\(\int \frac {1}{\arccos (a x)^{5/2}} \, dx\) [111]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F(-2)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 8, antiderivative size = 76 \[ \int \frac {1}{\arccos (a x)^{5/2}} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 x}{3 \sqrt {\arccos (a x)}}+\frac {4 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{3 a} \]

[Out]

4/3*FresnelS(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a+2/3*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(3/2)
+4/3*x/arccos(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4718, 4808, 4720, 3386, 3432} \[ \int \frac {1}{\arccos (a x)^{5/2}} \, dx=\frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{3 a}+\frac {4 x}{3 \sqrt {\arccos (a x)}} \]

[In]

Int[ArcCos[a*x]^(-5/2),x]

[Out]

(2*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^(3/2)) + (4*x)/(3*Sqrt[ArcCos[a*x]]) + (4*Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi
]*Sqrt[ArcCos[a*x]]])/(3*a)

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4718

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-Sqrt[1 - c^2*x^2])*((a + b*ArcCos[c*x])^(n +
1)/(b*c*(n + 1))), x] - Dist[c/(b*(n + 1)), Int[x*((a + b*ArcCos[c*x])^(n + 1)/Sqrt[1 - c^2*x^2]), x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4720

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[-(b*c)^(-1), Subst[Int[x^n*Sin[-a/b + x/b], x],
 x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 4808

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
(-(f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcCos[c*x])^(n + 1), x] + Dist[f*(m/(
b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcCos[c*x])^(n + 1), x], x] /
; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {1}{3} (2 a) \int \frac {x}{\sqrt {1-a^2 x^2} \arccos (a x)^{3/2}} \, dx \\ & = \frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 x}{3 \sqrt {\arccos (a x)}}-\frac {4}{3} \int \frac {1}{\sqrt {\arccos (a x)}} \, dx \\ & = \frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 x}{3 \sqrt {\arccos (a x)}}+\frac {4 \text {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\arccos (a x)\right )}{3 a} \\ & = \frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 x}{3 \sqrt {\arccos (a x)}}+\frac {8 \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\arccos (a x)}\right )}{3 a} \\ & = \frac {2 \sqrt {1-a^2 x^2}}{3 a \arccos (a x)^{3/2}}+\frac {4 x}{3 \sqrt {\arccos (a x)}}+\frac {4 \sqrt {2 \pi } \operatorname {FresnelS}\left (\sqrt {\frac {2}{\pi }} \sqrt {\arccos (a x)}\right )}{3 a} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\arccos (a x)^{5/2}} \, dx=-\frac {2 \left (-\sqrt {1-a^2 x^2}-e^{-i \arccos (a x)} \arccos (a x)-e^{i \arccos (a x)} \arccos (a x)+\sqrt {-i \arccos (a x)} \arccos (a x) \Gamma \left (\frac {1}{2},-i \arccos (a x)\right )+\sqrt {i \arccos (a x)} \arccos (a x) \Gamma \left (\frac {1}{2},i \arccos (a x)\right )\right )}{3 a \arccos (a x)^{3/2}} \]

[In]

Integrate[ArcCos[a*x]^(-5/2),x]

[Out]

(-2*(-Sqrt[1 - a^2*x^2] - ArcCos[a*x]/E^(I*ArcCos[a*x]) - E^(I*ArcCos[a*x])*ArcCos[a*x] + Sqrt[(-I)*ArcCos[a*x
]]*ArcCos[a*x]*Gamma[1/2, (-I)*ArcCos[a*x]] + Sqrt[I*ArcCos[a*x]]*ArcCos[a*x]*Gamma[1/2, I*ArcCos[a*x]]))/(3*a
*ArcCos[a*x]^(3/2))

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09

method result size
default \(\frac {\sqrt {2}\, \left (4 \arccos \left (a x \right )^{2} \pi \,\operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )+2 \arccos \left (a x \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, a x +\sqrt {2}\, \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{3 a \sqrt {\pi }\, \arccos \left (a x \right )^{2}}\) \(83\)

[In]

int(1/arccos(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/a*2^(1/2)/Pi^(1/2)*(4*arccos(a*x)^2*Pi*FresnelS(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))+2*arccos(a*x)^(3/2)*2^
(1/2)*Pi^(1/2)*a*x+2^(1/2)*arccos(a*x)^(1/2)*Pi^(1/2)*(-a^2*x^2+1)^(1/2))/arccos(a*x)^2

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(1/acos(a*x)**(5/2),x)

[Out]

Integral(acos(a*x)**(-5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\arccos (a x)^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

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

[In]

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

[Out]

integrate(arccos(a*x)^(-5/2), x)

Mupad [F(-1)]

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

[In]

int(1/acos(a*x)^(5/2),x)

[Out]

int(1/acos(a*x)^(5/2), x)