\(\int \frac {x^3 (1-c^2 x^2)^{5/2}}{a+b \arcsin (c x)} \, dx\) [305]

Optimal result

Integrand size = 28, antiderivative size = 245 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=-\frac {3 \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{128 b c^4}-\frac {\operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{32 b c^4}+\frac {3 \operatorname {CosIntegral}\left (\frac {7 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {7 a}{b}\right )}{256 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {9 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {9 a}{b}\right )}{256 b c^4}+\frac {3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{128 b c^4}+\frac {\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{32 b c^4}-\frac {3 \cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )}{256 b c^4}-\frac {\cos \left (\frac {9 a}{b}\right ) \text {Si}\left (\frac {9 (a+b \arcsin (c x))}{b}\right )}{256 b c^4} \] Output:

-3/128*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b/c^4-1/32*Ci(3*(a+b*arcsin(c*x))/ 
b)*sin(3*a/b)/b/c^4+3/256*Ci(7*(a+b*arcsin(c*x))/b)*sin(7*a/b)/b/c^4+1/256 
*Ci(9*(a+b*arcsin(c*x))/b)*sin(9*a/b)/b/c^4+3/128*cos(a/b)*Si((a+b*arcsin( 
c*x))/b)/b/c^4+1/32*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b/c^4-3/256*cos(7 
*a/b)*Si(7*(a+b*arcsin(c*x))/b)/b/c^4-1/256*cos(9*a/b)*Si(9*(a+b*arcsin(c* 
x))/b)/b/c^4
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\frac {-6 \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )-8 \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+3 \operatorname {CosIntegral}\left (7 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {7 a}{b}\right )+\operatorname {CosIntegral}\left (9 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {9 a}{b}\right )+6 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+8 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-3 \cos \left (\frac {7 a}{b}\right ) \text {Si}\left (7 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {9 a}{b}\right ) \text {Si}\left (9 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{256 b c^4} \] Input:

Integrate[(x^3*(1 - c^2*x^2)^(5/2))/(a + b*ArcSin[c*x]),x]
 

Output:

(-6*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] - 8*CosIntegral[3*(a/b + ArcSi 
n[c*x])]*Sin[(3*a)/b] + 3*CosIntegral[7*(a/b + ArcSin[c*x])]*Sin[(7*a)/b] 
+ CosIntegral[9*(a/b + ArcSin[c*x])]*Sin[(9*a)/b] + 6*Cos[a/b]*SinIntegral 
[a/b + ArcSin[c*x]] + 8*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])] - 
3*Cos[(7*a)/b]*SinIntegral[7*(a/b + ArcSin[c*x])] - Cos[(9*a)/b]*SinIntegr 
al[9*(a/b + ArcSin[c*x])])/(256*b*c^4)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5224, 25, 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.

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

\(\Big \downarrow \) 5224

\(\displaystyle \frac {\int -\frac {\cos ^6\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\cos ^6\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (-\frac {\sin \left (\frac {9 a}{b}-\frac {9 (a+b \arcsin (c x))}{b}\right )}{256 (a+b \arcsin (c x))}-\frac {3 \sin \left (\frac {7 a}{b}-\frac {7 (a+b \arcsin (c x))}{b}\right )}{256 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {3 \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{128 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {3}{128} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {1}{32} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {3}{256} \sin \left (\frac {7 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )+\frac {1}{256} \sin \left (\frac {9 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {9 (a+b \arcsin (c x))}{b}\right )+\frac {3}{128} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {1}{32} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {3}{256} \cos \left (\frac {7 a}{b}\right ) \text {Si}\left (\frac {7 (a+b \arcsin (c x))}{b}\right )-\frac {1}{256} \cos \left (\frac {9 a}{b}\right ) \text {Si}\left (\frac {9 (a+b \arcsin (c x))}{b}\right )}{b c^4}\)

Input:

Int[(x^3*(1 - c^2*x^2)^(5/2))/(a + b*ArcSin[c*x]),x]
 

Output:

((-3*CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b])/128 - (CosIntegral[(3*(a 
 + b*ArcSin[c*x]))/b]*Sin[(3*a)/b])/32 + (3*CosIntegral[(7*(a + b*ArcSin[c 
*x]))/b]*Sin[(7*a)/b])/256 + (CosIntegral[(9*(a + b*ArcSin[c*x]))/b]*Sin[( 
9*a)/b])/256 + (3*Cos[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/128 + (Cos[ 
(3*a)/b]*SinIntegral[(3*(a + b*ArcSin[c*x]))/b])/32 - (3*Cos[(7*a)/b]*SinI 
ntegral[(7*(a + b*ArcSin[c*x]))/b])/256 - (Cos[(9*a)/b]*SinIntegral[(9*(a 
+ b*ArcSin[c*x]))/b])/256)/(b*c^4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 = 0.19 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {8 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )-8 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )+6 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-6 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )-3 \,\operatorname {Si}\left (7 \arcsin \left (c x \right )+\frac {7 a}{b}\right ) \cos \left (\frac {7 a}{b}\right )+3 \,\operatorname {Ci}\left (7 \arcsin \left (c x \right )+\frac {7 a}{b}\right ) \sin \left (\frac {7 a}{b}\right )-\operatorname {Si}\left (9 \arcsin \left (c x \right )+\frac {9 a}{b}\right ) \cos \left (\frac {9 a}{b}\right )+\operatorname {Ci}\left (9 \arcsin \left (c x \right )+\frac {9 a}{b}\right ) \sin \left (\frac {9 a}{b}\right )}{256 c^{4} b}\)	\(185\)

Input:

int(x^3*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
 

Output:

1/256/c^4*(8*Si(3*arcsin(c*x)+3*a/b)*cos(3*a/b)-8*Ci(3*arcsin(c*x)+3*a/b)* 
sin(3*a/b)+6*Si(arcsin(c*x)+a/b)*cos(a/b)-6*Ci(arcsin(c*x)+a/b)*sin(a/b)-3 
*Si(7*arcsin(c*x)+7*a/b)*cos(7*a/b)+3*Ci(7*arcsin(c*x)+7*a/b)*sin(7*a/b)-S 
i(9*arcsin(c*x)+9*a/b)*cos(9*a/b)+Ci(9*arcsin(c*x)+9*a/b)*sin(9*a/b))/b
 

Fricas [F]

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

integrate(x^3*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")
 

Output:

integral((c^4*x^7 - 2*c^2*x^5 + x^3)*sqrt(-c^2*x^2 + 1)/(b*arcsin(c*x) + a 
), x)
 

Sympy [F]

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

integrate(x**3*(-c**2*x**2+1)**(5/2)/(a+b*asin(c*x)),x)
 

Output:

Integral(x**3*(-(c*x - 1)*(c*x + 1))**(5/2)/(a + b*asin(c*x)), x)
 

Maxima [F]

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

integrate(x^3*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")
 

Output:

integrate((-c^2*x^2 + 1)^(5/2)*x^3/(b*arcsin(c*x) + a), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (229) = 458\).

Time = 0.17 (sec) , antiderivative size = 746, normalized size of antiderivative = 3.04 \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\text {Too large to display} \] Input:

integrate(x^3*(-c^2*x^2+1)^(5/2)/(a+b*arcsin(c*x)),x, algorithm="giac")
 

Output:

cos(a/b)^8*cos_integral(9*a/b + 9*arcsin(c*x))*sin(a/b)/(b*c^4) - cos(a/b) 
^9*sin_integral(9*a/b + 9*arcsin(c*x))/(b*c^4) - 7/4*cos(a/b)^6*cos_integr 
al(9*a/b + 9*arcsin(c*x))*sin(a/b)/(b*c^4) + 3/4*cos(a/b)^6*cos_integral(7 
*a/b + 7*arcsin(c*x))*sin(a/b)/(b*c^4) + 9/4*cos(a/b)^7*sin_integral(9*a/b 
 + 9*arcsin(c*x))/(b*c^4) - 3/4*cos(a/b)^7*sin_integral(7*a/b + 7*arcsin(c 
*x))/(b*c^4) + 15/16*cos(a/b)^4*cos_integral(9*a/b + 9*arcsin(c*x))*sin(a/ 
b)/(b*c^4) - 15/16*cos(a/b)^4*cos_integral(7*a/b + 7*arcsin(c*x))*sin(a/b) 
/(b*c^4) - 27/16*cos(a/b)^5*sin_integral(9*a/b + 9*arcsin(c*x))/(b*c^4) + 
21/16*cos(a/b)^5*sin_integral(7*a/b + 7*arcsin(c*x))/(b*c^4) - 5/32*cos(a/ 
b)^2*cos_integral(9*a/b + 9*arcsin(c*x))*sin(a/b)/(b*c^4) + 9/32*cos(a/b)^ 
2*cos_integral(7*a/b + 7*arcsin(c*x))*sin(a/b)/(b*c^4) - 1/8*cos(a/b)^2*co 
s_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^4) + 15/32*cos(a/b)^3*sin_ 
integral(9*a/b + 9*arcsin(c*x))/(b*c^4) - 21/32*cos(a/b)^3*sin_integral(7* 
a/b + 7*arcsin(c*x))/(b*c^4) + 1/8*cos(a/b)^3*sin_integral(3*a/b + 3*arcsi 
n(c*x))/(b*c^4) + 1/256*cos_integral(9*a/b + 9*arcsin(c*x))*sin(a/b)/(b*c^ 
4) - 3/256*cos_integral(7*a/b + 7*arcsin(c*x))*sin(a/b)/(b*c^4) + 1/32*cos 
_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b*c^4) - 3/128*cos_integral(a/b 
 + arcsin(c*x))*sin(a/b)/(b*c^4) - 9/256*cos(a/b)*sin_integral(9*a/b + 9*a 
rcsin(c*x))/(b*c^4) + 21/256*cos(a/b)*sin_integral(7*a/b + 7*arcsin(c*x))/ 
(b*c^4) - 3/32*cos(a/b)*sin_integral(3*a/b + 3*arcsin(c*x))/(b*c^4) + 3...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (1-c^2 x^2\right )^{5/2}}{a+b \arcsin (c x)} \, dx=\int \frac {x^3\,{\left (1-c^2\,x^2\right )}^{5/2}}{a+b\,\mathrm {asin}\left (c\,x\right )} \,d x \] Input:

int((x^3*(1 - c^2*x^2)^(5/2))/(a + b*asin(c*x)),x)
 

Output:

int((x^3*(1 - c^2*x^2)^(5/2))/(a + b*asin(c*x)), x)
 

Reduce [F]

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

int(x^3*(-c^2*x^2+1)^(5/2)/(a+b*asin(c*x)),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt( - c**2*x**2 + 1)*x**7)/(asin(c*x)*b + a),x)*c**4 - 2*int((sqrt( 
 - c**2*x**2 + 1)*x**5)/(asin(c*x)*b + a),x)*c**2 + int((sqrt( - c**2*x**2 
 + 1)*x**3)/(asin(c*x)*b + a),x)