\(\int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx\) [372]

Optimal result

Integrand size = 28, antiderivative size = 204 \[ \int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=-\frac {x^5}{b c (a+b \arcsin (c x))}+\frac {5 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2 c^6}-\frac {15 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^6}+\frac {5 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^6} \] Output:

-x^5/b/c/(a+b*arcsin(c*x))+5/8*cos(a/b)*Ci((a+b*arcsin(c*x))/b)/b^2/c^6-15 
/16*cos(3*a/b)*Ci(3*(a+b*arcsin(c*x))/b)/b^2/c^6+5/16*cos(5*a/b)*Ci(5*(a+b 
*arcsin(c*x))/b)/b^2/c^6+5/8*sin(a/b)*Si((a+b*arcsin(c*x))/b)/b^2/c^6-15/1 
6*sin(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b^2/c^6+5/16*sin(5*a/b)*Si(5*(a+b*a 
rcsin(c*x))/b)/b^2/c^6
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=-\frac {x^5}{b c (a+b \arcsin (c x))}+\frac {5 \left (2 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )-3 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )-3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )\right )}{16 b^2 c^6} \] Input:

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

Output:

-(x^5/(b*c*(a + b*ArcSin[c*x]))) + (5*(2*Cos[a/b]*CosIntegral[a/b + ArcSin 
[c*x]] - 3*Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c*x])] + Cos[(5*a)/b]* 
CosIntegral[5*(a/b + ArcSin[c*x])] + 2*Sin[a/b]*SinIntegral[a/b + ArcSin[c 
*x]] - 3*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])] + Sin[(5*a)/b]*Si 
nIntegral[5*(a/b + ArcSin[c*x])]))/(16*b^2*c^6)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5222, 5146, 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.

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

\(\Big \downarrow \) 5222

\(\displaystyle \frac {5 \int \frac {x^4}{a+b \arcsin (c x)}dx}{b c}-\frac {x^5}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 5146

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

\(\Big \downarrow \) 4906

\(\displaystyle \frac {5 \int \left (\frac {\cos \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}-\frac {3 \cos \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {\cos \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c^6}-\frac {x^5}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 \left (\frac {1}{8} \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {3}{16} \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {1}{16} \cos \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {3}{16} \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {1}{16} \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )\right )}{b^2 c^6}-\frac {x^5}{b c (a+b \arcsin (c x))}\)

Input:

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

Output:

-(x^5/(b*c*(a + b*ArcSin[c*x]))) + (5*((Cos[a/b]*CosIntegral[(a + b*ArcSin 
[c*x])/b])/8 - (3*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]*Sin 
Integral[(a + b*ArcSin[c*x])/b])/8 - (3*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^6)
 

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 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]
 

rule 5222

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.67

method	result	size
default	\(-\frac {15 \arcsin \left (c x \right ) \operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b -10 \arcsin \left (c x \right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b -10 \arcsin \left (c x \right ) \operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b -5 \arcsin \left (c x \right ) \operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b -5 \arcsin \left (c x \right ) \operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b +15 \arcsin \left (c x \right ) \operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b +15 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a -10 \,\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a -10 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a -5 \,\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a -5 \,\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a +15 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a +10 x b c +\sin \left (5 \arcsin \left (c x \right )\right ) b -5 \sin \left (3 \arcsin \left (c x \right )\right ) b}{16 c^{6} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(340\)

Input:

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

Output:

-1/16/c^6*(15*arcsin(c*x)*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*b-10*arcsin(c 
*x)*Si(arcsin(c*x)+a/b)*sin(a/b)*b-10*arcsin(c*x)*Ci(arcsin(c*x)+a/b)*cos( 
a/b)*b-5*arcsin(c*x)*Si(5*arcsin(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+15*arcsin(c*x)*Si(3*arcsin(c*x)+3*a/b)* 
sin(3*a/b)*b+15*Ci(3*arcsin(c*x)+3*a/b)*cos(3*a/b)*a-10*Si(arcsin(c*x)+a/b 
)*sin(a/b)*a-10*Ci(arcsin(c*x)+a/b)*cos(a/b)*a-5*Si(5*arcsin(c*x)+5*a/b)*s 
in(5*a/b)*a-5*Ci(5*arcsin(c*x)+5*a/b)*cos(5*a/b)*a+15*Si(3*arcsin(c*x)+3*a 
/b)*sin(3*a/b)*a+10*x*b*c+sin(5*arcsin(c*x))*b-5*sin(3*arcsin(c*x))*b)/(a+ 
b*arcsin(c*x))/b^2
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-(x^5 - 5*(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integ 
rate(x^4/(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c), x))/( 
b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^5}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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