\(\int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{5/2}} \, dx\) [95]

Optimal result

Integrand size = 32, antiderivative size = 386 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{5/2}} \, dx=\frac {b^2 x}{3 d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {b (a+b \arcsin (c x))}{3 c d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2}}+\frac {2 x (a+b \arcsin (c x))^2}{3 d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {x (a+b \arcsin (c x))^2}{3 d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x} \left (1-c^2 x^2\right )}-\frac {2 i \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{3 c d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{3 c d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{3 c d^2 e^2 \sqrt {d+c d x} \sqrt {e-c e x}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 8.21 (sec) , antiderivative size = 722, normalized size of antiderivative = 1.87 \[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*a^2*c*x*(3 - 2*c^2*x^2) + b^2*(c*x + 6*c*x*ArcSin[c*x]^2 + (4*I)*Pi*Arc 
Sin[c*x]*Cos[3*ArcSin[c*x]] - (2*I)*ArcSin[c*x]^2*Cos[3*ArcSin[c*x]] + 8*P 
i*Cos[3*ArcSin[c*x]]*Log[1 + E^((-I)*ArcSin[c*x])] + 2*Pi*Cos[3*ArcSin[c*x 
]]*Log[1 - I*E^(I*ArcSin[c*x])] + 4*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 - 
 I*E^(I*ArcSin[c*x])] - 2*Pi*Cos[3*ArcSin[c*x]]*Log[1 + I*E^(I*ArcSin[c*x] 
)] + 4*ArcSin[c*x]*Cos[3*ArcSin[c*x]]*Log[1 + I*E^(I*ArcSin[c*x])] - 8*Pi* 
Cos[3*ArcSin[c*x]]*Log[Cos[ArcSin[c*x]/2]] + 2*Pi*Cos[3*ArcSin[c*x]]*Log[- 
Cos[(Pi + 2*ArcSin[c*x])/4]] + 2*Sqrt[1 - c^2*x^2]*((-3*I)*ArcSin[c*x]^2 + 
 ArcSin[c*x]*(-2 + (6*I)*Pi + 6*Log[1 - I*E^(I*ArcSin[c*x])] + 6*Log[1 + I 
*E^(I*ArcSin[c*x])]) + 3*Pi*(4*Log[1 + E^((-I)*ArcSin[c*x])] + Log[1 - I*E 
^(I*ArcSin[c*x])] - Log[1 + I*E^(I*ArcSin[c*x])] - 4*Log[Cos[ArcSin[c*x]/2 
]] + Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - Log[Sin[(Pi + 2*ArcSin[c*x])/4]]) 
) - 2*Pi*Cos[3*ArcSin[c*x]]*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (16*I)*(1 - 
 c^2*x^2)^(3/2)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (16*I)*(1 - c^2*x^2)^ 
(3/2)*PolyLog[2, I*E^(I*ArcSin[c*x])] + Sin[3*ArcSin[c*x]] + 2*ArcSin[c*x] 
^2*Sin[3*ArcSin[c*x]]) + 4*a*b*(Sqrt[1 - c^2*x^2]*(-1 + 2*Log[Cos[ArcSin[c 
*x]/2] - Sin[ArcSin[c*x]/2]] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/ 
2]] + 2*Cos[2*ArcSin[c*x]]*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + 
 Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])) + ArcSin[c*x]*(3*c*x + Sin 
[3*ArcSin[c*x]])))/(12*d^2*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(c - c^3...
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.60, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {5178, 5162, 5160, 5180, 3042, 4202, 2620, 2715, 2838, 5182, 208}

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.

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

\(\Big \downarrow \) 5178

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

\(\Big \downarrow \) 5162

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {2}{3} \int \frac {(a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 5160

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {2}{3} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-2 b c \int \frac {x (a+b \arcsin (c x))}{1-c^2 x^2}dx\right )+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 5180

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {2}{3} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 b \int \frac {c x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}d\arcsin (c x)}{c}\right )+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {2}{3} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 b \int (a+b \arcsin (c x)) \tan (\arcsin (c x))d\arcsin (c x)}{c}\right )+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (\frac {2}{3} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \int \frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1+e^{2 i \arcsin (c x)}}d\arcsin (c x)\right )}{c}\right )-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \left (-\frac {2}{3} b c \int \frac {x (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^2}dx+\frac {2}{3} \left (\frac {x (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (\frac {1}{2} i b \int \log \left (1+e^{2 i \arcsin (c x)}\right )d\arcsin (c x)-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )\right )}{c}\right )+\frac {x (a+b \arcsin (c x))^2}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (e-c e x)^{5/2}}\)

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 5182

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

((1 - c^2*x^2)^(5/2)*((x*(a + b*ArcSin[c*x])^2)/(3*(1 - c^2*x^2)^(3/2)) - 
(2*b*c*(-1/2*(b*x)/(c*Sqrt[1 - c^2*x^2]) + (a + b*ArcSin[c*x])/(2*c^2*(1 - 
 c^2*x^2))))/3 + (2*((x*(a + b*ArcSin[c*x])^2)/Sqrt[1 - c^2*x^2] - (2*b*(( 
(I/2)*(a + b*ArcSin[c*x])^2)/b - (2*I)*((-1/2*I)*(a + b*ArcSin[c*x])*Log[1 
 + E^((2*I)*ArcSin[c*x])] - (b*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/4)))/c) 
)/3))/((d + c*d*x)^(5/2)*(e - c*e*x)^(5/2))
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 5160

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[x*((a + b*ArcSin[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp[b 
*c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]   Int[x*((a + b*ArcSin[c*x 
])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d 
 + e, 0] && GtQ[n, 0]
 

rule 5162

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_ 
Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*d*(p + 1 
))), x] + (Simp[(2*p + 3)/(2*d*(p + 1))   Int[(d + e*x^2)^(p + 1)*(a + b*Ar 
cSin[c*x])^n, x], x] + Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2 
*x^2)^p]   Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x 
]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, 
 -1] && NeQ[p, -3/2]
 

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 5180

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), 
x_Symbol] :> Simp[-e^(-1)   Subst[Int[(a + b*x)^n*Tan[x], x], x, ArcSin[c*x 
]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]
 

rule 5182

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_ 
.), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 
1))), x] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   I 
nt[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, 
 b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2566 vs. \(2 (354 ) = 708\).

Time = 3.64 (sec) , antiderivative size = 2567, normalized size of antiderivative = 6.65

method	result	size
default	\(\text {Expression too large to display}\)	\(2567\)
parts	\(\text {Expression too large to display}\)	\(2567\)

Input:

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

Output:

a^2*(-1/3/c/d/e/(c*d*x+d)^(3/2)/(-c*e*x+e)^(3/2)+1/d*(-1/c/d/e/(c*d*x+d)^( 
1/2)/(-c*e*x+e)^(3/2)+2/d*(1/3/c/d/e/(-c*e*x+e)^(3/2)*(c*d*x+d)^(1/2)+1/3/ 
c/d/e^2/(-c*e*x+e)^(1/2)*(c*d*x+d)^(1/2))))+17/3*b^2*(-e*(c*x-1))^(1/2)*(d 
*(c*x+1))^(1/2)/e^3/d^3*c^2/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x 
)^2*x^3+2/3*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3*c^4/(3*c^6*x^ 
6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*x^2+1)*x^5-4/3*b^2*(-e*(c*x-1))^(1/2)*(d* 
(c*x+1))^(1/2)/e^3/d^3*c^2/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*x^2+1 
)*x^3+4/3*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3/c/(3*c^6*x^6-10 
*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-2*I*b^2*(-e*(c*x-1)) 
^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsi 
n(c*x)*x-4/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3/c/(3*c^6*x 
^6-10*c^4*x^4+11*c^2*x^2-4)*(-c^2*x^2+1)^(1/2)-2*b^2*(-e*(c*x-1))^(1/2)*(d 
*(c*x+1))^(1/2)/e^3/d^3*c^4/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x 
)^2*x^5+14/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3*c/(3*c^6*x 
^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)^2*(-c^2*x^2+1)^(1/2)*x^2+4/3*I*b^2 
*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3/d^3*c^4/(3*c^6*x^6-10*c^4*x^4+11 
*c^2*x^2-4)*arcsin(c*x)*(-c^2*x^2+1)*x^5-2*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c* 
x+1))^(1/2)/e^3/d^3*c^3/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)^2* 
(-c^2*x^2+1)^(1/2)*x^4-10/3*I*b^2*(-e*(c*x-1))^(1/2)*(d*(c*x+1))^(1/2)/e^3 
/d^3*c^2/(3*c^6*x^6-10*c^4*x^4+11*c^2*x^2-4)*arcsin(c*x)*(-c^2*x^2+1)*x...
 

Fricas [F]

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

integrate((a+b*arcsin(c*x))^2/(c*d*x+d)^(5/2)/(-c*e*x+e)^(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)*sq 
rt(-c*e*x + e)/(c^6*d^3*e^3*x^6 - 3*c^4*d^3*e^3*x^4 + 3*c^2*d^3*e^3*x^2 - 
d^3*e^3), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/3*a*b*c*(1/(c^4*d^(5/2)*e^(5/2)*x^2 - c^2*d^(5/2)*e^(5/2)) + 2*log(c*x + 
 1)/(c^2*d^(5/2)*e^(5/2)) + 2*log(c*x - 1)/(c^2*d^(5/2)*e^(5/2))) + 2/3*a* 
b*(x/((-c^2*d*e*x^2 + d*e)^(3/2)*d*e) + 2*x/(sqrt(-c^2*d*e*x^2 + d*e)*d^2* 
e^2))*arcsin(c*x) + 1/3*a^2*(x/((-c^2*d*e*x^2 + d*e)^(3/2)*d*e) + 2*x/(sqr 
t(-c^2*d*e*x^2 + d*e)*d^2*e^2)) + b^2*integrate(arctan2(c*x, sqrt(c*x + 1) 
*sqrt(-c*x + 1))^2/((c^4*d^2*e^2*x^4 - 2*c^2*d^2*e^2*x^2 + d^2*e^2)*sqrt(c 
*x + 1)*sqrt(-c*x + 1)), x)/(sqrt(d)*sqrt(e))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \arcsin (c x))^2}{(d+c d x)^{5/2} (e-c e x)^{5/2}} \, dx=\frac {6 \sqrt {c x +1}\, \sqrt {-c x +1}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, \sqrt {-c x +1}\, c^{4} x^{4}-2 \sqrt {c x +1}\, \sqrt {-c x +1}\, c^{2} x^{2}+\sqrt {c x +1}\, \sqrt {-c x +1}}d x \right ) a b \,c^{2} x^{2}-6 \sqrt {c x +1}\, \sqrt {-c x +1}\, \left (\int \frac {\mathit {asin} \left (c x \right )}{\sqrt {c x +1}\, \sqrt {-c x +1}\, c^{4} x^{4}-2 \sqrt {c x +1}\, \sqrt {-c x +1}\, c^{2} x^{2}+\sqrt {c x +1}\, \sqrt {-c x +1}}d x \right ) a b +3 \sqrt {c x +1}\, \sqrt {-c x +1}\, \left (\int \frac {\mathit {asin} \left (c x \right )^{2}}{\sqrt {c x +1}\, \sqrt {-c x +1}\, c^{4} x^{4}-2 \sqrt {c x +1}\, \sqrt {-c x +1}\, c^{2} x^{2}+\sqrt {c x +1}\, \sqrt {-c x +1}}d x \right ) b^{2} c^{2} x^{2}-3 \sqrt {c x +1}\, \sqrt {-c x +1}\, \left (\int \frac {\mathit {asin} \left (c x \right )^{2}}{\sqrt {c x +1}\, \sqrt {-c x +1}\, c^{4} x^{4}-2 \sqrt {c x +1}\, \sqrt {-c x +1}\, c^{2} x^{2}+\sqrt {c x +1}\, \sqrt {-c x +1}}d x \right ) b^{2}+2 a^{2} c^{2} x^{3}-3 a^{2} x}{3 \sqrt {e}\, \sqrt {d}\, \sqrt {c x +1}\, \sqrt {-c x +1}\, d^{2} e^{2} \left (c^{2} x^{2}-1\right )} \] Input:

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

Output:

(6*sqrt(c*x + 1)*sqrt( - c*x + 1)*int(asin(c*x)/(sqrt(c*x + 1)*sqrt( - c*x 
 + 1)*c**4*x**4 - 2*sqrt(c*x + 1)*sqrt( - c*x + 1)*c**2*x**2 + sqrt(c*x + 
1)*sqrt( - c*x + 1)),x)*a*b*c**2*x**2 - 6*sqrt(c*x + 1)*sqrt( - c*x + 1)*i 
nt(asin(c*x)/(sqrt(c*x + 1)*sqrt( - c*x + 1)*c**4*x**4 - 2*sqrt(c*x + 1)*s 
qrt( - c*x + 1)*c**2*x**2 + sqrt(c*x + 1)*sqrt( - c*x + 1)),x)*a*b + 3*sqr 
t(c*x + 1)*sqrt( - c*x + 1)*int(asin(c*x)**2/(sqrt(c*x + 1)*sqrt( - c*x + 
1)*c**4*x**4 - 2*sqrt(c*x + 1)*sqrt( - c*x + 1)*c**2*x**2 + sqrt(c*x + 1)* 
sqrt( - c*x + 1)),x)*b**2*c**2*x**2 - 3*sqrt(c*x + 1)*sqrt( - c*x + 1)*int 
(asin(c*x)**2/(sqrt(c*x + 1)*sqrt( - c*x + 1)*c**4*x**4 - 2*sqrt(c*x + 1)* 
sqrt( - c*x + 1)*c**2*x**2 + sqrt(c*x + 1)*sqrt( - c*x + 1)),x)*b**2 + 2*a 
**2*c**2*x**3 - 3*a**2*x)/(3*sqrt(e)*sqrt(d)*sqrt(c*x + 1)*sqrt( - c*x + 1 
)*d**2*e**2*(c**2*x**2 - 1))