3.1.69 \(\int \frac {x^2}{\text {ArcCos}(a x)^4} \, dx\) [69]

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

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4730, 4808, 4728, 3383, 4718, 4810} \begin {gather*} \frac {\text {CosIntegral}(\text {ArcCos}(a x))}{24 a^3}+\frac {9 \text {CosIntegral}(3 \text {ArcCos}(a x))}{8 a^3}-\frac {3 x^2 \sqrt {1-a^2 x^2}}{2 a \text {ArcCos}(a x)}+\frac {x^2 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}-\frac {x}{3 a^2 \text {ArcCos}(a x)^2}+\frac {\sqrt {1-a^2 x^2}}{3 a^3 \text {ArcCos}(a x)}+\frac {x^3}{2 \text {ArcCos}(a x)^2} \end {gather*}

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 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 4718

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] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

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]

Rule 4810

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(-(b*c^
(m + 1))^(-1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGt
Q[m, 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 \text {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 {\text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^3}+\frac {3 \text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^3}+\frac {9 \text {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*}

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Mathematica [A]
time = 0.09, size = 129, normalized size = 0.91 \begin {gather*} \frac {x^2 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}+\frac {-2 x+3 a^2 x^3}{6 a^2 \text {ArcCos}(a x)^2}-\frac {\sqrt {1-a^2 x^2} \left (-2+9 a^2 x^2\right )}{6 a^3 \text {ArcCos}(a x)}-\frac {10 \text {CosIntegral}(\text {ArcCos}(a x))}{3 a^3}-\frac {9 (-3 \text {CosIntegral}(\text {ArcCos}(a x))-\text {CosIntegral}(3 \text {ArcCos}(a x)))}{8 a^3} \end {gather*}

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 = 0.04, size = 117, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{12 \arccos \left (a x \right )^{3}}+\frac {a x}{24 \arccos \left (a x \right )^{2}}-\frac {\sqrt {-a^{2} x^{2}+1}}{24 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (\arccos \left (a x \right )\right )}{24}+\frac {\sin \left (3 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (3 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )^{2}}-\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )}+\frac {9 \cosineIntegral \left (3 \arccos \left (a x \right )\right )}{8}}{a^{3}}\)	\(117\)
default	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{12 \arccos \left (a x \right )^{3}}+\frac {a x}{24 \arccos \left (a x \right )^{2}}-\frac {\sqrt {-a^{2} x^{2}+1}}{24 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (\arccos \left (a x \right )\right )}{24}+\frac {\sin \left (3 \arccos \left (a x \right )\right )}{12 \arccos \left (a x \right )^{3}}+\frac {\cos \left (3 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )^{2}}-\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{8 \arccos \left (a x \right )}+\frac {9 \cosineIntegral \left (3 \arccos \left (a x \right )\right )}{8}}{a^{3}}\)	\(117\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/arccos(a*x)*(-a^2*x^2+1)^(1/2)+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 = 0.00, 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^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 = 0.00, 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^2/arccos(a*x)^4,x, algorithm="fricas")

[Out]

integral(x^2/arccos(a*x)^4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\operatorname {acos}^{4}{\left (a x \right )}}\, dx \end {gather*}

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 = 0.43, size = 121, normalized size = 0.86 \begin {gather*} \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 {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\mathrm {acos}\left (a\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/acos(a*x)^4, x)

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