\(\int \frac {1}{(a+b \arccos (-1+d x^2))^{5/2}} \, dx\) [76]

Optimal result

Integrand size = 16, antiderivative size = 221 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}} \, dx=\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}}+\frac {x}{3 b^2 \sqrt {a+b \arccos \left (-1+d x^2\right )}}+\frac {2 \left (\frac {1}{b}\right )^{5/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{3 d x}-\frac {2 \left (\frac {1}{b}\right )^{5/2} \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{3 d x} \] Output:

1/3*(-d^2*x^4+2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2-1))^(3/2)+1/3*x/b^2/( 
a+b*arccos(d*x^2-1))^(1/2)+2/3*(1/b)^(5/2)*Pi^(1/2)*cos(1/2*a/b)*cos(1/2*a 
rccos(d*x^2-1))*FresnelS((1/b)^(1/2)*(a+b*arccos(d*x^2-1))^(1/2)/Pi^(1/2)) 
/d/x-2/3*(1/b)^(5/2)*Pi^(1/2)*cos(1/2*arccos(d*x^2-1))*FresnelC((1/b)^(1/2 
)*(a+b*arccos(d*x^2-1))^(1/2)/Pi^(1/2))*sin(1/2*a/b)/d/x
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}} \, dx=\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\sqrt {\pi } \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )-\sqrt {\pi } \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {b} \left (\left (a+b \arccos \left (-1+d x^2\right )\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+b \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )\right )}{3 b^{5/2} d x \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \] Input:

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

Output:

(2*Cos[ArcCos[-1 + d*x^2]/2]*(Sqrt[Pi]*(a + b*ArcCos[-1 + d*x^2])^(3/2)*Co 
s[a/(2*b)]*FresnelS[Sqrt[a + b*ArcCos[-1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])] - S 
qrt[Pi]*(a + b*ArcCos[-1 + d*x^2])^(3/2)*FresnelC[Sqrt[a + b*ArcCos[-1 + d 
*x^2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)] + Sqrt[b]*((a + b*ArcCos[-1 + d*x^ 
2])*Cos[ArcCos[-1 + d*x^2]/2] + b*Sin[ArcCos[-1 + d*x^2]/2])))/(3*b^(5/2)* 
d*x*(a + b*ArcCos[-1 + d*x^2])^(3/2))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5328, 5320}

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.

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

\(\Big \downarrow \) 5328

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

\(\Big \downarrow \) 5320

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

Input:

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

Output:

Sqrt[2*d*x^2 - d^2*x^4]/(3*b*d*x*(a + b*ArcCos[-1 + d*x^2])^(3/2)) + x/(3* 
b^2*Sqrt[a + b*ArcCos[-1 + d*x^2]]) - ((-2*Sqrt[b^(-1)]*Sqrt[Pi]*Cos[a/(2* 
b)]*Cos[ArcCos[-1 + d*x^2]/2]*FresnelS[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 
+ d*x^2]])/Sqrt[Pi]])/(d*x) + (2*Sqrt[b^(-1)]*Sqrt[Pi]*Cos[ArcCos[-1 + d*x 
^2]/2]*FresnelC[(Sqrt[b^(-1)]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/Sqrt[Pi]]*Si 
n[a/(2*b)])/(d*x))/(3*b^2)
 

Defintions of rubi rules used

rule 5320

Int[1/Sqrt[(a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[2*Sq 
rt[Pi/b]*Sin[a/(2*b)]*Cos[ArcCos[-1 + d*x^2]/2]*(FresnelC[Sqrt[1/(Pi*b)]*Sq 
rt[a + b*ArcCos[-1 + d*x^2]]]/(d*x)), x] - Simp[2*Sqrt[Pi/b]*Cos[a/(2*b)]*C 
os[ArcCos[-1 + d*x^2]/2]*(FresnelS[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d* 
x^2]]]/(d*x)), x] /; FreeQ[{a, b, d}, x]
 

rule 5328

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate(1/(a+b*arccos(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}} \, dx=\text {too large to display} \] Input:

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

Output:

(sqrt(d)*acos(d*x**2 - 1)**2*int((sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x 
**2 + 2)*acos(d*x**2 - 1)*x)/(acos(d*x**2 - 1)**3*b**3*d*x**2 - 2*acos(d*x 
**2 - 1)**3*b**3 + 3*acos(d*x**2 - 1)**2*a*b**2*d*x**2 - 6*acos(d*x**2 - 1 
)**2*a*b**2 + 3*acos(d*x**2 - 1)*a**2*b*d*x**2 - 6*acos(d*x**2 - 1)*a**2*b 
 + a**3*d*x**2 - 2*a**3),x)*a*b**2*d + sqrt(d)*acos(d*x**2 - 1)**2*int((sq 
rt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x**2 + 2)*acos(d*x**2 - 1)**2*x)/(aco 
s(d*x**2 - 1)**3*b**3*d*x**2 - 2*acos(d*x**2 - 1)**3*b**3 + 3*acos(d*x**2 
- 1)**2*a*b**2*d*x**2 - 6*acos(d*x**2 - 1)**2*a*b**2 + 3*acos(d*x**2 - 1)* 
a**2*b*d*x**2 - 6*acos(d*x**2 - 1)*a**2*b + a**3*d*x**2 - 2*a**3),x)*b**3* 
d + acos(d*x**2 - 1)**2*int((sqrt(acos(d*x**2 - 1)*b + a)*acos(d*x**2 - 1) 
*x**2)/(acos(d*x**2 - 1)**3*b**3*d*x**2 - 2*acos(d*x**2 - 1)**3*b**3 + 3*a 
cos(d*x**2 - 1)**2*a*b**2*d*x**2 - 6*acos(d*x**2 - 1)**2*a*b**2 + 3*acos(d 
*x**2 - 1)*a**2*b*d*x**2 - 6*acos(d*x**2 - 1)*a**2*b + a**3*d*x**2 - 2*a** 
3),x)*b**3*d**2 - 2*acos(d*x**2 - 1)**2*int((sqrt(acos(d*x**2 - 1)*b + a)* 
acos(d*x**2 - 1))/(acos(d*x**2 - 1)**3*b**3*d*x**2 - 2*acos(d*x**2 - 1)**3 
*b**3 + 3*acos(d*x**2 - 1)**2*a*b**2*d*x**2 - 6*acos(d*x**2 - 1)**2*a*b**2 
 + 3*acos(d*x**2 - 1)*a**2*b*d*x**2 - 6*acos(d*x**2 - 1)*a**2*b + a**3*d*x 
**2 - 2*a**3),x)*b**3*d - sqrt(d)*sqrt(acos(d*x**2 - 1)*b + a)*sqrt( - d*x 
**2 + 2)*acos(d*x**2 - 1) + 2*sqrt(d)*acos(d*x**2 - 1)*int((sqrt(acos(d*x* 
*2 - 1)*b + a)*sqrt( - d*x**2 + 2)*acos(d*x**2 - 1)*x)/(acos(d*x**2 - 1...