\(\int \frac {1}{(a+b \arccos (1+d x^2))^3} \, dx\) [56]

Optimal result

Integrand size = 14, antiderivative size = 173 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^3} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{4 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {x}{8 b^2 \left (a+b \arccos \left (1+d x^2\right )\right )}-\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}}-\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )}{8 \sqrt {2} b^3 \sqrt {-d x^2}} \] Output:

1/4*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))^2+1/8*x/b^2/(a+b* 
arccos(d*x^2+1))-1/16*x*cos(1/2*a/b)*Ci(1/2*(a+b*arccos(d*x^2+1))/b)*2^(1/ 
2)/b^3/(-d*x^2)^(1/2)-1/16*x*sin(1/2*a/b)*Si(1/2*(a+b*arccos(d*x^2+1))/b)* 
2^(1/2)/b^3/(-d*x^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^3} \, dx=\frac {\frac {2 b^2 \sqrt {-d x^2 \left (2+d x^2\right )}}{d \left (a+b \arccos \left (1+d x^2\right )\right )^2}+\frac {b x^2}{a+b \arccos \left (1+d x^2\right )}+\frac {\sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (\cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )+\sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (1+d x^2\right )}{2 b}\right )\right )}{d}}{8 b^3 x} \] Input:

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

Output:

((2*b^2*Sqrt[-(d*x^2*(2 + d*x^2))])/(d*(a + b*ArcCos[1 + d*x^2])^2) + (b*x 
^2)/(a + b*ArcCos[1 + d*x^2]) + (Sin[ArcCos[1 + d*x^2]/2]*(Cos[a/(2*b)]*Co 
sIntegral[(a + b*ArcCos[1 + d*x^2])/(2*b)] + Sin[a/(2*b)]*SinIntegral[(a + 
 b*ArcCos[1 + d*x^2])/(2*b)]))/d)/(8*b^3*x)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.01, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5328, 5316}

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.

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

\(\Big \downarrow \) 5328

\(\displaystyle -\frac {\int \frac {1}{a+b \arccos \left (d x^2+1\right )}dx}{8 b^2}+\frac {x}{8 b^2 \left (a+b \arccos \left (d x^2+1\right )\right )}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^2}\)

\(\Big \downarrow \) 5316

\(\displaystyle -\frac {\frac {x \cos \left (\frac {a}{2 b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arccos \left (d x^2+1\right )}{2 b}\right )}{\sqrt {2} b \sqrt {-d x^2}}+\frac {x \sin \left (\frac {a}{2 b}\right ) \text {Si}\left (\frac {a+b \arccos \left (d x^2+1\right )}{2 b}\right )}{\sqrt {2} b \sqrt {-d x^2}}}{8 b^2}+\frac {x}{8 b^2 \left (a+b \arccos \left (d x^2+1\right )\right )}+\frac {\sqrt {-d^2 x^4-2 d x^2}}{4 b d x \left (a+b \arccos \left (d x^2+1\right )\right )^2}\)

Input:

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

Output:

Sqrt[-2*d*x^2 - d^2*x^4]/(4*b*d*x*(a + b*ArcCos[1 + d*x^2])^2) + x/(8*b^2* 
(a + b*ArcCos[1 + d*x^2])) - ((x*Cos[a/(2*b)]*CosIntegral[(a + b*ArcCos[1 
+ d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[-(d*x^2)]) + (x*Sin[a/(2*b)]*SinIntegral 
[(a + b*ArcCos[1 + d*x^2])/(2*b)])/(Sqrt[2]*b*Sqrt[-(d*x^2)]))/(8*b^2)
 

Defintions of rubi rules used

rule 5316

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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