\(\int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx\) [85]

Optimal result

Integrand size = 32, antiderivative size = 897 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {8 b^2 e^3 \left (1-c^2 x^2\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b^2 e^3 x \left (1-c^2 x^2\right )}{4 d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 e^3 \sqrt {1-c^2 x^2} \arcsin (c x)}{4 c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {8 b e^3 x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b c e^3 x^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{2 d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {8 e^3 (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {8 e^3 x (a+b \arcsin (c x))^2}{d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {8 i e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {4 e^3 \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {e^3 x \left (1-c^2 x^2\right ) (a+b \arcsin (c x))^2}{2 d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {5 e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{2 b c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {32 i b e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {16 b e^3 \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {16 i b^2 e^3 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {16 i b^2 e^3 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {8 i b^2 e^3 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c d \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

8*b^2*e^3*(-c^2*x^2+1)/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/4*b^2*e^3*x* 
(-c^2*x^2+1)/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/4*b^2*e^3*(-c^2*x^2+1)^( 
1/2)*arcsin(c*x)/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+8*b*e^3*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)-1/2*b*c*e^3* 
x^2*(-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 
)-8*e^3*(a+b*arcsin(c*x))^2/c/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+8*e^3*x*( 
a+b*arcsin(c*x))^2/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-8*I*e^3*(-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)-4*e^3*(-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)+1/2*e^3* 
x*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/d/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-5/2* 
e^3*(-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)-32*I*b*e^3*(-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)+16*b*e^3*(-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^3*(-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)-16*I*b^2*e^3*(-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)+16*I*b^2*e^3*(-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)
 

Mathematica [A] (verified)

Time = 19.58 (sec) , antiderivative size = 1642, normalized size of antiderivative = 1.83 \[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(e^2*(12*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2]*(-24 - 7*c* 
x + c^2*x^2)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + 180*a^2*Sqrt[d]*S 
qrt[e]*(1 + c*x)*Sqrt[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]) - 24*a*b*(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[ArcSi 
n[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]) - 8*b^2*(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[Arc 
Sin[c*x]/2]) + ArcSin[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*Arc 
Sin[c*x])/4]])*(Cos[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])) - 96*a* 
b*(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[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) + ArcSin[c*x]*((2 + Sq 
rt[1 - c^2*x^2])*Cos[ArcSin[c*x]/2] + (-2 + Sqrt[1 - c^2*x^2])*Sin[ArcSin[ 
c*x]/2])) - 16*b^2*(1 + c*x)*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(2*ArcSin[...
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 411, 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.

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

\(\Big \downarrow \) 5178

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {e^4 (1-c x)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^4 \left (1-c^2 x^2\right )^{3/2} \int \frac {(1-c x)^4 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 5274

\(\displaystyle \frac {e^4 \left (1-c^2 x^2\right )^{3/2} \int \left (-\frac {c^2 x^2 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {4 c x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {7 (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}+\frac {8 (1-c x) (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}\right )dx}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (1-c^2 x^2\right )^{3/2} \left (-\frac {32 i b \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+\frac {1}{2} x \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2-\frac {4 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c}+\frac {8 x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {8 (a+b \arcsin (c x))^2}{c \sqrt {1-c^2 x^2}}-\frac {1}{2} b c x^2 (a+b \arcsin (c x))-\frac {5 (a+b \arcsin (c x))^3}{2 b c}-\frac {8 i (a+b \arcsin (c x))^2}{c}+\frac {16 b \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))}{c}+8 a b x+\frac {16 i b^2 \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c}-\frac {16 i b^2 \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c}-\frac {8 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c}+8 b^2 x \arcsin (c x)+\frac {b^2 \arcsin (c x)}{4 c}-\frac {1}{4} b^2 x \sqrt {1-c^2 x^2}+\frac {8 b^2 \sqrt {1-c^2 x^2}}{c}\right )}{(c d x+d)^{3/2} (e-c e x)^{3/2}}\)

Input:

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

Output:

(e^4*(1 - c^2*x^2)^(3/2)*(8*a*b*x + (8*b^2*Sqrt[1 - c^2*x^2])/c - (b^2*x*S 
qrt[1 - c^2*x^2])/4 + (b^2*ArcSin[c*x])/(4*c) + 8*b^2*x*ArcSin[c*x] - (b*c 
*x^2*(a + b*ArcSin[c*x]))/2 - ((8*I)*(a + b*ArcSin[c*x])^2)/c - (8*(a + b* 
ArcSin[c*x])^2)/(c*Sqrt[1 - c^2*x^2]) + (8*x*(a + b*ArcSin[c*x])^2)/Sqrt[1 
 - c^2*x^2] - (4*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/c + (x*Sqrt[1 - 
c^2*x^2]*(a + b*ArcSin[c*x])^2)/2 - (5*(a + b*ArcSin[c*x])^3)/(2*b*c) - (( 
32*I)*b*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/c + (16*b*(a + b*Ar 
cSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/c + ((16*I)*b^2*PolyLog[2, (-I) 
*E^(I*ArcSin[c*x])])/c - ((16*I)*b^2*PolyLog[2, I*E^(I*ArcSin[c*x])])/c - 
((8*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 = 3.93 (sec) , antiderivative size = 1078, normalized size of antiderivative = 1.20

method	result	size
default	\(\text {Expression too large to display}\)	\(1078\)

Input:

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

Output:

5/2*(-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^2/b+1/32*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2 
)*(4*c^3*x^3-2*c^2*x^2-4*I*x^2*c^2*(-c^2*x^2+1)^(1/2)-3*c*x+2*I*(-c^2*x^2+ 
1)^(1/2)*c*x+1+I*(-c^2*x^2+1)^(1/2))*(2*I*b^2*arcsin(c*x)+2*arcsin(c*x)^2* 
b^2+2*I*a*b+4*arcsin(c*x)*a*b+2*a^2-b^2)*e^2/(c*x+1)/(c*x-1)/d^2/c-(-e*(c* 
x-1))^(1/2)*(d*(c*x+1))^(1/2)*(-I*(-c^2*x^2+1)^(1/2)+c*x-1)*(2*I*b^2*arcsi 
n(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)/ 
(c*x-1)/d^2/c-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)/(c*x-1)/d^2/c+1/32*(-e*(c*x-1))^(1/2)*(d*(c 
*x+1))^(1/2)*(2*I*(-c^2*x^2+1)^(1/2)*c*x+2*c^2*x^2-I*(-c^2*x^2+1)^(1/2)-c* 
x-1)*(-2*I*b^2*arcsin(c*x)+2*arcsin(c*x)^2*b^2-2*I*b*a+4*arcsin(c*x)*a*b+2 
*a^2-b^2)*e^2/(c*x+1)/(c*x-1)/d^2/c-8*e^2*(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)+16*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+2*I*a*ln(I*c*x+(-c^2*x^2+1)^(1/2)+I)-2*I*a*ln(I 
*c*x+(-c^2*x^2+1)^(1/2))+2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*b)*e^2- 
1/8*(I*(-c^2*x^2+1)^(1/2)+c*x-1)*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)*(15* 
I*b^2*arcsin(c*x)+8*arcsin(c*x)^2*b^2+15*I*a*b+16*arcsin(c*x)*a*b+8*a^2...
 

Fricas [F]

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

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e-c e x)^{5/2} (a+b \arcsin (c x))^2}{(d+c d x)^{3/2}} \, dx=\frac {\sqrt {e}\, e^{2} \left (30 \sqrt {c x +1}\, \mathit {asin} \left (\frac {\sqrt {-c x +1}}{\sqrt {2}}\right ) a^{2}+\sqrt {-c x +1}\, a^{2} c^{2} x^{2}-7 \sqrt {-c x +1}\, a^{2} c x -24 \sqrt {-c x +1}\, a^{2}+4 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x^{2}}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b \,c^{3}-8 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right ) x}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b \,c^{2}+4 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) a b c +2 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2} x^{2}}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c^{3}-4 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2} x}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c^{2}+2 \sqrt {c x +1}\, \left (\int \frac {\sqrt {-c x +1}\, \mathit {asin} \left (c x \right )^{2}}{\sqrt {c x +1}\, c x +\sqrt {c x +1}}d x \right ) b^{2} c \right )}{2 \sqrt {d}\, \sqrt {c x +1}\, c d} \] Input:

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

Output:

(sqrt(e)*e**2*(30*sqrt(c*x + 1)*asin(sqrt( - c*x + 1)/sqrt(2))*a**2 + sqrt 
( - c*x + 1)*a**2*c**2*x**2 - 7*sqrt( - c*x + 1)*a**2*c*x - 24*sqrt( - c*x 
 + 1)*a**2 + 4*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)*x**2)/(sqrt(c 
*x + 1)*c*x + sqrt(c*x + 1)),x)*a*b*c**3 - 8*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 + 4*sq 
rt(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 + 2*sqrt(c*x + 1)*int((sqrt( - c*x + 1)*asin(c*x)**2*x**2) 
/(sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b**2*c**3 - 4*sqrt(c*x + 1)*int((s 
qrt( - c*x + 1)*asin(c*x)**2*x)/(sqrt(c*x + 1)*c*x + sqrt(c*x + 1)),x)*b** 
2*c**2 + 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))/(2*sqrt(d)*sqrt(c*x + 1)*c*d)