3.111 \(\int \frac {1}{\cos ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{3 a}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}} \]

[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)

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Rubi [A]  time = 0.09, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4622, 4720, 4624, 3305, 3351} \[ \frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{3 a}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[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 3305

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 3351

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

Rule 4622

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] /; Fre
eQ[{a, b, c}, x] && LtQ[n, -1]

Rule 4624

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

Rule 4720

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

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{-1}(a x)^{5/2}} \, dx &=\frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {1}{3} (2 a) \int \frac {x}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}} \, dx\\ &=\frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}}-\frac {4}{3} \int \frac {1}{\sqrt {\cos ^{-1}(a x)}} \, dx\\ &=\frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {4 \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{3 a}\\ &=\frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {8 \operatorname {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{3 a}\\ &=\frac {2 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}+\frac {4 x}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{3 a}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 122, normalized size = 1.61 \[ -\frac {2 \left (-\sqrt {1-a^2 x^2}-e^{-i \cos ^{-1}(a x)} \cos ^{-1}(a x)-e^{i \cos ^{-1}(a x)} \cos ^{-1}(a x)+\sqrt {-i \cos ^{-1}(a x)} \cos ^{-1}(a x) \Gamma \left (\frac {1}{2},-i \cos ^{-1}(a x)\right )+\sqrt {i \cos ^{-1}(a x)} \cos ^{-1}(a x) \Gamma \left (\frac {1}{2},i \cos ^{-1}(a x)\right )\right )}{3 a \cos ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[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))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\arccos \left (a x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(a*x)^(-5/2), x)

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maple [A]  time = 0.17, size = 83, normalized size = 1.09 \[ \frac {\sqrt {2}\, \left (4 \pi \arccos \left (a x \right )^{2} \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )+2 \arccos \left (a x \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, x a +\sqrt {2}\, \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }\, \sqrt {-a^{2} x^{2}+1}\right )}{3 a \sqrt {\pi }\, \arccos \left (a x \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {acos}\left (a\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/acos(a*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(a*x)**(-5/2), x)

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