\(\int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx\) [161]

Optimal result

Integrand size = 16, antiderivative size = 250 \[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=-\frac {2 x^2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arcsin (c x)}}-\frac {\sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}+\frac {\sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )}{b^{3/2} c^3}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{b^{3/2} c^3}-\frac {\sqrt {\frac {3 \pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{b^{3/2} c^3} \] Output:

-2*x^2*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arcsin(c*x))^(1/2)-1/2*2^(1/2)*Pi^(1/2) 
*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))/b^(3/ 
2)/c^3+1/2*6^(1/2)*Pi^(1/2)*cos(3*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arcs 
in(c*x))^(1/2)/b^(1/2))/b^(3/2)/c^3+1/2*2^(1/2)*Pi^(1/2)*FresnelC(2^(1/2)/ 
Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(a/b)/b^(3/2)/c^3-1/2*6^(1/2) 
*Pi^(1/2)*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arcsin(c*x))^(1/2)/b^(1/2))*sin(3 
*a/b)/b^(3/2)/c^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.37 \[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\frac {e^{-\frac {3 i (a+b \arcsin (c x))}{b}} \left (e^{\frac {3 i a}{b}}-e^{\frac {3 i a}{b}+2 i \arcsin (c x)}-e^{\frac {3 i a}{b}+4 i \arcsin (c x)}+e^{\frac {3 i (a+2 b \arcsin (c x))}{b}}+e^{\frac {2 i a}{b}+3 i \arcsin (c x)} \sqrt {-\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {i (a+b \arcsin (c x))}{b}\right )+e^{\frac {4 i a}{b}+3 i \arcsin (c x)} \sqrt {\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {i (a+b \arcsin (c x))}{b}\right )-\sqrt {3} e^{3 i \arcsin (c x)} \sqrt {-\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i (a+b \arcsin (c x))}{b}\right )-\sqrt {3} e^{3 i \left (\frac {2 a}{b}+\arcsin (c x)\right )} \sqrt {\frac {i (a+b \arcsin (c x))}{b}} \Gamma \left (\frac {1}{2},\frac {3 i (a+b \arcsin (c x))}{b}\right )\right )}{4 b c^3 \sqrt {a+b \arcsin (c x)}} \] Input:

Integrate[x^2/(a + b*ArcSin[c*x])^(3/2),x]
 

Output:

(E^(((3*I)*a)/b) - E^(((3*I)*a)/b + (2*I)*ArcSin[c*x]) - E^(((3*I)*a)/b + 
(4*I)*ArcSin[c*x]) + E^(((3*I)*(a + 2*b*ArcSin[c*x]))/b) + E^(((2*I)*a)/b 
+ (3*I)*ArcSin[c*x])*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((-I)*( 
a + b*ArcSin[c*x]))/b] + E^(((4*I)*a)/b + (3*I)*ArcSin[c*x])*Sqrt[(I*(a + 
b*ArcSin[c*x]))/b]*Gamma[1/2, (I*(a + b*ArcSin[c*x]))/b] - Sqrt[3]*E^((3*I 
)*ArcSin[c*x])*Sqrt[((-I)*(a + b*ArcSin[c*x]))/b]*Gamma[1/2, ((-3*I)*(a + 
b*ArcSin[c*x]))/b] - Sqrt[3]*E^((3*I)*((2*a)/b + ArcSin[c*x]))*Sqrt[(I*(a 
+ b*ArcSin[c*x]))/b]*Gamma[1/2, ((3*I)*(a + b*ArcSin[c*x]))/b])/(4*b*c^3*E 
^(((3*I)*(a + b*ArcSin[c*x]))/b)*Sqrt[a + b*ArcSin[c*x]])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.

\(\displaystyle \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx\)

\(\Big \downarrow \) 5142

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 \left (\frac {1}{2} \sqrt {\frac {\pi }{2}} \sqrt {b} \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )-\frac {1}{2} \sqrt {\frac {3 \pi }{2}} \sqrt {b} \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \sqrt {b} \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )+\frac {1}{2} \sqrt {\frac {3 \pi }{2}} \sqrt {b} \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \arcsin (c x)}}{\sqrt {b}}\right )\right )}{b^2 c^3}-\frac {2 x^2 \sqrt {1-c^2 x^2}}{b c \sqrt {a+b \arcsin (c x)}}\)

Input:

Int[x^2/(a + b*ArcSin[c*x])^(3/2),x]
 

Output:

(-2*x^2*Sqrt[1 - c^2*x^2])/(b*c*Sqrt[a + b*ArcSin[c*x]]) + (2*(-1/2*(Sqrt[ 
b]*Sqrt[Pi/2]*Cos[a/b]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[ 
b]]) + (Sqrt[b]*Sqrt[(3*Pi)/2]*Cos[(3*a)/b]*FresnelS[(Sqrt[6/Pi]*Sqrt[a + 
b*ArcSin[c*x]])/Sqrt[b]])/2 + (Sqrt[b]*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sqr 
t[a + b*ArcSin[c*x]])/Sqrt[b]]*Sin[a/b])/2 - (Sqrt[b]*Sqrt[(3*Pi)/2]*Fresn 
elC[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c*x]])/Sqrt[b]]*Sin[(3*a)/b])/2))/(b^2*c 
^3)
 

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 = 0.13 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.20

method	result	size
default	\(-\frac {-\sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {a}{b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )-\sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {a}{b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )+\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \cos \left (\frac {3 a}{b}\right ) \operatorname {FresnelS}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {-\frac {3}{b}}+\sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}\, \sin \left (\frac {3 a}{b}\right ) \operatorname {FresnelC}\left (\frac {3 \sqrt {2}\, \sqrt {a +b \arcsin \left (c x \right )}}{\sqrt {\pi }\, \sqrt {-\frac {3}{b}}\, b}\right ) \sqrt {-\frac {3}{b}}+\cos \left (-\frac {a +b \arcsin \left (c x \right )}{b}+\frac {a}{b}\right )-\cos \left (-\frac {3 \left (a +b \arcsin \left (c x \right )\right )}{b}+\frac {3 a}{b}\right )}{2 c^{3} b \sqrt {a +b \arcsin \left (c x \right )}}\)	\(301\)

Input:

int(x^2/(a+b*arcsin(c*x))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/c^3/b/(a+b*arcsin(c*x))^(1/2)*(-(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*ar 
csin(c*x))^(1/2)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcs 
in(c*x))^(1/2)/b)-(-1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*si 
n(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)+P 
i^(1/2)*2^(1/2)*(a+b*arcsin(c*x))^(1/2)*cos(3*a/b)*FresnelS(3*2^(1/2)/Pi^( 
1/2)/(-3/b)^(1/2)*(a+b*arcsin(c*x))^(1/2)/b)*(-3/b)^(1/2)+Pi^(1/2)*2^(1/2) 
*(a+b*arcsin(c*x))^(1/2)*sin(3*a/b)*FresnelC(3*2^(1/2)/Pi^(1/2)/(-3/b)^(1/ 
2)*(a+b*arcsin(c*x))^(1/2)/b)*(-3/b)^(1/2)+cos(-(a+b*arcsin(c*x))/b+a/b)-c 
os(-3*(a+b*arcsin(c*x))/b+3*a/b))
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^2/(a+b*arcsin(c*x))^(3/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\int \frac {x^{2}}{\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**2/(a+b*asin(c*x))**(3/2),x)
 

Output:

Integral(x**2/(a + b*asin(c*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2/(a+b*arcsin(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^2/(b*arcsin(c*x) + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\int { \frac {x^{2}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^2/(a+b*arcsin(c*x))^(3/2),x, algorithm="giac")
 

Output:

integrate(x^2/(b*arcsin(c*x) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\int \frac {x^2}{{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^{3/2}} \,d x \] Input:

int(x^2/(a + b*asin(c*x))^(3/2),x)
 

Output:

int(x^2/(a + b*asin(c*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^2}{(a+b \arcsin (c x))^{3/2}} \, dx=\int \frac {\sqrt {\mathit {asin} \left (c x \right ) b +a}\, x^{2}}{\mathit {asin} \left (c x \right )^{2} b^{2}+2 \mathit {asin} \left (c x \right ) a b +a^{2}}d x \] Input:

int(x^2/(a+b*asin(c*x))^(3/2),x)
 

Output:

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