\(\int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx\) [373]

Optimal result

Integrand size = 28, antiderivative size = 141 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=-\frac {x^4}{b c (a+b \arcsin (c x))}-\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{b^2 c^5}+\frac {\operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {4 a}{b}\right )}{2 b^2 c^5}+\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )}{b^2 c^5}-\frac {\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )}{2 b^2 c^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.83 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=\frac {-\frac {2 b c^4 x^4}{a+b \arcsin (c x)}-2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{2 b^2 c^5} \] Input:

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

Output:

((-2*b*c^4*x^4)/(a + b*ArcSin[c*x]) - 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])])/(2*b^2*c^5)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5222, 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.

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

\(\Big \downarrow \) 5222

\(\displaystyle \frac {4 \int \frac {x^3}{a+b \arcsin (c x)}dx}{b c}-\frac {x^4}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 5146

\(\displaystyle \frac {4 \int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c^5}-\frac {x^4}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {4 \int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin ^3\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b^2 c^5}-\frac {x^4}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {4 \int \left (\frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right )}{4 (a+b \arcsin (c x))}-\frac {\sin \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c^5}-\frac {x^4}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 \left (-\frac {1}{4} \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )+\frac {1}{8} \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )+\frac {1}{4} \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arcsin (c x))}{b}\right )-\frac {1}{8} \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 (a+b \arcsin (c x))}{b}\right )\right )}{b^2 c^5}-\frac {x^4}{b c (a+b \arcsin (c x))}\)

Input:

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

Output:

-(x^4/(b*c*(a + b*ArcSin[c*x]))) + (4*(-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^2*c^5)
 

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]
 

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 = 0.47 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.77

method	result	size
default	\(-\frac {4 \arcsin \left (c x \right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b -4 \arcsin \left (c x \right ) \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) b -8 \arcsin \left (c x \right ) \operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b +8 \arcsin \left (c x \right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b +4 \,\operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a -4 \sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) a -8 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a +8 \,\operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a +\cos \left (4 \arcsin \left (c x \right )\right ) b -4 \cos \left (2 \arcsin \left (c x \right )\right ) b +3 b}{8 c^{5} \left (a +b \arcsin \left (c x \right )\right ) b^{2}}\)	\(250\)

Input:

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

Output:

-1/8/c^5*(4*arcsin(c*x)*Si(4*arcsin(c*x)+4*a/b)*cos(4*a/b)*b-4*arcsin(c*x) 
*sin(4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*b-8*arcsin(c*x)*Si(2*arcsin(c*x)+2*a/b 
)*cos(2*a/b)*b+8*arcsin(c*x)*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*b+4*Si(4*a 
rcsin(c*x)+4*a/b)*cos(4*a/b)*a-4*sin(4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*a-8*Si 
(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*a+8*Ci(2*arcsin(c*x)+2*a/b)*sin(2*a/b)*a+ 
cos(4*arcsin(c*x))*b-4*cos(2*arcsin(c*x))*b+3*b)/(a+b*arcsin(c*x))/b^2
 

Fricas [F]

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

integrate(x^4/(-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^4/(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

-(x^4 - 4*(b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + a*b*c)*integ 
rate(x^3/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 876 vs. \(2 (137) = 274\).

Time = 0.23 (sec) , antiderivative size = 876, normalized size of antiderivative = 6.21 \[ \int \frac {x^4}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

4*b*arcsin(c*x)*cos(a/b)^3*cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b 
^3*c^5*arcsin(c*x) + a*b^2*c^5) - 4*b*arcsin(c*x)*cos(a/b)^4*sin_integral( 
4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 4*a*cos(a/b)^3* 
cos_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2* 
c^5) - 4*a*cos(a/b)^4*sin_integral(4*a/b + 4*arcsin(c*x))/(b^3*c^5*arcsin( 
c*x) + a*b^2*c^5) - 2*b*arcsin(c*x)*cos(a/b)*cos_integral(4*a/b + 4*arcsin 
(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*b*arcsin(c*x)*cos(a/ 
b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b 
^2*c^5) + 4*b*arcsin(c*x)*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x))/( 
b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 2*b*arcsin(c*x)*cos(a/b)^2*sin_integral 
(2*a/b + 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*a*cos(a/b)*c 
os_integral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c 
^5) - 2*a*cos(a/b)*cos_integral(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b^3*c^5*a 
rcsin(c*x) + a*b^2*c^5) + 4*a*cos(a/b)^2*sin_integral(4*a/b + 4*arcsin(c*x 
))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 2*a*cos(a/b)^2*sin_integral(2*a/b + 
 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - (c^2*x^2 - 1)^2*b/(b^3 
*c^5*arcsin(c*x) + a*b^2*c^5) - 1/2*b*arcsin(c*x)*sin_integral(4*a/b + 4*a 
rcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - b*arcsin(c*x)*sin_integral 
(2*a/b + 2*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 2*(c^2*x^2 - 1 
)*b/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 1/2*a*sin_integral(4*a/b + 4*ar...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**4/(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)