3.1.54 \(\int \frac {x^3}{\text {ArcCos}(a x)^2} \, dx\) [54]

Optimal. Leaf size=56 \[ \frac {x^3 \sqrt {1-a^2 x^2}}{a \text {ArcCos}(a x)}-\frac {\text {CosIntegral}(2 \text {ArcCos}(a x))}{2 a^4}-\frac {\text {CosIntegral}(4 \text {ArcCos}(a x))}{2 a^4} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4728, 3383} \begin {gather*} -\frac {\text {CosIntegral}(2 \text {ArcCos}(a x))}{2 a^4}-\frac {\text {CosIntegral}(4 \text {ArcCos}(a x))}{2 a^4}+\frac {x^3 \sqrt {1-a^2 x^2}}{a \text {ArcCos}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/ArcCos[a*x]^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 50, normalized size = 0.89 \begin {gather*} -\frac {-\frac {2 a^3 x^3 \sqrt {1-a^2 x^2}}{\text {ArcCos}(a x)}+\text {CosIntegral}(2 \text {ArcCos}(a x))+\text {CosIntegral}(4 \text {ArcCos}(a x))}{2 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcCos[a*x]^2,x]

[Out]

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

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Maple [A]
time = 0.05, size = 54, normalized size = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}-\frac {\cosineIntegral \left (2 \arccos \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )}-\frac {\cosineIntegral \left (4 \arccos \left (a x \right )\right )}{2}}{a^{4}}\)	\(54\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}-\frac {\cosineIntegral \left (2 \arccos \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )}-\frac {\cosineIntegral \left (4 \arccos \left (a x \right )\right )}{2}}{a^{4}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arccos(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4/arccos(a*x)*sin(2*arccos(a*x))-1/2*Ci(2*arccos(a*x))+1/8/arccos(a*x)*sin(4*arccos(a*x))-1/2*Ci(4*ar
ccos(a*x)))

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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)^2,x, algorithm="maxima")

[Out]

(sqrt(a*x + 1)*sqrt(-a*x + 1)*x^3 - a*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)*integrate((4*a^2*x^4 - 3*x^2)
*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))/(a*arctan2(sqrt(
a*x + 1)*sqrt(-a*x + 1), a*x))

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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)^2,x, algorithm="fricas")

[Out]

integral(x^3/arccos(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\operatorname {acos}^{2}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/acos(a*x)**2,x)

[Out]

Integral(x**3/acos(a*x)**2, x)

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Giac [A]
time = 0.44, size = 50, normalized size = 0.89 \begin {gather*} \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{a \arccos \left (a x\right )} - \frac {\operatorname {Ci}\left (4 \, \arccos \left (a x\right )\right )}{2 \, a^{4}} - \frac {\operatorname {Ci}\left (2 \, \arccos \left (a x\right )\right )}{2 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arccos(a*x)^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^3}{{\mathrm {acos}\left (a\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/acos(a*x)^2,x)

[Out]

int(x^3/acos(a*x)^2, x)

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