\(\int (a+b \arccos (-1+d x^2))^{3/2} \, dx\) [72]

Optimal result

Integrand size = 16, antiderivative size = 207 \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \, dx=-\frac {3 b \sqrt {2 d x^2-d^2 x^4} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{d x}+x \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}+\frac {6 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{3/2} d x}-\frac {6 \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{\left (\frac {1}{b}\right )^{3/2} d x} \] Output:

-3*b*(-d^2*x^4+2*d*x^2)^(1/2)*(a+b*arccos(d*x^2-1))^(1/2)/d/x+x*(a+b*arcco 
s(d*x^2-1))^(3/2)+6*Pi^(1/2)*cos(1/2*a/b)*cos(1/2*arccos(d*x^2-1))*Fresnel 
S((1/b)^(1/2)*(a+b*arccos(d*x^2-1))^(1/2)/Pi^(1/2))/(1/b)^(3/2)/d/x-6*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)/(1/b)^(3/2)/d/x
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.93 \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \, dx=\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (3 b^{3/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )-3 b^{3/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {a+b \arccos \left (-1+d x^2\right )} \left (a \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+b \arccos \left (-1+d x^2\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )-3 b \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )\right )}{d x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5314, 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.

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

\(\Big \downarrow \) 5314

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

\(\Big \downarrow \) 5320

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

Input:

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

Output:

(-3*b*Sqrt[2*d*x^2 - d^2*x^4]*Sqrt[a + b*ArcCos[-1 + d*x^2]])/(d*x) + x*(a 
 + b*ArcCos[-1 + d*x^2])^(3/2) - 3*b^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]]*S 
in[a/(2*b)])/(d*x))
 

Defintions of rubi rules used

rule 5314

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

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

integrate((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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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