3.1.68 \(\int \frac {x^3}{\text {ArcCos}(a x)^4} \, dx\) [68]

Optimal. Leaf size=143 \[ \frac {x^3 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}-\frac {x^2}{2 a^2 \text {ArcCos}(a x)^2}+\frac {2 x^4}{3 \text {ArcCos}(a x)^2}+\frac {x \sqrt {1-a^2 x^2}}{a^3 \text {ArcCos}(a x)}-\frac {8 x^3 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)}+\frac {\text {CosIntegral}(2 \text {ArcCos}(a x))}{3 a^4}+\frac {4 \text {CosIntegral}(4 \text {ArcCos}(a x))}{3 a^4} \]

[Out]

-1/2*x^2/a^2/arccos(a*x)^2+2/3*x^4/arccos(a*x)^2+1/3*Ci(2*arccos(a*x))/a^4+4/3*Ci(4*arccos(a*x))/a^4+1/3*x^3*(
-a^2*x^2+1)^(1/2)/a/arccos(a*x)^3+x*(-a^2*x^2+1)^(1/2)/a^3/arccos(a*x)-8/3*x^3*(-a^2*x^2+1)^(1/2)/a/arccos(a*x
)

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4730, 4808, 4728, 3383} \begin {gather*} \frac {\text {CosIntegral}(2 \text {ArcCos}(a x))}{3 a^4}+\frac {4 \text {CosIntegral}(4 \text {ArcCos}(a x))}{3 a^4}-\frac {x^2}{2 a^2 \text {ArcCos}(a x)^2}-\frac {8 x^3 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)}+\frac {x^3 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}+\frac {x \sqrt {1-a^2 x^2}}{a^3 \text {ArcCos}(a x)}+\frac {2 x^4}{3 \text {ArcCos}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/ArcCos[a*x]^4,x]

[Out]

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

Rule 3383

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 4728

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

Rule 4730

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(-x^m)*Sqrt[1 - c^2*x^2]*((a + b*Arc
Cos[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 4808

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.18, size = 107, normalized size = 0.75 \begin {gather*} \frac {\frac {a x \left (2 a^2 x^2 \sqrt {1-a^2 x^2}+a x \left (-3+4 a^2 x^2\right ) \text {ArcCos}(a x)-2 \sqrt {1-a^2 x^2} \left (-3+8 a^2 x^2\right ) \text {ArcCos}(a x)^2\right )}{\text {ArcCos}(a x)^3}+2 \text {CosIntegral}(2 \text {ArcCos}(a x))+8 \text {CosIntegral}(4 \text {ArcCos}(a x))}{6 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcCos[a*x]^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 114, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (2 \arccos \left (a x \right )\right )}{3}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{24 \arccos \left (a x \right )^{3}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (4 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {4 \cosineIntegral \left (4 \arccos \left (a x \right )\right )}{3}}{a^{4}}\)	\(114\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (2 \arccos \left (a x \right )\right )}{6 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (2 \arccos \left (a x \right )\right )}{3}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{24 \arccos \left (a x \right )^{3}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{2}}-\frac {\sin \left (4 \arccos \left (a x \right )\right )}{3 \arccos \left (a x \right )}+\frac {4 \cosineIntegral \left (4 \arccos \left (a x \right )\right )}{3}}{a^{4}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arccos(a*x)^4,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/12/arccos(a*x)^3*sin(2*arccos(a*x))+1/12/arccos(a*x)^2*cos(2*arccos(a*x))-1/6/arccos(a*x)*sin(2*arcco
s(a*x))+1/3*Ci(2*arccos(a*x))+1/24/arccos(a*x)^3*sin(4*arccos(a*x))+1/12/arccos(a*x)^2*cos(4*arccos(a*x))-1/3/
arccos(a*x)*sin(4*arccos(a*x))+4/3*Ci(4*arccos(a*x)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/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(1/3*(32*a^4*x^4 - 30*a^2*x^2 + 3)*sqrt(a*x +
1)*sqrt(-a*x + 1)/((a^5*x^2 - a^3)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)), x) + 2*(a^2*x^3 - (8*a^2*x^3
- 3*x)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)*sqrt(a*x + 1)*sqrt(-a*x + 1) + (4*a^3*x^4 - 3*a*x^2)*arct
an2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x))/(a^3*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^3)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arccos(a*x)^4,x, algorithm="fricas")

[Out]

integral(x^3/arccos(a*x)^4, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\operatorname {acos}^{4}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/acos(a*x)**4,x)

[Out]

Integral(x**3/acos(a*x)**4, x)

________________________________________________________________________________________

Giac [A]
time = 0.45, size = 125, normalized size = 0.87 \begin {gather*} \frac {2 \, x^{4}}{3 \, \arccos \left (a x\right )^{2}} - \frac {8 \, \sqrt {-a^{2} x^{2} + 1} x^{3}}{3 \, a \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{3 \, a \arccos \left (a x\right )^{3}} - \frac {x^{2}}{2 \, a^{2} \arccos \left (a x\right )^{2}} + \frac {\sqrt {-a^{2} x^{2} + 1} x}{a^{3} \arccos \left (a x\right )} + \frac {4 \, \operatorname {Ci}\left (4 \, \arccos \left (a x\right )\right )}{3 \, a^{4}} + \frac {\operatorname {Ci}\left (2 \, \arccos \left (a x\right )\right )}{3 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arccos(a*x)^4,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\mathrm {acos}\left (a\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/acos(a*x)^4,x)

[Out]

int(x^3/acos(a*x)^4, x)

________________________________________________________________________________________