\(\int \frac {x^3}{(a+b \arcsin (c x))^2} \, dx\) [47]

Optimal result

Integrand size = 14, antiderivative size = 160 \[ \int \frac {x^3}{(a+b \arcsin (c x))^2} \, dx=-\frac {x^3 \sqrt {1-c^2 x^2}}{b c (a+b \arcsin (c x))}+\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c^4}-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c^4}+\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c^4}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c^4} \] Output:

-x^3*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arcsin(c*x))+1/2*cos(2*a/b)*Ci(2*(a+b*arc 
sin(c*x))/b)/b^2/c^4-1/2*cos(4*a/b)*Ci(4*(a+b*arcsin(c*x))/b)/b^2/c^4+1/2* 
sin(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b^2/c^4-1/2*sin(4*a/b)*Si(4*(a+b*arcs 
in(c*x))/b)/b^2/c^4
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.26 \[ \int \frac {x^3}{(a+b \arcsin (c x))^2} \, dx=\frac {-2 b c^3 x^3 \sqrt {1-c^2 x^2}+(a+b \arcsin (c x)) \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-(a+b \arcsin (c x)) \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+a \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+b \arcsin (c x) \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-a \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-b \arcsin (c x) \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{2 b^2 c^4 (a+b \arcsin (c x))} \] Input:

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

Output:

(-2*b*c^3*x^3*Sqrt[1 - c^2*x^2] + (a + b*ArcSin[c*x])*Cos[(2*a)/b]*CosInte 
gral[2*(a/b + ArcSin[c*x])] - (a + b*ArcSin[c*x])*Cos[(4*a)/b]*CosIntegral 
[4*(a/b + ArcSin[c*x])] + a*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x]) 
] + b*ArcSin[c*x]*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c*x])] - a*Sin[ 
(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] - b*ArcSin[c*x]*Sin[(4*a)/b]*S 
inIntegral[4*(a/b + ArcSin[c*x])])/(2*b^2*c^4*(a + b*ArcSin[c*x]))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5142, 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.

\(\displaystyle \int \frac {x^3}{(a+b \arcsin (c x))^2} \, dx\)

\(\Big \downarrow \) 5142

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-((x^3*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcSin[c*x]))) + ((Cos[(2*a)/b]*CosI 
ntegral[(2*(a + b*ArcSin[c*x]))/b])/2 - (Cos[(4*a)/b]*CosIntegral[(4*(a + 
b*ArcSin[c*x]))/b])/2 + (Sin[(2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/ 
b])/2 - (Sin[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c*x]))/b])/2)/(b^2*c^4)
 

Defintions of rubi rules used

rule 2009

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

rule 5142

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x 
^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp 
[1/(b^2*c^(m + 1)*(n + 1))   Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[-a/b 
 + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c* 
x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (4 \arcsin \left (c x \right )\right )}{8 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{2 b^{2}}-\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{4 \left (a +b \arcsin \left (c x \right )\right ) b}+\frac {\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 b^{2}}}{c^{4}}\)	\(150\)
default	\(\frac {\frac {\sin \left (4 \arcsin \left (c x \right )\right )}{8 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{2 b^{2}}-\frac {\sin \left (2 \arcsin \left (c x \right )\right )}{4 \left (a +b \arcsin \left (c x \right )\right ) b}+\frac {\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 b^{2}}}{c^{4}}\)	\(150\)

Input:

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

Output:

1/c^4*(1/8*sin(4*arcsin(c*x))/(a+b*arcsin(c*x))/b-1/2*(Si(4*arcsin(c*x)+4* 
a/b)*sin(4*a/b)+Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/b))/b^2-1/4*sin(2*arcsin(c 
*x))/(a+b*arcsin(c*x))/b+1/2*(Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)+Ci(2*arcs 
in(c*x)+2*a/b)*cos(2*a/b))/b^2)
 

Fricas [F]

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

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

Output:

integral(x^3/(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^3 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqr 
t(-c*x + 1)) + a*b*c)*integrate((4*c^2*x^4 - 3*x^2)*sqrt(c*x + 1)*sqrt(-c* 
x + 1)/(a*b*c^3*x^2 - a*b*c + (b^2*c^3*x^2 - b^2*c)*arctan2(c*x, sqrt(c*x 
+ 1)*sqrt(-c*x + 1))), 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. 854 vs. \(2 (150) = 300\).

Time = 0.16 (sec) , antiderivative size = 854, normalized size of antiderivative = 5.34 \[ \int \frac {x^3}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-4*b*arcsin(c*x)*cos(a/b)^4*cos_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^4*a 
rcsin(c*x) + a*b^2*c^4) - 4*b*arcsin(c*x)*cos(a/b)^3*sin(a/b)*sin_integral 
(4*a/b + 4*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 4*a*cos(a/b)^4 
*cos_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 4 
*a*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^4*arcsin 
(c*x) + a*b^2*c^4) + 4*b*arcsin(c*x)*cos(a/b)^2*cos_integral(4*a/b + 4*arc 
sin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + b*arcsin(c*x)*cos(a/b)^2*cos 
_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 2*b*a 
rcsin(c*x)*cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^4* 
arcsin(c*x) + a*b^2*c^4) + b*arcsin(c*x)*cos(a/b)*sin(a/b)*sin_integral(2* 
a/b + 2*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + (-c^2*x^2 + 1)^(3 
/2)*b*c*x/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 4*a*cos(a/b)^2*cos_integral( 
4*a/b + 4*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + a*cos(a/b)^2*co 
s_integral(2*a/b + 2*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) + 2*a* 
cos(a/b)*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^4*arcsin(c*x) 
 + a*b^2*c^4) + a*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arcsin(c*x))/(b 
^3*c^4*arcsin(c*x) + a*b^2*c^4) - sqrt(-c^2*x^2 + 1)*b*c*x/(b^3*c^4*arcsin 
(c*x) + a*b^2*c^4) - 1/2*b*arcsin(c*x)*cos_integral(4*a/b + 4*arcsin(c*x)) 
/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 1/2*b*arcsin(c*x)*cos_integral(2*a/b 
+ 2*arcsin(c*x))/(b^3*c^4*arcsin(c*x) + a*b^2*c^4) - 1/2*a*cos_integral...
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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