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

Optimal. Leaf size=97 \[ \frac{2 \text{CosIntegral}\left (2 \cos ^{-1}(a x)\right )}{3 a^2}-\frac{2 x \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)}+\frac{x \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{1}{6 a^2 \cos ^{-1}(a x)^2}+\frac{x^2}{3 \cos ^{-1}(a x)^2} \]

[Out]

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

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Rubi [A]  time = 0.166116, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {4634, 4720, 4632, 3302, 4642} \[ \frac{2 \text{CosIntegral}\left (2 \cos ^{-1}(a x)\right )}{3 a^2}-\frac{2 x \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)}+\frac{x \sqrt{1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac{1}{6 a^2 \cos ^{-1}(a x)^2}+\frac{x^2}{3 \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4642

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

Rubi steps

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

Mathematica [A]  time = 0.103208, size = 86, normalized size = 0.89 \[ \frac{2 a x \sqrt{1-a^2 x^2}-4 a x \sqrt{1-a^2 x^2} \cos ^{-1}(a x)^2+\left (2 a^2 x^2-1\right ) \cos ^{-1}(a x)+4 \cos ^{-1}(a x)^3 \text{CosIntegral}\left (2 \cos ^{-1}(a x)\right )}{6 a^2 \cos ^{-1}(a x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 60, normalized size = 0.6 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{\sin \left ( 2\,\arccos \left ( ax \right ) \right ) }{6\, \left ( \arccos \left ( ax \right ) \right ) ^{3}}}+{\frac{\cos \left ( 2\,\arccos \left ( ax \right ) \right ) }{6\, \left ( \arccos \left ( ax \right ) \right ) ^{2}}}-{\frac{\sin \left ( 2\,\arccos \left ( ax \right ) \right ) }{3\,\arccos \left ( ax \right ) }}+{\frac{2\,{\it Ci} \left ( 2\,\arccos \left ( ax \right ) \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(1/6/arccos(a*x)^3*sin(2*arccos(a*x))+1/6/arccos(a*x)^2*cos(2*arccos(a*x))-1/3/arccos(a*x)*sin(2*arccos(
a*x))+2/3*Ci(2*arccos(a*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{4 \, a^{2} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{3} \int \frac{{\left (2 \, a^{2} x^{2} - 1\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} - 2 \,{\left (2 \, a x \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )^{2} - a x\right )} \sqrt{a x + 1} \sqrt{-a x + 1} +{\left (2 \, a^{2} x^{2} - 1\right )} \arctan \left (\sqrt{a x + 1} \sqrt{-a x + 1}, a x\right )}{6 \, a^{2} \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/arccos(a*x)^4,x, algorithm="maxima")

[Out]

1/6*(6*a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3*integrate(2/3*(2*a^2*x^2 - 1)*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*(2*a*x*arctan2(sqrt(a*x + 1)*sqrt(-a*x
+ 1), a*x)^2 - a*x)*sqrt(a*x + 1)*sqrt(-a*x + 1) + (2*a^2*x^2 - 1)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)
)/(a^2*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}{\arccos \left (a x\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arccos(a*x)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/acos(a*x)**4, x)

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Giac [A]  time = 1.15565, size = 112, normalized size = 1.15 \begin{align*} \frac{x^{2}}{3 \, \arccos \left (a x\right )^{2}} - \frac{2 \, \sqrt{-a^{2} x^{2} + 1} x}{3 \, a \arccos \left (a x\right )} + \frac{2 \, \operatorname{Ci}\left (2 \, \arccos \left (a x\right )\right )}{3 \, a^{2}} + \frac{\sqrt{-a^{2} x^{2} + 1} x}{3 \, a \arccos \left (a x\right )^{3}} - \frac{1}{6 \, a^{2} \arccos \left (a x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^2/arccos(a*x)^2 - 2/3*sqrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)) + 2/3*cos_integral(2*arccos(a*x))/a^2 + 1/3*s
qrt(-a^2*x^2 + 1)*x/(a*arccos(a*x)^3) - 1/6/(a^2*arccos(a*x)^2)