\(\int \frac {x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx\) [290]

Optimal result

Integrand size = 28, antiderivative size = 206 \[ \int \frac {x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx=-\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c^5}-\frac {\cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 b c^5}+\frac {\cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c^5}+\frac {\log (a+b \arcsin (c x))}{16 b c^5}-\frac {\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 b c^5}-\frac {\sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 b c^5}+\frac {\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 b c^5} \] Output:

-1/32*cos(2*a/b)*Ci(2*(a+b*arcsin(c*x))/b)/b/c^5-1/16*cos(4*a/b)*Ci(4*(a+b 
*arcsin(c*x))/b)/b/c^5+1/32*cos(6*a/b)*Ci(6*(a+b*arcsin(c*x))/b)/b/c^5+1/1 
6*ln(a+b*arcsin(c*x))/b/c^5-1/32*sin(2*a/b)*Si(2*(a+b*arcsin(c*x))/b)/b/c^ 
5-1/16*sin(4*a/b)*Si(4*(a+b*arcsin(c*x))/b)/b/c^5+1/32*sin(6*a/b)*Si(6*(a+ 
b*arcsin(c*x))/b)/b/c^5
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.74 \[ \int \frac {x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx=-\frac {\cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+2 \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-2 \log (a+b \arcsin (c x))+\sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+2 \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\sin \left (\frac {6 a}{b}\right ) \text {Si}\left (6 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{32 b c^5} \] Input:

Integrate[(x^4*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]
 

Output:

-1/32*(Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSin[c*x])] + 2*Cos[(4*a)/b]*Co 
sIntegral[4*(a/b + ArcSin[c*x])] - Cos[(6*a)/b]*CosIntegral[6*(a/b + ArcSi 
n[c*x])] - 2*Log[a + b*ArcSin[c*x]] + Sin[(2*a)/b]*SinIntegral[2*(a/b + Ar 
cSin[c*x])] + 2*Sin[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])] - Sin[(6*a 
)/b]*SinIntegral[6*(a/b + ArcSin[c*x])])/(b*c^5)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {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.

\(\displaystyle \int \frac {x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx\)

\(\Big \downarrow \) 5224

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

\(\Big \downarrow \) 4906

\(\displaystyle \frac {\int \left (\frac {\cos \left (\frac {6 a}{b}-\frac {6 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}-\frac {\cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{32 (a+b \arcsin (c x))}+\frac {1}{16 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b c^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{32} \cos \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} \cos \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \cos \left (\frac {6 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} \sin \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{32} \sin \left (\frac {6 a}{b}\right ) \text {Si}\left (\frac {6 (a+b \arcsin (c x))}{b}\right )+\frac {1}{16} \log (a+b \arcsin (c x))}{b c^5}\)

Input:

Int[(x^4*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]),x]
 

Output:

(-1/32*(Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b]) - (Cos[(4*a)/ 
b]*CosIntegral[(4*(a + b*ArcSin[c*x]))/b])/16 + (Cos[(6*a)/b]*CosIntegral[ 
(6*(a + b*ArcSin[c*x]))/b])/32 + Log[a + b*ArcSin[c*x]]/16 - (Sin[(2*a)/b] 
*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/32 - (Sin[(4*a)/b]*SinIntegral[(4 
*(a + b*ArcSin[c*x]))/b])/16 + (Sin[(6*a)/b]*SinIntegral[(6*(a + b*ArcSin[ 
c*x]))/b])/32)/(b*c^5)
 

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 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 = 0.22 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )+2 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right )+\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )-\operatorname {Si}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \sin \left (\frac {6 a}{b}\right )-\operatorname {Ci}\left (6 \arcsin \left (c x \right )+\frac {6 a}{b}\right ) \cos \left (\frac {6 a}{b}\right )+2 \,\operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right )-2 \ln \left (a +b \arcsin \left (c x \right )\right )}{32 c^{5} b}\)	\(157\)

Input:

int(x^4*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
 

Output:

-1/32/c^5*(Si(2*arcsin(c*x)+2*a/b)*sin(2*a/b)+2*Si(4*arcsin(c*x)+4*a/b)*si 
n(4*a/b)+Ci(2*arcsin(c*x)+2*a/b)*cos(2*a/b)-Si(6*arcsin(c*x)+6*a/b)*sin(6* 
a/b)-Ci(6*arcsin(c*x)+6*a/b)*cos(6*a/b)+2*Ci(4*arcsin(c*x)+4*a/b)*cos(4*a/ 
b)-2*ln(a+b*arcsin(c*x)))/b
 

Fricas [F]

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

integrate(x^4*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**4*(-c**2*x**2+1)**(1/2)/(a+b*asin(c*x)),x)
 

Output:

Integral(x**4*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*asin(c*x)), x)
 

Maxima [F]

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

integrate(x^4*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="maxima")
 

Output:

integrate(sqrt(-c^2*x^2 + 1)*x^4/(b*arcsin(c*x) + a), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (192) = 384\).

Time = 0.15 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.29 \[ \int \frac {x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)} \, dx =\text {Too large to display} \] Input:

integrate(x^4*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x)),x, algorithm="giac")
 

Output:

cos(a/b)^6*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^5) + cos(a/b)^5*sin(a/ 
b)*sin_integral(6*a/b + 6*arcsin(c*x))/(b*c^5) - 3/2*cos(a/b)^4*cos_integr 
al(6*a/b + 6*arcsin(c*x))/(b*c^5) - 1/2*cos(a/b)^4*cos_integral(4*a/b + 4* 
arcsin(c*x))/(b*c^5) - cos(a/b)^3*sin(a/b)*sin_integral(6*a/b + 6*arcsin(c 
*x))/(b*c^5) - 1/2*cos(a/b)^3*sin(a/b)*sin_integral(4*a/b + 4*arcsin(c*x)) 
/(b*c^5) + 9/16*cos(a/b)^2*cos_integral(6*a/b + 6*arcsin(c*x))/(b*c^5) + 1 
/2*cos(a/b)^2*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) - 1/16*cos(a/b)^ 
2*cos_integral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 3/16*cos(a/b)*sin(a/b)*sin 
_integral(6*a/b + 6*arcsin(c*x))/(b*c^5) + 1/4*cos(a/b)*sin(a/b)*sin_integ 
ral(4*a/b + 4*arcsin(c*x))/(b*c^5) - 1/16*cos(a/b)*sin(a/b)*sin_integral(2 
*a/b + 2*arcsin(c*x))/(b*c^5) - 1/32*cos_integral(6*a/b + 6*arcsin(c*x))/( 
b*c^5) - 1/16*cos_integral(4*a/b + 4*arcsin(c*x))/(b*c^5) + 1/32*cos_integ 
ral(2*a/b + 2*arcsin(c*x))/(b*c^5) + 1/16*log(b*arcsin(c*x) + a)/(b*c^5)
 

Mupad [F(-1)]

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

int((x^4*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)),x)
 

Output:

int((x^4*(1 - c^2*x^2)^(1/2))/(a + b*asin(c*x)), x)
 

Reduce [F]

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

int(x^4*(-c^2*x^2+1)^(1/2)/(a+b*asin(c*x)),x)
                                                                                    
                                                                                    
 

Output:

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