\(\int (a-b \arcsin (1-d x^2))^{5/2} \, dx\) [299]

Optimal result

Integrand size = 18, antiderivative size = 299 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=-15 b^2 x \sqrt {a-b \arcsin \left (1-d x^2\right )}+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}+\frac {15 \sqrt {\pi } x \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}-\frac {15 \sqrt {\pi } x \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )}{\left (-\frac {1}{b}\right )^{5/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Output:

-15*b^2*x*(a+b*arcsin(d*x^2-1))^(1/2)+5*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*ar 
csin(d*x^2-1))^(3/2)/d/x+x*(a+b*arcsin(d*x^2-1))^(5/2)+15*Pi^(1/2)*x*Fresn 
elC((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))*(cos(1/2*a/b)-sin(1 
/2*a/b))/(-1/b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2*arcsin(d*x^2-1)))- 
15*Pi^(1/2)*x*FresnelS((-1/b)^(1/2)*(a+b*arcsin(d*x^2-1))^(1/2)/Pi^(1/2))* 
(cos(1/2*a/b)+sin(1/2*a/b))/(-1/b)^(5/2)/(cos(1/2*arcsin(d*x^2-1))+sin(1/2 
*arcsin(d*x^2-1)))
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.98 \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\frac {5 b \sqrt {-d x^2 \left (-2+d x^2\right )} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}+\frac {15 b x \left (-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right )+\sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cos \left (\frac {a}{2 b}\right )+\sin \left (\frac {a}{2 b}\right )\right )+\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )\right )}{\left (-\frac {1}{b}\right )^{3/2} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )} \] Input:

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

Output:

(5*b*Sqrt[-(d*x^2*(-2 + d*x^2))]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x) + 
x*(a - b*ArcSin[1 - d*x^2])^(5/2) + (15*b*x*(-(Sqrt[Pi]*FresnelC[(Sqrt[-b^ 
(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] - Sin[a/(2*b) 
])) + Sqrt[Pi]*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt 
[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]) + Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - 
d*x^2]]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))/((-b^(-1)) 
^(3/2)*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2]))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5313, 5310}

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.

\(\displaystyle \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx\)

\(\Big \downarrow \) 5313

\(\displaystyle -15 b^2 \int \sqrt {a-b \arcsin \left (1-d x^2\right )}dx+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}\)

\(\Big \downarrow \) 5310

\(\displaystyle -15 b^2 \left (-\frac {\sqrt {\pi } x \left (\cos \left (\frac {a}{2 b}\right )-\sin \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+\frac {\sqrt {\pi } x \left (\sin \left (\frac {a}{2 b}\right )+\cos \left (\frac {a}{2 b}\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {-\frac {1}{b}} \sqrt {a-b \arcsin \left (1-d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {-\frac {1}{b}} \left (\cos \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )-\sin \left (\frac {1}{2} \arcsin \left (1-d x^2\right )\right )\right )}+x \sqrt {a-b \arcsin \left (1-d x^2\right )}\right )+\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a-b \arcsin \left (1-d x^2\right )\right )^{3/2}}{d x}+x \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2}\)

Input:

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

Output:

(5*b*Sqrt[2*d*x^2 - d^2*x^4]*(a - b*ArcSin[1 - d*x^2])^(3/2))/(d*x) + x*(a 
 - b*ArcSin[1 - d*x^2])^(5/2) - 15*b^2*(x*Sqrt[a - b*ArcSin[1 - d*x^2]] - 
(Sqrt[Pi]*x*FresnelC[(Sqrt[-b^(-1)]*Sqrt[a - b*ArcSin[1 - d*x^2]])/Sqrt[Pi 
]]*(Cos[a/(2*b)] - Sin[a/(2*b)]))/(Sqrt[-b^(-1)]*(Cos[ArcSin[1 - d*x^2]/2] 
 - Sin[ArcSin[1 - d*x^2]/2])) + (Sqrt[Pi]*x*FresnelS[(Sqrt[-b^(-1)]*Sqrt[a 
 - b*ArcSin[1 - d*x^2]])/Sqrt[Pi]]*(Cos[a/(2*b)] + Sin[a/(2*b)]))/(Sqrt[-b 
^(-1)]*(Cos[ArcSin[1 - d*x^2]/2] - Sin[ArcSin[1 - d*x^2]/2])))
 

Defintions of rubi rules used

rule 5310

Int[Sqrt[(a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[x*Sq 
rt[a + b*ArcSin[c + d*x^2]], x] + (-Simp[Sqrt[Pi]*x*(Cos[a/(2*b)] + c*Sin[a 
/(2*b)])*(FresnelC[Sqrt[c/(Pi*b)]*Sqrt[a + b*ArcSin[c + d*x^2]]]/(Sqrt[c/b] 
*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + d*x^2]/2]))), x] + Simp[Sqrt[ 
Pi]*x*(Cos[a/(2*b)] - c*Sin[a/(2*b)])*(FresnelS[Sqrt[c/(Pi*b)]*Sqrt[a + b*A 
rcSin[c + d*x^2]]]/(Sqrt[c/b]*(Cos[ArcSin[c + d*x^2]/2] - c*Sin[ArcSin[c + 
d*x^2]/2]))), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^2, 1]
 

rule 5313

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)^2]*(b_.))^(n_), x_Symbol] :> Simp[x*( 
a + b*ArcSin[c + d*x^2])^n, x] + (Simp[2*b*n*Sqrt[-2*c*d*x^2 - d^2*x^4]*((a 
 + b*ArcSin[c + d*x^2])^(n - 1)/(d*x)), x] - Simp[4*b^2*n*(n - 1)   Int[(a 
+ b*ArcSin[c + d*x^2])^(n - 2), x], x]) /; FreeQ[{a, b, c, d}, x] && EqQ[c^ 
2, 1] && GtQ[n, 1]
 

Maple [F]

\[\int {\left (a +b \arcsin \left (d \,x^{2}-1\right )\right )}^{\frac {5}{2}}d x\]

Input:

int((a+b*arcsin(d*x^2-1))^(5/2),x)
 

Output:

int((a+b*arcsin(d*x^2-1))^(5/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \left (a-b \arcsin \left (1-d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)
 

Giac [F]

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

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

Output:

integrate((b*arcsin(d*x^2 - 1) + a)^(5/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(asin(d*x**2 - 1)*b + a),x)*a**2 + 2*int(sqrt(asin(d*x**2 - 1)*b + 
 a)*asin(d*x**2 - 1),x)*a*b + int(sqrt(asin(d*x**2 - 1)*b + a)*asin(d*x**2 
 - 1)**2,x)*b**2