\(\int \frac {x^3}{a+b \arcsin (c x)} \, dx\) [40]

Optimal result

Integrand size = 14, antiderivative size = 125 \[ \int \frac {x^3}{a+b \arcsin (c x)} \, dx=-\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b c^4}+\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b c^4}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{4 b c^4}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 b c^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {x^3}{a+b \arcsin (c x)} \, dx=\frac {-2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{8 b c^4} \] Input:

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

Output:

(-2*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] + CosIntegral[4*(a/b + 
 ArcSin[c*x])]*Sin[(4*a)/b] + 2*Cos[(2*a)/b]*SinIntegral[2*(a/b + ArcSin[c 
*x])] - Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])])/(8*b*c^4)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5146, 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.

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

\(\Big \downarrow \) 5146

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {\int \left (\frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{4 (a+b \arcsin (c x))}-\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^4}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/4*(CosIntegral[(2*(a + b*ArcSin[c*x]))/b]*Sin[(2*a)/b]) + (CosIntegral 
[(4*(a + b*ArcSin[c*x]))/b]*Sin[(4*a)/b])/8 + (Cos[(2*a)/b]*SinIntegral[(2 
*(a + b*ArcSin[c*x]))/b])/4 - (Cos[(4*a)/b]*SinIntegral[(4*(a + b*ArcSin[c 
*x]))/b])/8)/(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 5146

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {-\frac {\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{8 b}+\frac {\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b}+\frac {\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{4 b}-\frac {\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b}}{c^{4}}\)	\(110\)
default	\(\frac {-\frac {\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )}{8 b}+\frac {\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{8 b}+\frac {\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{4 b}-\frac {\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{4 b}}{c^{4}}\)	\(110\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(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. 237 vs. \(2 (117) = 234\).

Time = 0.13 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.90 \[ \int \frac {x^3}{a+b \arcsin (c x)} \, dx=\frac {\cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{4}} - \frac {\cos \left (\frac {a}{b}\right )^{4} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{4}} - \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, b c^{4}} - \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{2 \, b c^{4}} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{b c^{4}} + \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{2 \, b c^{4}} - \frac {\operatorname {Si}\left (\frac {4 \, a}{b} + 4 \, \arcsin \left (c x\right )\right )}{8 \, b c^{4}} - \frac {\operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arcsin \left (c x\right )\right )}{4 \, b c^{4}} \] Input:

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

Output:

cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b*c^4) - cos(a/b) 
^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^4) - 1/2*cos(a/b)*cos_integral 
(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b*c^4) - 1/2*cos(a/b)*cos_integral(2*a/b 
 + 2*arcsin(c*x))*sin(a/b)/(b*c^4) + cos(a/b)^2*sin_integral(4*a/b + 4*arc 
sin(c*x))/(b*c^4) + 1/2*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b* 
c^4) - 1/8*sin_integral(4*a/b + 4*arcsin(c*x))/(b*c^4) - 1/4*sin_integral( 
2*a/b + 2*arcsin(c*x))/(b*c^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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