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

Optimal. Leaf size=70 \[ -\frac {5 \text {Ci}\left (2 \cos ^{-1}(a x)\right )}{16 a^6}-\frac {\text {Ci}\left (4 \cos ^{-1}(a x)\right )}{2 a^6}-\frac {3 \text {Ci}\left (6 \cos ^{-1}(a x)\right )}{16 a^6}+\frac {x^5 \sqrt {1-a^2 x^2}}{a \cos ^{-1}(a x)} \]

[Out]

-5/16*Ci(2*arccos(a*x))/a^6-1/2*Ci(4*arccos(a*x))/a^6-3/16*Ci(6*arccos(a*x))/a^6+x^5*(-a^2*x^2+1)^(1/2)/a/arcc
os(a*x)

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Rubi [A]  time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4632, 3302} \[ -\frac {5 \text {CosIntegral}\left (2 \cos ^{-1}(a x)\right )}{16 a^6}-\frac {\text {CosIntegral}\left (4 \cos ^{-1}(a x)\right )}{2 a^6}-\frac {3 \text {CosIntegral}\left (6 \cos ^{-1}(a x)\right )}{16 a^6}+\frac {x^5 \sqrt {1-a^2 x^2}}{a \cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^5*Sqrt[1 - a^2*x^2])/(a*ArcCos[a*x]) - (5*CosIntegral[2*ArcCos[a*x]])/(16*a^6) - CosIntegral[4*ArcCos[a*x]]
/(2*a^6) - (3*CosIntegral[6*ArcCos[a*x]])/(16*a^6)

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 4632

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1
), Cos[x]^(m - 1)*(m - (m + 1)*Cos[x]^2), x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] &&
GeQ[n, -2] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x^5}{\cos ^{-1}(a x)^2} \, dx &=\frac {x^5 \sqrt {1-a^2 x^2}}{a \cos ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \left (-\frac {5 \cos (2 x)}{16 x}-\frac {\cos (4 x)}{2 x}-\frac {3 \cos (6 x)}{16 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^6}\\ &=\frac {x^5 \sqrt {1-a^2 x^2}}{a \cos ^{-1}(a x)}-\frac {3 \operatorname {Subst}\left (\int \frac {\cos (6 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^6}-\frac {5 \operatorname {Subst}\left (\int \frac {\cos (2 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^6}-\frac {\operatorname {Subst}\left (\int \frac {\cos (4 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^6}\\ &=\frac {x^5 \sqrt {1-a^2 x^2}}{a \cos ^{-1}(a x)}-\frac {5 \text {Ci}\left (2 \cos ^{-1}(a x)\right )}{16 a^6}-\frac {\text {Ci}\left (4 \cos ^{-1}(a x)\right )}{2 a^6}-\frac {3 \text {Ci}\left (6 \cos ^{-1}(a x)\right )}{16 a^6}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 63, normalized size = 0.90 \[ -\frac {-\frac {16 a^5 x^5 \sqrt {1-a^2 x^2}}{\cos ^{-1}(a x)}+5 \text {Ci}\left (2 \cos ^{-1}(a x)\right )+8 \text {Ci}\left (4 \cos ^{-1}(a x)\right )+3 \text {Ci}\left (6 \cos ^{-1}(a x)\right )}{16 a^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((-16*a^5*x^5*Sqrt[1 - a^2*x^2])/ArcCos[a*x] + 5*CosIntegral[2*ArcCos[a*x]] + 8*CosIntegral[4*ArcCos[a*x
]] + 3*CosIntegral[6*ArcCos[a*x]])/a^6

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{5}}{\arccos \left (a x\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^5/arccos(a*x)^2, x)

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giac [A]  time = 0.19, size = 62, normalized size = 0.89 \[ \frac {\sqrt {-a^{2} x^{2} + 1} x^{5}}{a \arccos \left (a x\right )} - \frac {3 \, \operatorname {Ci}\left (6 \, \arccos \left (a x\right )\right )}{16 \, a^{6}} - \frac {\operatorname {Ci}\left (4 \, \arccos \left (a x\right )\right )}{2 \, a^{6}} - \frac {5 \, \operatorname {Ci}\left (2 \, \arccos \left (a x\right )\right )}{16 \, a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a^2*x^2 + 1)*x^5/(a*arccos(a*x)) - 3/16*cos_integral(6*arccos(a*x))/a^6 - 1/2*cos_integral(4*arccos(a*x)
)/a^6 - 5/16*cos_integral(2*arccos(a*x))/a^6

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maple [A]  time = 0.15, size = 78, normalized size = 1.11 \[ \frac {\frac {5 \sin \left (2 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}-\frac {5 \Ci \left (2 \arccos \left (a x \right )\right )}{16}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )}-\frac {\Ci \left (4 \arccos \left (a x \right )\right )}{2}+\frac {\sin \left (6 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}-\frac {3 \Ci \left (6 \arccos \left (a x \right )\right )}{16}}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^6*(5/32/arccos(a*x)*sin(2*arccos(a*x))-5/16*Ci(2*arccos(a*x))+1/8/arccos(a*x)*sin(4*arccos(a*x))-1/2*Ci(4*
arccos(a*x))+1/32/arccos(a*x)*sin(6*arccos(a*x))-3/16*Ci(6*arccos(a*x)))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^5/acos(a*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**5/acos(a*x)**2, x)

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