\(\int \frac {1}{(a+b \arccos (-1+d x^2))^{5/2}} \, dx\) [99]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 221, 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 = {4913, 4905} \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}} \, dx=\frac {x}{3 b^2 \sqrt {a+b \arccos \left (d x^2-1\right )}}+\frac {\sqrt {2 d x^2-d^2 x^4}}{3 b d x \left (a+b \arccos \left (d x^2-1\right )\right )^{3/2}}-\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{3 d x}+\frac {2 \sqrt {\pi } \left (\frac {1}{b}\right )^{5/2} \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{3 d x} \]

[In]

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

[Out]

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

Rule 4905

Int[1/Sqrt[(a_.) + ArcCos[-1 + (d_.)*(x_)^2]*(b_.)], x_Symbol] :> Simp[2*Sqrt[Pi/b]*Sin[a/(2*b)]*Cos[ArcCos[-1
+ d*x^2]/2]*(FresnelC[Sqrt[1/(Pi*b)]*Sqrt[a + b*ArcCos[-1 + d*x^2]]]/(d*x)), x] - Simp[2*Sqrt[Pi/b]*Cos[a/(2*
b)]*Cos[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 4913

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] + (-Dist[1/(4*b^2*(n + 1)*(n + 2)), Int[(a + b*ArcCos[c + d*x^2])^(n + 2), x], x]
- Simp[Sqrt[-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]) /; FreeQ[{a, b, c
, d}, x] && EqQ[c^2, 1] && LtQ[n, -1] && NeQ[n, -2]

Rubi steps \begin{align*} \text {integral}& = \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 {\int \frac {1}{\sqrt {a+b \arccos \left (-1+d x^2\right )}} \, dx}{3 b^2} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}} \, dx=\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (\sqrt {\pi } \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )-\sqrt {\pi } \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2} \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {b} \left (\left (a+b \arccos \left (-1+d x^2\right )\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+b \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )\right )}{3 b^{5/2} d x \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}} \]

[In]

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

[Out]

(2*Cos[ArcCos[-1 + d*x^2]/2]*(Sqrt[Pi]*(a + b*ArcCos[-1 + d*x^2])^(3/2)*Cos[a/(2*b)]*FresnelS[Sqrt[a + b*ArcCo
s[-1 + d*x^2]]/(Sqrt[b]*Sqrt[Pi])] - Sqrt[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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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