3.1.61 \(\int \frac {x^3}{\text {ArcCos}(a x)^3} \, dx\) [61]

Optimal. Leaf size=83 \[ \frac {x^3 \sqrt {1-a^2 x^2}}{2 a \text {ArcCos}(a x)^2}-\frac {3 x^2}{2 a^2 \text {ArcCos}(a x)}+\frac {2 x^4}{\text {ArcCos}(a x)}+\frac {\text {Si}(2 \text {ArcCos}(a x))}{2 a^4}+\frac {\text {Si}(4 \text {ArcCos}(a x))}{a^4} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4730, 4808, 4732, 4491, 3380, 12} \begin {gather*} \frac {\text {Si}(2 \text {ArcCos}(a x))}{2 a^4}+\frac {\text {Si}(4 \text {ArcCos}(a x))}{a^4}-\frac {3 x^2}{2 a^2 \text {ArcCos}(a x)}+\frac {x^3 \sqrt {1-a^2 x^2}}{2 a \text {ArcCos}(a x)^2}+\frac {2 x^4}{\text {ArcCos}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

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 4732

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

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

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Mathematica [A]
time = 0.12, size = 70, normalized size = 0.84 \begin {gather*} \frac {\frac {a^2 x^2 \left (a x \sqrt {1-a^2 x^2}+\left (-3+4 a^2 x^2\right ) \text {ArcCos}(a x)\right )}{\text {ArcCos}(a x)^2}+\text {Si}(2 \text {ArcCos}(a x))+2 \text {Si}(4 \text {ArcCos}(a x))}{2 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 82, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )^{2}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}+\frac {\sinIntegral \left (2 \arccos \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{16 \arccos \left (a x \right )^{2}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}+\sinIntegral \left (4 \arccos \left (a x \right )\right )}{a^{4}}\)	\(82\)
default	\(\frac {\frac {\sin \left (2 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )^{2}}+\frac {\cos \left (2 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}+\frac {\sinIntegral \left (2 \arccos \left (a x \right )\right )}{2}+\frac {\sin \left (4 \arccos \left (a x \right )\right )}{16 \arccos \left (a x \right )^{2}}+\frac {\cos \left (4 \arccos \left (a x \right )\right )}{4 \arccos \left (a x \right )}+\sinIntegral \left (4 \arccos \left (a x \right )\right )}{a^{4}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/8/arccos(a*x)^2*sin(2*arccos(a*x))+1/4/arccos(a*x)*cos(2*arccos(a*x))+1/2*Si(2*arccos(a*x))+1/16/arcc
os(a*x)^2*sin(4*arccos(a*x))+1/4/arccos(a*x)*cos(4*arccos(a*x))+Si(4*arccos(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)^3,x, algorithm="maxima")

[Out]

1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^3 - 2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate((8*a^2*x^3
- 3*x)/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x), x) + (4*a^2*x^4 - 3*x^2)*arctan2(sqrt(a*x + 1)*sqrt(-a*x +
1), a*x))/(a^2*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)

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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)^3,x, algorithm="fricas")

[Out]

integral(x^3/arccos(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/acos(a*x)**3, x)

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Giac [A]
time = 0.44, size = 75, normalized size = 0.90 \begin {gather*} \frac {2 \, x^{4}}{\arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{3}}{2 \, a \arccos \left (a x\right )^{2}} - \frac {3 \, x^{2}}{2 \, a^{2} \arccos \left (a x\right )} + \frac {\operatorname {Si}\left (4 \, \arccos \left (a x\right )\right )}{a^{4}} + \frac {\operatorname {Si}\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)^3,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3/acos(a*x)^3, x)

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