\(\int \frac {(c-a^2 c x^2)^2}{\arcsin (a x)^2} \, dx\) [23]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 78 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)^2} \, dx=-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arcsin (a x)}-\frac {5 c^2 \text {Si}(\arcsin (a x))}{8 a}-\frac {15 c^2 \text {Si}(3 \arcsin (a x))}{16 a}-\frac {5 c^2 \text {Si}(5 \arcsin (a x))}{16 a} \] Output:

-c^2*(-a^2*x^2+1)^(5/2)/a/arcsin(a*x)-5/8*c^2*Si(arcsin(a*x))/a-15/16*c^2* 
Si(3*arcsin(a*x))/a-5/16*c^2*Si(5*arcsin(a*x))/a
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)^2} \, dx=-\frac {c^2 \left (16 \left (1-a^2 x^2\right )^{5/2}+10 \arcsin (a x) \text {Si}(\arcsin (a x))+15 \arcsin (a x) \text {Si}(3 \arcsin (a x))+5 \arcsin (a x) \text {Si}(5 \arcsin (a x))\right )}{16 a \arcsin (a x)} \] Input:

Integrate[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]
 

Output:

-1/16*(c^2*(16*(1 - a^2*x^2)^(5/2) + 10*ArcSin[a*x]*SinIntegral[ArcSin[a*x 
]] + 15*ArcSin[a*x]*SinIntegral[3*ArcSin[a*x]] + 5*ArcSin[a*x]*SinIntegral 
[5*ArcSin[a*x]]))/(a*ArcSin[a*x])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5166, 5224, 4906, 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.

\(\displaystyle \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)^2} \, dx\)

\(\Big \downarrow \) 5166

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

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {5 c^2 \int \left (\frac {a x}{8 \arcsin (a x)}+\frac {3 \sin (3 \arcsin (a x))}{16 \arcsin (a x)}+\frac {\sin (5 \arcsin (a x))}{16 \arcsin (a x)}\right )d\arcsin (a x)}{a}-\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arcsin (a x)}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {c^2 \left (1-a^2 x^2\right )^{5/2}}{a \arcsin (a x)}-\frac {5 c^2 \left (\frac {1}{8} \text {Si}(\arcsin (a x))+\frac {3}{16} \text {Si}(3 \arcsin (a x))+\frac {1}{16} \text {Si}(5 \arcsin (a x))\right )}{a}\)

Input:

Int[(c - a^2*c*x^2)^2/ArcSin[a*x]^2,x]
 

Output:

-((c^2*(1 - a^2*x^2)^(5/2))/(a*ArcSin[a*x])) - (5*c^2*(SinIntegral[ArcSin[ 
a*x]]/8 + (3*SinIntegral[3*ArcSin[a*x]])/16 + SinIntegral[5*ArcSin[a*x]]/1 
6))/a
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5166
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[Sqrt[1 - c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSin[c*x])^(n + 1 
)/(b*c*(n + 1))), x] + Simp[c*((2*p + 1)/(b*(n + 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, p}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]
 

rule 5224
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x 
^2)^p]   Subst[Int[x^n*Sin[-a/b + x/b]^m*Cos[-a/b + x/b]^(2*p + 1), x], x, 
a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] 
 && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
 
Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06

method result size
derivativedivides \(-\frac {c^{2} \left (10 \,\operatorname {Si}\left (\arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+15 \,\operatorname {Si}\left (3 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+5 \,\operatorname {Si}\left (5 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cos \left (3 \arcsin \left (a x \right )\right )+\cos \left (5 \arcsin \left (a x \right )\right )\right )}{16 a \arcsin \left (a x \right )}\) \(83\)
default \(-\frac {c^{2} \left (10 \,\operatorname {Si}\left (\arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+15 \,\operatorname {Si}\left (3 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+5 \,\operatorname {Si}\left (5 \arcsin \left (a x \right )\right ) \arcsin \left (a x \right )+10 \sqrt {-a^{2} x^{2}+1}+5 \cos \left (3 \arcsin \left (a x \right )\right )+\cos \left (5 \arcsin \left (a x \right )\right )\right )}{16 a \arcsin \left (a x \right )}\) \(83\)

Input:

int((-a^2*c*x^2+c)^2/arcsin(a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

-1/16/a*c^2*(10*Si(arcsin(a*x))*arcsin(a*x)+15*Si(3*arcsin(a*x))*arcsin(a* 
x)+5*Si(5*arcsin(a*x))*arcsin(a*x)+10*(-a^2*x^2+1)^(1/2)+5*cos(3*arcsin(a* 
x))+cos(5*arcsin(a*x)))/arcsin(a*x)
 

Fricas [F]

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

integrate((-a^2*c*x^2+c)^2/arcsin(a*x)^2,x, algorithm="fricas")
 

Output:

integral((a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)/arcsin(a*x)^2, x)
 

Sympy [F]

\[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)^2} \, dx=c^{2} \left (\int \left (- \frac {2 a^{2} x^{2}}{\operatorname {asin}^{2}{\left (a x \right )}}\right )\, dx + \int \frac {a^{4} x^{4}}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asin}^{2}{\left (a x \right )}}\, dx\right ) \] Input:

integrate((-a**2*c*x**2+c)**2/asin(a*x)**2,x)
 

Output:

c**2*(Integral(-2*a**2*x**2/asin(a*x)**2, x) + Integral(a**4*x**4/asin(a*x 
)**2, x) + Integral(asin(a*x)**(-2), x))
 

Maxima [F]

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

integrate((-a^2*c*x^2+c)^2/arcsin(a*x)^2,x, algorithm="maxima")
 

Output:

(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))*integrate(5*(a^3*c^2*x^3 - a 
*c^2*x)*sqrt(a*x + 1)*sqrt(-a*x + 1)/arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x 
+ 1)), x) - (a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2)*sqrt(a*x + 1)*sqrt(-a*x + 
1))/(a*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1)))
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04 \[ \int \frac {\left (c-a^2 c x^2\right )^2}{\arcsin (a x)^2} \, dx=-\frac {{\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} c^{2}}{a \arcsin \left (a x\right )} - \frac {5 \, c^{2} \operatorname {Si}\left (5 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac {15 \, c^{2} \operatorname {Si}\left (3 \, \arcsin \left (a x\right )\right )}{16 \, a} - \frac {5 \, c^{2} \operatorname {Si}\left (\arcsin \left (a x\right )\right )}{8 \, a} \] Input:

integrate((-a^2*c*x^2+c)^2/arcsin(a*x)^2,x, algorithm="giac")
 

Output:

-(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*c^2/(a*arcsin(a*x)) - 5/16*c^2*sin_int 
egral(5*arcsin(a*x))/a - 15/16*c^2*sin_integral(3*arcsin(a*x))/a - 5/8*c^2 
*sin_integral(arcsin(a*x))/a
 

Mupad [F(-1)]

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

int((c - a^2*c*x^2)^2/asin(a*x)^2,x)
 

Output:

int((c - a^2*c*x^2)^2/asin(a*x)^2, x)
 

Reduce [F]

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

int((-a^2*c*x^2+c)^2/asin(a*x)^2,x)
                                                                                    
                                                                                    
 

Output:

c**2*(int(x**4/asin(a*x)**2,x)*a**4 - 2*int(x**2/asin(a*x)**2,x)*a**2 + in 
t(1/asin(a*x)**2,x))