3.53 \(\int \frac{x^4}{\cos ^{-1}(a x)^2} \, dx\)

Optimal. Leaf size=68 \[ -\frac{\text{CosIntegral}\left (\cos ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{CosIntegral}\left (5 \cos ^{-1}(a x)\right )}{16 a^5}+\frac{x^4 \sqrt{1-a^2 x^2}}{a \cos ^{-1}(a x)} \]

[Out]

(x^4*Sqrt[1 - a^2*x^2])/(a*ArcCos[a*x]) - CosIntegral[ArcCos[a*x]]/(8*a^5) - (9*CosIntegral[3*ArcCos[a*x]])/(1
6*a^5) - (5*CosIntegral[5*ArcCos[a*x]])/(16*a^5)

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Rubi [A]  time = 0.062514, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4632, 3302} \[ -\frac{\text{CosIntegral}\left (\cos ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{CosIntegral}\left (5 \cos ^{-1}(a x)\right )}{16 a^5}+\frac{x^4 \sqrt{1-a^2 x^2}}{a \cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCos[a*x]^2,x]

[Out]

(x^4*Sqrt[1 - a^2*x^2])/(a*ArcCos[a*x]) - CosIntegral[ArcCos[a*x]]/(8*a^5) - (9*CosIntegral[3*ArcCos[a*x]])/(1
6*a^5) - (5*CosIntegral[5*ArcCos[a*x]])/(16*a^5)

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]

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]

Rubi steps

\begin{align*} \int \frac{x^4}{\cos ^{-1}(a x)^2} \, dx &=\frac{x^4 \sqrt{1-a^2 x^2}}{a \cos ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \left (-\frac{\cos (x)}{8 x}-\frac{9 \cos (3 x)}{16 x}-\frac{5 \cos (5 x)}{16 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^5}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{a \cos ^{-1}(a x)}-\frac{\operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^5}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos (5 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^5}-\frac{9 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^5}\\ &=\frac{x^4 \sqrt{1-a^2 x^2}}{a \cos ^{-1}(a x)}-\frac{\text{Ci}\left (\cos ^{-1}(a x)\right )}{8 a^5}-\frac{9 \text{Ci}\left (3 \cos ^{-1}(a x)\right )}{16 a^5}-\frac{5 \text{Ci}\left (5 \cos ^{-1}(a x)\right )}{16 a^5}\\ \end{align*}

Mathematica [A]  time = 0.160174, size = 61, normalized size = 0.9 \[ -\frac{-\frac{16 a^4 x^4 \sqrt{1-a^2 x^2}}{\cos ^{-1}(a x)}+2 \text{CosIntegral}\left (\cos ^{-1}(a x)\right )+9 \text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )+5 \text{CosIntegral}\left (5 \cos ^{-1}(a x)\right )}{16 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcCos[a*x]^2,x]

[Out]

-((-16*a^4*x^4*Sqrt[1 - a^2*x^2])/ArcCos[a*x] + 2*CosIntegral[ArcCos[a*x]] + 9*CosIntegral[3*ArcCos[a*x]] + 5*
CosIntegral[5*ArcCos[a*x]])/(16*a^5)

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Maple [A]  time = 0.05, size = 81, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{5}} \left ({\frac{3\,\sin \left ( 3\,\arccos \left ( ax \right ) \right ) }{16\,\arccos \left ( ax \right ) }}-{\frac{9\,{\it Ci} \left ( 3\,\arccos \left ( ax \right ) \right ) }{16}}+{\frac{\sin \left ( 5\,\arccos \left ( ax \right ) \right ) }{16\,\arccos \left ( ax \right ) }}-{\frac{5\,{\it Ci} \left ( 5\,\arccos \left ( ax \right ) \right ) }{16}}+{\frac{1}{8\,\arccos \left ( ax \right ) }\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{{\it Ci} \left ( \arccos \left ( ax \right ) \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arccos(a*x)^2,x)

[Out]

1/a^5*(3/16/arccos(a*x)*sin(3*arccos(a*x))-9/16*Ci(3*arccos(a*x))+1/16/arccos(a*x)*sin(5*arccos(a*x))-5/16*Ci(
5*arccos(a*x))+1/8*(-a^2*x^2+1)^(1/2)/arccos(a*x)-1/8*Ci(arccos(a*x)))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{4}}{\arccos \left (a x\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^4/arccos(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\operatorname{acos}^{2}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/acos(a*x)**2,x)

[Out]

Integral(x**4/acos(a*x)**2, x)

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Giac [A]  time = 1.17492, size = 81, normalized size = 1.19 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1} x^{4}}{a \arccos \left (a x\right )} - \frac{5 \, \operatorname{Ci}\left (5 \, \arccos \left (a x\right )\right )}{16 \, a^{5}} - \frac{9 \, \operatorname{Ci}\left (3 \, \arccos \left (a x\right )\right )}{16 \, a^{5}} - \frac{\operatorname{Ci}\left (\arccos \left (a x\right )\right )}{8 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^2,x, algorithm="giac")

[Out]

sqrt(-a^2*x^2 + 1)*x^4/(a*arccos(a*x)) - 5/16*cos_integral(5*arccos(a*x))/a^5 - 9/16*cos_integral(3*arccos(a*x
))/a^5 - 1/8*cos_integral(arccos(a*x))/a^5