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

Optimal. Leaf size=141 \[ \frac{\text{CosIntegral}\left (\cos ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )}{8 a^3}-\frac{3 x^2 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)}+\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac{\sqrt{1-a^2 x^2}}{3 a^3 \cos ^{-1}(a x)}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2} \]

[Out]

(x^2*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) - x/(3*a^2*ArcCos[a*x]^2) + x^3/(2*ArcCos[a*x]^2) + Sqrt[1 - a^2*x
^2]/(3*a^3*ArcCos[a*x]) - (3*x^2*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]) + CosIntegral[ArcCos[a*x]]/(24*a^3) + (9
*CosIntegral[3*ArcCos[a*x]])/(8*a^3)

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Rubi [A]  time = 0.312028, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4634, 4720, 4632, 3302, 4622, 4724} \[ \frac{\text{CosIntegral}\left (\cos ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )}{8 a^3}-\frac{3 x^2 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)}+\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}+\frac{\sqrt{1-a^2 x^2}}{3 a^3 \cos ^{-1}(a x)}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) - x/(3*a^2*ArcCos[a*x]^2) + x^3/(2*ArcCos[a*x]^2) + Sqrt[1 - a^2*x
^2]/(3*a^3*ArcCos[a*x]) - (3*x^2*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]) + CosIntegral[ArcCos[a*x]]/(24*a^3) + (9
*CosIntegral[3*ArcCos[a*x]])/(8*a^3)

Rule 4634

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[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcCos[c*x])^(n + 1
))/Sqrt[1 - c^2*x^2], x], x] + Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcCos[c*x])^(n + 1))/Sqrt[1 - c^2*
x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

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]

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]

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 4724

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Cos[x]^m*Sin[x]^(2*p + 1), x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin{align*} \int \frac{x^2}{\cos ^{-1}(a x)^4} \, dx &=\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{2 \int \frac{x}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3} \, dx}{3 a}+a \int \frac{x^3}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)^3} \, dx\\ &=\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2}-\frac{3}{2} \int \frac{x^2}{\cos ^{-1}(a x)^2} \, dx+\frac{\int \frac{1}{\cos ^{-1}(a x)^2} \, dx}{3 a^2}\\ &=\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2}+\frac{\sqrt{1-a^2 x^2}}{3 a^3 \cos ^{-1}(a x)}-\frac{3 x^2 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)}-\frac{3 \operatorname{Subst}\left (\int \left (-\frac{\cos (x)}{4 x}-\frac{3 \cos (3 x)}{4 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{2 a^3}+\frac{\int \frac{x}{\sqrt{1-a^2 x^2} \cos ^{-1}(a x)} \, dx}{3 a}\\ &=\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2}+\frac{\sqrt{1-a^2 x^2}}{3 a^3 \cos ^{-1}(a x)}-\frac{3 x^2 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)}-\frac{\operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^3}\\ &=\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{x}{3 a^2 \cos ^{-1}(a x)^2}+\frac{x^3}{2 \cos ^{-1}(a x)^2}+\frac{\sqrt{1-a^2 x^2}}{3 a^3 \cos ^{-1}(a x)}-\frac{3 x^2 \sqrt{1-a^2 x^2}}{2 a \cos ^{-1}(a x)}+\frac{\text{Ci}\left (\cos ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{Ci}\left (3 \cos ^{-1}(a x)\right )}{8 a^3}\\ \end{align*}

Mathematica [A]  time = 0.119853, size = 129, normalized size = 0.91 \[ -\frac{10 \text{CosIntegral}\left (\cos ^{-1}(a x)\right )}{3 a^3}-\frac{9 \left (-3 \text{CosIntegral}\left (\cos ^{-1}(a x)\right )-\text{CosIntegral}\left (3 \cos ^{-1}(a x)\right )\right )}{8 a^3}+\frac{x^2 \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{\sqrt{1-a^2 x^2} \left (9 a^2 x^2-2\right )}{6 a^3 \cos ^{-1}(a x)}+\frac{3 a^2 x^3-2 x}{6 a^2 \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) + (-2*x + 3*a^2*x^3)/(6*a^2*ArcCos[a*x]^2) - (Sqrt[1 - a^2*x^2]*(-
2 + 9*a^2*x^2))/(6*a^3*ArcCos[a*x]) - (10*CosIntegral[ArcCos[a*x]])/(3*a^3) - (9*(-3*CosIntegral[ArcCos[a*x]]
- CosIntegral[3*ArcCos[a*x]]))/(8*a^3)

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Maple [A]  time = 0.056, size = 117, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{3}} \left ({\frac{1}{12\, \left ( \arccos \left ( ax \right ) \right ) ^{3}}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{ax}{24\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}-{\frac{1}{24\,\arccos \left ( ax \right ) }\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{{\it Ci} \left ( \arccos \left ( ax \right ) \right ) }{24}}+{\frac{\sin \left ( 3\,\arccos \left ( ax \right ) \right ) }{12\, \left ( \arccos \left ( ax \right ) \right ) ^{3}}}+{\frac{\cos \left ( 3\,\arccos \left ( ax \right ) \right ) }{8\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\sin \left ( 3\,\arccos \left ( ax \right ) \right ) }{8\,\arccos \left ( ax \right ) }}+{\frac{9\,{\it Ci} \left ( 3\,\arccos \left ( ax \right ) \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*(1/12/arccos(a*x)^3*(-a^2*x^2+1)^(1/2)+1/24*a*x/arccos(a*x)^2-1/24*(-a^2*x^2+1)^(1/2)/arccos(a*x)+1/24*C
i(arccos(a*x))+1/12/arccos(a*x)^3*sin(3*arccos(a*x))+1/8/arccos(a*x)^2*cos(3*arccos(a*x))-3/8/arccos(a*x)*sin(
3*arccos(a*x))+9/8*Ci(3*arccos(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{3} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3} \int \frac{{\left (27 \, a^{2} x^{3} - 20 \, x\right )} \sqrt{a x + 1} \sqrt{-a x + 1}}{{\left (a^{3} x^{2} - a\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}\,{d x} +{\left (2 \, a^{2} x^{2} -{\left (9 \, a^{2} x^{2} - 2\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2}\right )} \sqrt{a x + 1} \sqrt{-a x + 1} +{\left (3 \, a^{3} x^{3} - 2 \, a x\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{6 \, a^{3} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a^3*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3*integrate(1/6*(27*a^2*x^3 - 20*x)*sqrt(a*x + 1)*sqrt(-
a*x + 1)/((a^3*x^2 - a)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x) + (2*a^2*x^2 - (9*a^2*x^2 - 2)*arctan2
(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)*sqrt(a*x + 1)*sqrt(-a*x + 1) + (3*a^3*x^3 - 2*a*x)*arctan2(sqrt(a*x + 1
)*sqrt(-a*x + 1), a*x))/(a^3*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/arccos(a*x)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/acos(a*x)**4, x)

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Giac [A]  time = 1.19292, size = 163, normalized size = 1.16 \begin{align*} \frac{x^{3}}{2 \, \arccos \left (a x\right )^{2}} - \frac{3 \, \sqrt{-a^{2} x^{2} + 1} x^{2}}{2 \, a \arccos \left (a x\right )} + \frac{\sqrt{-a^{2} x^{2} + 1} x^{2}}{3 \, a \arccos \left (a x\right )^{3}} + \frac{9 \, \operatorname{Ci}\left (3 \, \arccos \left (a x\right )\right )}{8 \, a^{3}} + \frac{\operatorname{Ci}\left (\arccos \left (a x\right )\right )}{24 \, a^{3}} - \frac{x}{3 \, a^{2} \arccos \left (a x\right )^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{3 \, a^{3} \arccos \left (a x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^3/arccos(a*x)^2 - 3/2*sqrt(-a^2*x^2 + 1)*x^2/(a*arccos(a*x)) + 1/3*sqrt(-a^2*x^2 + 1)*x^2/(a*arccos(a*x)
^3) + 9/8*cos_integral(3*arccos(a*x))/a^3 + 1/24*cos_integral(arccos(a*x))/a^3 - 1/3*x/(a^2*arccos(a*x)^2) + 1
/3*sqrt(-a^2*x^2 + 1)/(a^3*arccos(a*x))