

integral((a^2*c^2*e^2*x^2 - 2*a^2*c*e^2*x + a^2*e^2 + (b^2*c^2*e^2*x^2 - 2 
*b^2*c*e^2*x + b^2*e^2)*arcsin(c*x)^2 + 2*(a*b*c^2*e^2*x^2 - 2*a*b*c*e^2*x 
 + a*b*e^2)*arcsin(c*x))*sqrt(c*d*x + d)*sqrt(-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(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((-c*e*x+e)^(5/2)*(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(e)*e**2*( - 30*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2*c*x 
 - 30*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2 + 3*sqrt( - c*x + 
1)*a**2*c**2*x**2 + 34*sqrt( - c*x + 1)*a**2*c*x + 23*sqrt( - c*x + 1)*a** 
2 + 6*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x**2)/(sqrt(c*x + 1)*c 
**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c**4*x + 6*sqrt(c*x 
 + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x**2)/(sqrt(c*x + 1)*c**2*x**2 + 2*s 
qrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c**3 - 12*sqrt(c*x + 1)*int((sqrt 
( - c*x + 1)*asin(c*x)*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*x - 12*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asi 
n(c*x)*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 + 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*sqr 
t(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x))/(sqrt(c*x + 1)*c**2*x**2 + 2*s 
qrt(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*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**4*x + 3*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*a 
sin(c*x)**2*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 - 6*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2* 
x)/(sqrt(c*x + 1)*c**2*x**2 + 2*sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2 
*c**3*x - 6*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2*x)/(sqrt(c...