\(\int \frac {x^4}{(a+b \arcsin (c x))^2} \, dx\) [46]

Optimal result

Integrand size = 14, antiderivative size = 218 \[ \int \frac {x^4}{(a+b \arcsin (c x))^2} \, dx=-\frac {x^4 \sqrt {1-c^2 x^2}}{b c (a+b \arcsin (c x))}+\frac {\operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b^2 c^5}-\frac {9 \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16 b^2 c^5}+\frac {5 \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right ) \sin \left (\frac {5 a}{b}\right )}{16 b^2 c^5}-\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{8 b^2 c^5}+\frac {9 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^5}-\frac {5 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 b^2 c^5} \] Output:

-x^4*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arcsin(c*x))+1/8*Ci((a+b*arcsin(c*x))/b)* 
sin(a/b)/b^2/c^5-9/16*Ci(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b^2/c^5+5/16*Ci 
(5*(a+b*arcsin(c*x))/b)*sin(5*a/b)/b^2/c^5-1/8*cos(a/b)*Si((a+b*arcsin(c*x 
))/b)/b^2/c^5+9/16*cos(3*a/b)*Si(3*(a+b*arcsin(c*x))/b)/b^2/c^5-5/16*cos(5 
*a/b)*Si(5*(a+b*arcsin(c*x))/b)/b^2/c^5
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.78 \[ \int \frac {x^4}{(a+b \arcsin (c x))^2} \, dx=\frac {-\frac {16 b c^4 x^4 \sqrt {1-c^2 x^2}}{a+b \arcsin (c x)}+2 \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )-9 \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {3 a}{b}\right )+5 \operatorname {CosIntegral}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {5 a}{b}\right )-2 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )+9 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-5 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\arcsin (c x)\right )\right )}{16 b^2 c^5} \] Input:

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

Output:

((-16*b*c^4*x^4*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]) + 2*CosIntegral[a/b 
 + ArcSin[c*x]]*Sin[a/b] - 9*CosIntegral[3*(a/b + ArcSin[c*x])]*Sin[(3*a)/ 
b] + 5*CosIntegral[5*(a/b + ArcSin[c*x])]*Sin[(5*a)/b] - 2*Cos[a/b]*SinInt 
egral[a/b + ArcSin[c*x]] + 9*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x] 
)] - 5*Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c*x])])/(16*b^2*c^5)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5142, 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.

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

\(\Big \downarrow \) 5142

\(\displaystyle \frac {\int \left (\frac {5 \sin \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}-\frac {9 \sin \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right )}{16 (a+b \arcsin (c x))}+\frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{8 (a+b \arcsin (c x))}\right )d(a+b \arcsin (c x))}{b^2 c^5}-\frac {x^4 \sqrt {1-c^2 x^2}}{b c (a+b \arcsin (c x))}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{8} \sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )-\frac {9}{16} \sin \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )+\frac {5}{16} \sin \left (\frac {5 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )-\frac {1}{8} \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )+\frac {9}{16} \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 (a+b \arcsin (c x))}{b}\right )-\frac {5}{16} \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 (a+b \arcsin (c x))}{b}\right )}{b^2 c^5}-\frac {x^4 \sqrt {1-c^2 x^2}}{b c (a+b \arcsin (c x))}\)

Input:

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

Output:

-((x^4*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcSin[c*x]))) + ((CosIntegral[(a + 
b*ArcSin[c*x])/b]*Sin[a/b])/8 - (9*CosIntegral[(3*(a + b*ArcSin[c*x]))/b]* 
Sin[(3*a)/b])/16 + (5*CosIntegral[(5*(a + b*ArcSin[c*x]))/b]*Sin[(5*a)/b]) 
/16 - (Cos[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/8 + (9*Cos[(3*a)/b]*Si 
nIntegral[(3*(a + b*ArcSin[c*x]))/b])/16 - (5*Cos[(5*a)/b]*SinIntegral[(5* 
(a + b*ArcSin[c*x]))/b])/16)/(b^2*c^5)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5142

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x 
^m*Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^(n + 1)/(b*c*(n + 1))), x] - Simp 
[1/(b^2*c^(m + 1)*(n + 1))   Subst[Int[ExpandTrigReduce[x^(n + 1), Sin[-a/b 
 + x/b]^(m - 1)*(m - (m + 1)*Sin[-a/b + x/b]^2), x], x], x, a + b*ArcSin[c* 
x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {3 \cos \left (3 \arcsin \left (c x \right )\right )}{16 \left (a +b \arcsin \left (c x \right )\right ) b}+\frac {\frac {9 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{16}-\frac {9 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16}}{b^{2}}-\frac {\cos \left (5 \arcsin \left (c x \right )\right )}{16 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {5 \left (\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )\right )}{16 b^{2}}-\frac {\sqrt {-c^{2} x^{2}+1}}{8 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b^{2}}}{c^{5}}\)	\(222\)
default	\(\frac {\frac {3 \cos \left (3 \arcsin \left (c x \right )\right )}{16 \left (a +b \arcsin \left (c x \right )\right ) b}+\frac {\frac {9 \,\operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right )}{16}-\frac {9 \,\operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right )}{16}}{b^{2}}-\frac {\cos \left (5 \arcsin \left (c x \right )\right )}{16 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {5 \left (\operatorname {Si}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right )-\operatorname {Ci}\left (5 \arcsin \left (c x \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right )\right )}{16 b^{2}}-\frac {\sqrt {-c^{2} x^{2}+1}}{8 \left (a +b \arcsin \left (c x \right )\right ) b}-\frac {\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{8 b^{2}}}{c^{5}}\)	\(222\)

Input:

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

Output:

1/c^5*(3/16*cos(3*arcsin(c*x))/(a+b*arcsin(c*x))/b+9/16*(Si(3*arcsin(c*x)+ 
3*a/b)*cos(3*a/b)-Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b))/b^2-1/16*cos(5*arcsi 
n(c*x))/(a+b*arcsin(c*x))/b-5/16*(Si(5*arcsin(c*x)+5*a/b)*cos(5*a/b)-Ci(5* 
arcsin(c*x)+5*a/b)*sin(5*a/b))/b^2-1/8*(-c^2*x^2+1)^(1/2)/(a+b*arcsin(c*x) 
)/b-1/8*(Si(arcsin(c*x)+a/b)*cos(a/b)-Ci(arcsin(c*x)+a/b)*sin(a/b))/b^2)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^4 - (b^2*c*arctan2(c*x, sqrt(c*x + 1)*sqr 
t(-c*x + 1)) + a*b*c)*integrate((5*c^2*x^5 - 4*x^3)*sqrt(c*x + 1)*sqrt(-c* 
x + 1)/(a*b*c^3*x^2 - a*b*c + (b^2*c^3*x^2 - b^2*c)*arctan2(c*x, sqrt(c*x 
+ 1)*sqrt(-c*x + 1))), 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. 1299 vs. \(2 (204) = 408\).

Time = 0.17 (sec) , antiderivative size = 1299, normalized size of antiderivative = 5.96 \[ \int \frac {x^4}{(a+b \arcsin (c x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

5*b*arcsin(c*x)*cos(a/b)^4*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b 
^3*c^5*arcsin(c*x) + a*b^2*c^5) - 5*b*arcsin(c*x)*cos(a/b)^5*sin_integral( 
5*a/b + 5*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 5*a*cos(a/b)^4* 
cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2* 
c^5) - 5*a*cos(a/b)^5*sin_integral(5*a/b + 5*arcsin(c*x))/(b^3*c^5*arcsin( 
c*x) + a*b^2*c^5) - 15/4*b*arcsin(c*x)*cos(a/b)^2*cos_integral(5*a/b + 5*a 
rcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 9/4*b*arcsin(c*x) 
*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c 
*x) + a*b^2*c^5) + 25/4*b*arcsin(c*x)*cos(a/b)^3*sin_integral(5*a/b + 5*ar 
csin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 9/4*b*arcsin(c*x)*cos(a/b)^ 
3*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) - 
15/4*a*cos(a/b)^2*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b^3*c^5*ar 
csin(c*x) + a*b^2*c^5) - 9/4*a*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(c* 
x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 25/4*a*cos(a/b)^3*sin_int 
egral(5*a/b + 5*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 9/4*a*cos 
(a/b)^3*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^5*arcsin(c*x) + a*b^2*c 
^5) + 5/16*b*arcsin(c*x)*cos_integral(5*a/b + 5*arcsin(c*x))*sin(a/b)/(b^3 
*c^5*arcsin(c*x) + a*b^2*c^5) + 9/16*b*arcsin(c*x)*cos_integral(3*a/b + 3* 
arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2*c^5) + 1/8*b*arcsin(c*x 
)*cos_integral(a/b + arcsin(c*x))*sin(a/b)/(b^3*c^5*arcsin(c*x) + a*b^2...
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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