\(\int \frac {1}{(a+b \arccos (1+d x^2))^{7/2}} \, dx\) [70]

Optimal result

Integrand size = 16, antiderivative size = 269 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {\sqrt {-2 d x^2-d^2 x^4}}{5 b d x \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}}+\frac {x}{15 b^2 \left (a+b \arccos \left (1+d x^2\right )\right )^{3/2}}-\frac {\sqrt {-2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt {a+b \arccos \left (1+d x^2\right )}}-\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{15 d x}+\frac {2 \left (\frac {1}{b}\right )^{7/2} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )}{15 d x} \] Output:

1/5*(-d^2*x^4-2*d*x^2)^(1/2)/b/d/x/(a+b*arccos(d*x^2+1))^(5/2)+1/15*x/b^2/ 
(a+b*arccos(d*x^2+1))^(3/2)-1/15*(-d^2*x^4-2*d*x^2)^(1/2)/b^3/d/x/(a+b*arc 
cos(d*x^2+1))^(1/2)-2/15*(1/b)^(7/2)*Pi^(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 
/15*(1/b)^(7/2)*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 = 0.63 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\frac {2 \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right ) \left (-\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+\sqrt {\pi } \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2} \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {b} \left (-\left (\left (-3 b^2+\left (a+b \arccos \left (1+d x^2\right )\right )^2\right ) \cos \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )-b \left (a+b \arccos \left (1+d x^2\right )\right ) \sin \left (\frac {1}{2} \arccos \left (1+d x^2\right )\right )\right )\right )}{15 b^{7/2} d x \left (a+b \arccos \left (1+d x^2\right )\right )^{5/2}} \] Input:

Integrate[(a + b*ArcCos[1 + d*x^2])^(-7/2),x]
 

Output:

(2*Sin[ArcCos[1 + d*x^2]/2]*(-(Sqrt[Pi]*(a + b*ArcCos[1 + d*x^2])^(5/2)*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])^(5/2)*FresnelC[Sqrt[a + b*ArcCos[1 + d*x 
^2]]/(Sqrt[b]*Sqrt[Pi])]*Sin[a/(2*b)] + Sqrt[b]*(-((-3*b^2 + (a + b*ArcCos 
[1 + d*x^2])^2)*Cos[ArcCos[1 + d*x^2]/2]) - b*(a + b*ArcCos[1 + d*x^2])*Si 
n[ArcCos[1 + d*x^2]/2])))/(15*b^(7/2)*d*x*(a + b*ArcCos[1 + d*x^2])^(5/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5328, 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.

\(\displaystyle \int \frac {1}{\left (a+b \arccos \left (d x^2+1\right )\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 5328

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

\(\Big \downarrow \) 5322

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

Input:

Int[(a + b*ArcCos[1 + d*x^2])^(-7/2),x]
 

Output:

Sqrt[-2*d*x^2 - d^2*x^4]/(5*b*d*x*(a + b*ArcCos[1 + d*x^2])^(5/2)) + x/(15 
*b^2*(a + b*ArcCos[1 + d*x^2])^(3/2)) - (Sqrt[-2*d*x^2 - d^2*x^4]/(b*d*x*S 
qrt[a + b*ArcCos[1 + d*x^2]]) + (2*(b^(-1))^(3/2)*Sqrt[Pi]*Cos[a/(2*b)]*Fr 
esnelS[(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)*Sqrt[Pi]*FresnelC[(Sqrt[b^(-1)]*Sq 
rt[a + b*ArcCos[1 + d*x^2]])/Sqrt[Pi]]*Sin[a/(2*b)]*Sin[ArcCos[1 + d*x^2]/ 
2])/(d*x))/(15*b^2)
 

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]
 

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]

\[\int \frac {1}{{\left (a +b \arccos \left (d \,x^{2}+1\right )\right )}^{\frac {7}{2}}}d x\]

Input:

int(1/(a+b*arccos(d*x^2+1))^(7/2),x)
 

Output:

int(1/(a+b*arccos(d*x^2+1))^(7/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {acos}{\left (d x^{2} + 1 \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(a+b*acos(d*x**2+1))**(7/2),x)
 

Output:

Integral((a + b*acos(d*x**2 + 1))**(-7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),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]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int { \frac {1}{{\left (b \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/(a+b*arccos(d*x^2+1))^(7/2),x, algorithm="giac")
 

Output:

integrate((b*arccos(d*x^2 + 1) + a)^(-7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {acos}\left (d\,x^2+1\right )\right )}^{7/2}} \,d x \] Input:

int(1/(a + b*acos(d*x^2 + 1))^(7/2),x)
 

Output:

int(1/(a + b*acos(d*x^2 + 1))^(7/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b \arccos \left (1+d x^2\right )\right )^{7/2}} \, dx=\text {too large to display} \] Input:

int(1/(a+b*acos(d*x^2+1))^(7/2),x)
 

Output:

(sqrt(d)*acos(d*x**2 + 1)**3*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)**4*b**4*d*x**2 + 2*acos(d*x 
**2 + 1)**4*b**4 + 4*acos(d*x**2 + 1)**3*a*b**3*d*x**2 + 8*acos(d*x**2 + 1 
)**3*a*b**3 + 6*acos(d*x**2 + 1)**2*a**2*b**2*d*x**2 + 12*acos(d*x**2 + 1) 
**2*a**2*b**2 + 4*acos(d*x**2 + 1)*a**3*b*d*x**2 + 8*acos(d*x**2 + 1)*a**3 
*b + a**4*d*x**2 + 2*a**4),x)*a*b**3*d + sqrt(d)*acos(d*x**2 + 1)**3*int(( 
sqrt(acos(d*x**2 + 1)*b + a)*sqrt( - d*x**2 - 2)*acos(d*x**2 + 1)**2*x)/(a 
cos(d*x**2 + 1)**4*b**4*d*x**2 + 2*acos(d*x**2 + 1)**4*b**4 + 4*acos(d*x** 
2 + 1)**3*a*b**3*d*x**2 + 8*acos(d*x**2 + 1)**3*a*b**3 + 6*acos(d*x**2 + 1 
)**2*a**2*b**2*d*x**2 + 12*acos(d*x**2 + 1)**2*a**2*b**2 + 4*acos(d*x**2 + 
 1)*a**3*b*d*x**2 + 8*acos(d*x**2 + 1)*a**3*b + a**4*d*x**2 + 2*a**4),x)*b 
**4*d + 3*acos(d*x**2 + 1)**3*int((sqrt(acos(d*x**2 + 1)*b + a)*acos(d*x** 
2 + 1)*x**2)/(acos(d*x**2 + 1)**4*b**4*d*x**2 + 2*acos(d*x**2 + 1)**4*b**4 
 + 4*acos(d*x**2 + 1)**3*a*b**3*d*x**2 + 8*acos(d*x**2 + 1)**3*a*b**3 + 6* 
acos(d*x**2 + 1)**2*a**2*b**2*d*x**2 + 12*acos(d*x**2 + 1)**2*a**2*b**2 + 
4*acos(d*x**2 + 1)*a**3*b*d*x**2 + 8*acos(d*x**2 + 1)*a**3*b + a**4*d*x**2 
 + 2*a**4),x)*b**4*d**2 + 6*acos(d*x**2 + 1)**3*int((sqrt(acos(d*x**2 + 1) 
*b + a)*acos(d*x**2 + 1))/(acos(d*x**2 + 1)**4*b**4*d*x**2 + 2*acos(d*x**2 
 + 1)**4*b**4 + 4*acos(d*x**2 + 1)**3*a*b**3*d*x**2 + 8*acos(d*x**2 + 1)** 
3*a*b**3 + 6*acos(d*x**2 + 1)**2*a**2*b**2*d*x**2 + 12*acos(d*x**2 + 1)...