\(\int \frac {x (1-c^2 x^2)^{3/2}}{(a+b \arcsin (c x))^2} \, dx\) [360]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}+\frac {\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2 c^2}+\frac {9 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^2}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^2}+\frac {\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2 c^2}+\frac {9 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^2}+\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^2} \] Output:

-x*(-c^2*x^2+1)^2/b/c/(a+b*arcsin(c*x))+1/8*cos(a/b)*Ci((a+b*arcsin(c*x))/ 
b)/b^2/c^2+9/16*cos(3*a/b)*Ci(3*(a+b*arcsin(c*x))/b)/b^2/c^2+5/16*cos(5*a/ 
b)*Ci(5*(a+b*arcsin(c*x))/b)/b^2/c^2+1/8*sin(a/b)*Si((a+b*arcsin(c*x))/b)/ 
b^2/c^2+9/16*sin(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b^2/c^2+5/16*sin(5*a/b)* 
Si(5*(a+b*arcsin(c*x))/b)/b^2/c^2
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.38 \[ \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\frac {-16 b c x+32 b c^3 x^3-16 b c^5 x^5+2 (a+b \arcsin (c x)) \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )+9 (a+b \arcsin (c x)) \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+5 a \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+5 b \arcsin (c x) \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+2 a \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+2 b \arcsin (c x) \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+9 a \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+9 b \arcsin (c x) \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+5 a \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+5 b \arcsin (c x) \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{16 b^2 c^2 (a+b \arcsin (c x))} \] Input:

Integrate[(x*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x])^2,x]
 

Output:

(-16*b*c*x + 32*b*c^3*x^3 - 16*b*c^5*x^5 + 2*(a + b*ArcSin[c*x])*Cos[a/b]* 
CosIntegral[a/b + ArcSin[c*x]] + 9*(a + b*ArcSin[c*x])*Cos[(3*a)/b]*CosInt 
egral[3*(a/b + ArcSin[c*x])] + 5*a*Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSi 
n[c*x])] + 5*b*ArcSin[c*x]*Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSin[c*x])] 
 + 2*a*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]] + 2*b*ArcSin[c*x]*Sin[a/b]* 
SinIntegral[a/b + ArcSin[c*x]] + 9*a*Sin[(3*a)/b]*SinIntegral[3*(a/b + Arc 
Sin[c*x])] + 9*b*ArcSin[c*x]*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x] 
)] + 5*a*Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])] + 5*b*ArcSin[c*x] 
*Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b^2*c^2*(a + b*ArcSi 
n[c*x]))
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.36, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5214, 5168, 3042, 3793, 2009, 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.

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

\(\Big \downarrow \) 5214

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

\(\Big \downarrow \) 5168

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

\(\displaystyle \frac {\int \left (\frac {\cos \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{4 (a+b \arcsin (c x))}+\frac {3 \cos \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{4 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c^2}-\frac {5 c \int \frac {x^2 \left (1-c^2 x^2\right )}{a+b \arcsin (c x)}dx}{b}-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {5 c \int \frac {x^2 \left (1-c^2 x^2\right )}{a+b \arcsin (c x)}dx}{b}+\frac {\frac {3}{4} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {3}{4} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{b^2 c^2}-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 5224

\(\displaystyle -\frac {5 \int \frac {\cos ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^2\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c^2}+\frac {\frac {3}{4} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {3}{4} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{b^2 c^2}-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {5 \int \left (-\frac {\cos \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {\cos \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c^2}+\frac {\frac {3}{4} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {3}{4} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{b^2 c^2}-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3}{4} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {3}{4} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{4} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{b^2 c^2}-\frac {5 \left (\frac {1}{8} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {1}{16} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {1}{16} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )\right )}{b^2 c^2}-\frac {x \left (1-c^2 x^2\right )^2}{b c (a+b \arcsin (c x))}\)

Input:

Int[(x*(1 - c^2*x^2)^(3/2))/(a + b*ArcSin[c*x])^2,x]
 

Output:

-((x*(1 - c^2*x^2)^2)/(b*c*(a + b*ArcSin[c*x]))) + ((3*Cos[a/b]*CosIntegra 
l[(a + b*ArcSin[c*x])/b])/4 + (Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcSin[c 
*x]))/b])/4 + (3*Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/4 + (Sin[(3* 
a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/4)/(b^2*c^2) - (5*((Cos[a/b] 
*CosIntegral[(a + b*ArcSin[c*x])/b])/8 - (Cos[(3*a)/b]*CosIntegral[(3*(a + 
 b*ArcSin[c*x]))/b])/16 - (Cos[(5*a)/b]*CosIntegral[(5*(a + b*ArcSin[c*x]) 
)/b])/16 + (Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/8 - (Sin[(3*a)/b] 
*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/16 - (Sin[(5*a)/b]*SinIntegral[(5 
*(a + b*ArcSin[c*x]))/b])/16))/(b^2*c^2)
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

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 5168

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[(1/(b*c))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p]   Subst[Int[ 
x^n*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, 0]
 

rule 5214

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

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 = 0.40 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {9 \arcsin \left (c x \right ) \operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b +9 \arcsin \left (c x \right ) \operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b +2 \arcsin \left (c x \right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +2 \arcsin \left (c x \right ) \operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b +5 \arcsin \left (c x \right ) \operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b +5 \arcsin \left (c x \right ) \operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b +9 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a +9 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a +2 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +2 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a +5 \,\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a +5 \,\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a -2 x b c -3 \sin \left (3 \arcsin \left (c x \right )\right ) b -\sin \left (5 \arcsin \left (c x \right )\right ) b}{16 c^{2} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(341\)

Input:

int(x*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x))^2,x,method=_RETURNVERBOSE)
 

Output:

1/16/c^2*(9*arcsin(c*x)*Si(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*b+9*arcsin(c*x) 
*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*b+2*arcsin(c*x)*Si(arcsin(c*x)+a/b)*si 
n(a/b)*b+2*arcsin(c*x)*Ci(arcsin(c*x)+a/b)*cos(a/b)*b+5*arcsin(c*x)*Si(5*a 
rcsin(c*x)+5*a/b)*sin(5*a/b)*b+5*arcsin(c*x)*Ci(5*arcsin(c*x)+5*a/b)*cos(5 
*a/b)*b+9*Si(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*a+9*Ci(3*arcsin(c*x)+3*a/b)*c 
os(3*a/b)*a+2*Si(arcsin(c*x)+a/b)*sin(a/b)*a+2*Ci(arcsin(c*x)+a/b)*cos(a/b 
)*a+5*Si(5*arcsin(c*x)+5*a/b)*sin(5*a/b)*a+5*Ci(5*arcsin(c*x)+5*a/b)*cos(5 
*a/b)*a-2*x*b*c-3*sin(3*arcsin(c*x))*b-sin(5*arcsin(c*x))*b)/(a+b*arcsin(c 
*x))/b^2
 

Fricas [F]

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

integrate(x*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x))^2,x, algorithm="fricas")
 

Output:

integral(-(c^2*x^3 - x)*sqrt(-c^2*x^2 + 1)/(b^2*arcsin(c*x)^2 + 2*a*b*arcs 
in(c*x) + a^2), x)
 

Sympy [F]

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

integrate(x*(-c**2*x**2+1)**(3/2)/(a+b*asin(c*x))**2,x)
 

Output:

Integral(x*(-(c*x - 1)*(c*x + 1))**(3/2)/(a + b*asin(c*x))**2, x)
 

Maxima [F]

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

integrate(x*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x))^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-(c^4*x^5 - 2*c^2*x^3 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) 
+ a*b*c)*integrate((5*c^4*x^4 - 6*c^2*x^2 + 1)/(b^2*c*arctan2(c*x, sqrt(c* 
x + 1)*sqrt(-c*x + 1)) + a*b*c), x) + x)/(b^2*c*arctan2(c*x, sqrt(c*x + 1) 
*sqrt(-c*x + 1)) + a*b*c)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1215 vs. \(2 (201) = 402\).

Time = 0.24 (sec) , antiderivative size = 1215, normalized size of antiderivative = 5.68 \[ \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

integrate(x*(-c^2*x^2+1)^(3/2)/(a+b*arcsin(c*x))^2,x, algorithm="giac")
 

Output:

5*b*arcsin(c*x)*cos(a/b)^5*cos_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^2*ar 
csin(c*x) + a*b^2*c^2) + 5*b*arcsin(c*x)*cos(a/b)^4*sin(a/b)*sin_integral( 
5*a/b + 5*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 5*a*cos(a/b)^5* 
cos_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 5* 
a*cos(a/b)^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^2*arcsin( 
c*x) + a*b^2*c^2) - 25/4*b*arcsin(c*x)*cos(a/b)^3*cos_integral(5*a/b + 5*a 
rcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 9/4*b*arcsin(c*x)*cos(a/b) 
^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) - 
 15/4*b*arcsin(c*x)*cos(a/b)^2*sin(a/b)*sin_integral(5*a/b + 5*arcsin(c*x) 
)/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 9/4*b*arcsin(c*x)*cos(a/b)^2*sin(a/b 
)*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) - 
(c^2*x^2 - 1)^2*b*c*x/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) - 25/4*a*cos(a/b)^ 
3*cos_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 
9/4*a*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^2*arcsin(c*x) 
+ a*b^2*c^2) - 15/4*a*cos(a/b)^2*sin(a/b)*sin_integral(5*a/b + 5*arcsin(c* 
x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 9/4*a*cos(a/b)^2*sin(a/b)*sin_inte 
gral(3*a/b + 3*arcsin(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 25/16*b*ar 
csin(c*x)*cos(a/b)*cos_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^2*arcsin(c*x 
) + a*b^2*c^2) - 27/16*b*arcsin(c*x)*cos(a/b)*cos_integral(3*a/b + 3*arcsi 
n(c*x))/(b^3*c^2*arcsin(c*x) + a*b^2*c^2) + 1/8*b*arcsin(c*x)*cos(a/b)*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (1-c^2 x^2\right )^{3/2}}{(a+b \arcsin (c x))^2} \, dx=\int \frac {x\,{\left (1-c^2\,x^2\right )}^{3/2}}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2} \,d x \] Input:

int((x*(1 - c^2*x^2)^(3/2))/(a + b*asin(c*x))^2,x)
 

Output:

int((x*(1 - c^2*x^2)^(3/2))/(a + b*asin(c*x))^2, x)
 

Reduce [F]

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

int(x*(-c^2*x^2+1)^(3/2)/(a+b*asin(c*x))^2,x)
 

Output:

 - int((sqrt( - c**2*x**2 + 1)*x**3)/(asin(c*x)**2*b**2 + 2*asin(c*x)*a*b 
+ a**2),x)*c**2 + int((sqrt( - c**2*x**2 + 1)*x)/(asin(c*x)**2*b**2 + 2*as 
in(c*x)*a*b + a**2),x)