\(\int \frac {1}{(a+b \arccos (1+d x^2))^{3/2}} \, dx\) [68]

Optimal result

Integrand size = 16, antiderivative size = 190 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{b d x \sqrt {a+b \arccos \left (1+d x^2\right )}}+\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x}-\frac {2 \left (\frac {1}{b}\right )^{3/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{d x} \] Output:

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

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\frac {\frac {\sqrt {b} \sqrt {-d x^2 \left (2+d x^2\right )}}{\sqrt {a+b \arccos \left (1+d x^2\right )}}+2 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )-2 \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{b^{3/2} d x} \] Input:

Integrate[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]
 

Output:

((Sqrt[b]*Sqrt[-(d*x^2*(2 + d*x^2))])/Sqrt[a + b*ArcCos[1 + d*x^2]] + 2*Sq 
rt[Pi]*Cos[a/(2*b)]*FresnelS[Sqrt[a + b*ArcCos[1 + d*x^2]]/(Sqrt[b]*Sqrt[P 
i])]*Sin[ArcCos[1 + d*x^2]/2] - 2*Sqrt[Pi]*FresnelC[Sqrt[a + b*ArcCos[1 + 
d*x^2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2])/(b^(3/2 
)*d*x)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5322}

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.

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

\(\Big \downarrow \) 5322

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

Input:

Int[(a + b*ArcCos[1 + d*x^2])^(-3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 5322

Int[((a_.) + ArcCos[1 + (d_.)*(x_)^2]*(b_.))^(-3/2), x_Symbol] :> Simp[Sqrt 
[-2*d*x^2 - d^2*x^4]/(b*d*x*Sqrt[a + b*ArcCos[1 + d*x^2]]), x] + (-Simp[2*( 
1/b)^(3/2)*Sqrt[Pi]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/2]*(FresnelC[Sqrt[1/ 
(Pi*b)]*Sqrt[a + b*ArcCos[1 + d*x^2]]]/(d*x)), x] + Simp[2*(1/b)^(3/2)*Sqrt 
[Pi]*Cos[a/(2*b)]*Sin[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]
 

Maple [F]

\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )}^{\frac {3}{2}}}d x\]

Input:

int(1/(a+b*arccos(d*x^2+1))^(3/2),x)
 

Output:

int(1/(a+b*arccos(d*x^2+1))^(3/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(a+b*acos(d*x**2+1))**(3/2),x)
 

Output:

Integral((a + b*acos(d*x**2 + 1))**(-3/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: sign: argument cannot be imagi 
nary; found sqrt((-_SAGE_VAR_d*_SAGE_VAR_x^2)-2)
 

Giac [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*arccos(d*x^2 + 1) + a)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{3/2}} \,d x \] Input:

int(1/(a + b*acos(d*x^2 + 1))^(3/2),x)
 

Output:

int(1/(a + b*acos(d*x^2 + 1))^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*acos(d*x^2+1))^(3/2),x)
 

Output:

( - sqrt(d)*sqrt(acos(d*x**2 + 1)*b + a)*sqrt( - d*x**2 - 2)*acos(d*x**2 + 
 1) + 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)**2*b**2*d*x**2 + 2*acos(d 
*x**2 + 1)**2*b**2 + 2*acos(d*x**2 + 1)*a*b*d*x**2 + 4*acos(d*x**2 + 1)*a* 
b + a**2*d*x**2 + 2*a**2),x)*a*b*d + sqrt(d)*acos(d*x**2 + 1)*int((sqrt(ac 
os(d*x**2 + 1)*b + a)*sqrt( - d*x**2 - 2)*acos(d*x**2 + 1)**2*x)/(acos(d*x 
**2 + 1)**2*b**2*d*x**2 + 2*acos(d*x**2 + 1)**2*b**2 + 2*acos(d*x**2 + 1)* 
a*b*d*x**2 + 4*acos(d*x**2 + 1)*a*b + a**2*d*x**2 + 2*a**2),x)*b**2*d - ac 
os(d*x**2 + 1)*int((sqrt(acos(d*x**2 + 1)*b + a)*acos(d*x**2 + 1)*x**2)/(a 
cos(d*x**2 + 1)**2*b**2*d*x**2 + 2*acos(d*x**2 + 1)**2*b**2 + 2*acos(d*x** 
2 + 1)*a*b*d*x**2 + 4*acos(d*x**2 + 1)*a*b + a**2*d*x**2 + 2*a**2),x)*b**2 
*d**2 - 2*acos(d*x**2 + 1)*int((sqrt(acos(d*x**2 + 1)*b + a)*acos(d*x**2 + 
 1))/(acos(d*x**2 + 1)**2*b**2*d*x**2 + 2*acos(d*x**2 + 1)**2*b**2 + 2*aco 
s(d*x**2 + 1)*a*b*d*x**2 + 4*acos(d*x**2 + 1)*a*b + a**2*d*x**2 + 2*a**2), 
x)*b**2*d + sqrt(d)*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)**2*b**2*d*x**2 + 2*acos(d*x**2 + 1)* 
*2*b**2 + 2*acos(d*x**2 + 1)*a*b*d*x**2 + 4*acos(d*x**2 + 1)*a*b + a**2*d* 
x**2 + 2*a**2),x)*a**2*d + sqrt(d)*int((sqrt(acos(d*x**2 + 1)*b + a)*sqrt( 
 - d*x**2 - 2)*acos(d*x**2 + 1)**2*x)/(acos(d*x**2 + 1)**2*b**2*d*x**2 + 2 
*acos(d*x**2 + 1)**2*b**2 + 2*acos(d*x**2 + 1)*a*b*d*x**2 + 4*acos(d*x*...