3.1.67 \(\int \frac {x^4}{\text {ArcCos}(a x)^4} \, dx\) [67]

Optimal. Leaf size=158 \[ \frac {x^4 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}-\frac {2 x^3}{3 a^2 \text {ArcCos}(a x)^2}+\frac {5 x^5}{6 \text {ArcCos}(a x)^2}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{a^3 \text {ArcCos}(a x)}-\frac {25 x^4 \sqrt {1-a^2 x^2}}{6 a \text {ArcCos}(a x)}+\frac {\text {CosIntegral}(\text {ArcCos}(a x))}{48 a^5}+\frac {27 \text {CosIntegral}(3 \text {ArcCos}(a x))}{32 a^5}+\frac {125 \text {CosIntegral}(5 \text {ArcCos}(a x))}{96 a^5} \]

[Out]

-2/3*x^3/a^2/arccos(a*x)^2+5/6*x^5/arccos(a*x)^2+1/48*Ci(arccos(a*x))/a^5+27/32*Ci(3*arccos(a*x))/a^5+125/96*C
i(5*arccos(a*x))/a^5+1/3*x^4*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^3+2*x^2*(-a^2*x^2+1)^(1/2)/a^3/arccos(a*x)-25/6*
x^4*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)

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Rubi [A]
time = 0.23, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 12, 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}(\text {ArcCos}(a x))}{48 a^5}+\frac {27 \text {CosIntegral}(3 \text {ArcCos}(a x))}{32 a^5}+\frac {125 \text {CosIntegral}(5 \text {ArcCos}(a x))}{96 a^5}-\frac {2 x^3}{3 a^2 \text {ArcCos}(a x)^2}-\frac {25 x^4 \sqrt {1-a^2 x^2}}{6 a \text {ArcCos}(a x)}+\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^3}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{a^3 \text {ArcCos}(a x)}+\frac {5 x^5}{6 \text {ArcCos}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^3) - (2*x^3)/(3*a^2*ArcCos[a*x]^2) + (5*x^5)/(6*ArcCos[a*x]^2) + (2*x
^2*Sqrt[1 - a^2*x^2])/(a^3*ArcCos[a*x]) - (25*x^4*Sqrt[1 - a^2*x^2])/(6*a*ArcCos[a*x]) + CosIntegral[ArcCos[a*
x]]/(48*a^5) + (27*CosIntegral[3*ArcCos[a*x]])/(32*a^5) + (125*CosIntegral[5*ArcCos[a*x]])/(96*a^5)

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^4}{\cos ^{-1}(a x)^4} \, dx &=\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac {4 \int \frac {x^3}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^3} \, dx}{3 a}+\frac {1}{3} (5 a) \int \frac {x^5}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^3} \, dx\\ &=\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac {2 x^3}{3 a^2 \cos ^{-1}(a x)^2}+\frac {5 x^5}{6 \cos ^{-1}(a x)^2}-\frac {25}{6} \int \frac {x^4}{\cos ^{-1}(a x)^2} \, dx+\frac {2 \int \frac {x^2}{\cos ^{-1}(a x)^2} \, dx}{a^2}\\ &=\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac {2 x^3}{3 a^2 \cos ^{-1}(a x)^2}+\frac {5 x^5}{6 \cos ^{-1}(a x)^2}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{a^3 \cos ^{-1}(a x)}-\frac {25 x^4 \sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {2 \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 )}{a^5}-\frac {25 \text {Subst}\left (\int \left (-\frac {\cos (x)}{8 x}-\frac {9 \cos (3 x)}{16 x}-\frac {5 \cos (5 x)}{16 x}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{6 a^5}\\ &=\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac {2 x^3}{3 a^2 \cos ^{-1}(a x)^2}+\frac {5 x^5}{6 \cos ^{-1}(a x)^2}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{a^3 \cos ^{-1}(a x)}-\frac {25 x^4 \sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}+\frac {25 \text {Subst}\left (\int \frac {\cos (x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{48 a^5}+\frac {125 \text {Subst}\left (\int \frac {\cos (5 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{96 a^5}-\frac {3 \text {Subst}\left (\int \frac {\cos (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{2 a^5}+\frac {75 \text {Subst}\left (\int \frac {\cos (3 x)}{x} \, dx,x,\cos ^{-1}(a x)\right )}{32 a^5}\\ &=\frac {x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^3}-\frac {2 x^3}{3 a^2 \cos ^{-1}(a x)^2}+\frac {5 x^5}{6 \cos ^{-1}(a x)^2}+\frac {2 x^2 \sqrt {1-a^2 x^2}}{a^3 \cos ^{-1}(a x)}-\frac {25 x^4 \sqrt {1-a^2 x^2}}{6 a \cos ^{-1}(a x)}+\frac {\text {Ci}\left (\cos ^{-1}(a x)\right )}{48 a^5}+\frac {27 \text {Ci}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac {125 \text {Ci}\left (5 \cos ^{-1}(a x)\right )}{96 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 159, normalized size = 1.01 \begin {gather*} \frac {32 a^4 x^4 \sqrt {1-a^2 x^2}-64 a^3 x^3 \text {ArcCos}(a x)+80 a^5 x^5 \text {ArcCos}(a x)+192 a^2 x^2 \sqrt {1-a^2 x^2} \text {ArcCos}(a x)^2-400 a^4 x^4 \sqrt {1-a^2 x^2} \text {ArcCos}(a x)^2+2 \text {ArcCos}(a x)^3 \text {CosIntegral}(\text {ArcCos}(a x))+81 \text {ArcCos}(a x)^3 \text {CosIntegral}(3 \text {ArcCos}(a x))+125 \text {ArcCos}(a x)^3 \text {CosIntegral}(5 \text {ArcCos}(a x))}{96 a^5 \text {ArcCos}(a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a^4*x^4*Sqrt[1 - a^2*x^2] - 64*a^3*x^3*ArcCos[a*x] + 80*a^5*x^5*ArcCos[a*x] + 192*a^2*x^2*Sqrt[1 - a^2*x^2
]*ArcCos[a*x]^2 - 400*a^4*x^4*Sqrt[1 - a^2*x^2]*ArcCos[a*x]^2 + 2*ArcCos[a*x]^3*CosIntegral[ArcCos[a*x]] + 81*
ArcCos[a*x]^3*CosIntegral[3*ArcCos[a*x]] + 125*ArcCos[a*x]^3*CosIntegral[5*ArcCos[a*x]])/(96*a^5*ArcCos[a*x]^3
)

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Maple [A]
time = 0.14, size = 171, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{24 \arccos \left (a x \right )^{3}}+\frac {a x}{48 \arccos \left (a x \right )^{2}}-\frac {\sqrt {-a^{2} x^{2}+1}}{48 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (\arccos \left (a x \right )\right )}{48}+\frac {\sin \left (3 \arccos \left (a x \right )\right )}{16 \arccos \left (a x \right )^{3}}+\frac {3 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}-\frac {9 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \cosineIntegral \left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{48 \arccos \left (a x \right )^{3}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{96 \arccos \left (a x \right )^{2}}-\frac {25 \sin \left (5 \arccos \left (a x \right )\right )}{96 \arccos \left (a x \right )}+\frac {125 \cosineIntegral \left (5 \arccos \left (a x \right )\right )}{96}}{a^{5}}\)	\(171\)
default	\(\frac {\frac {\sqrt {-a^{2} x^{2}+1}}{24 \arccos \left (a x \right )^{3}}+\frac {a x}{48 \arccos \left (a x \right )^{2}}-\frac {\sqrt {-a^{2} x^{2}+1}}{48 \arccos \left (a x \right )}+\frac {\cosineIntegral \left (\arccos \left (a x \right )\right )}{48}+\frac {\sin \left (3 \arccos \left (a x \right )\right )}{16 \arccos \left (a x \right )^{3}}+\frac {3 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}-\frac {9 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \cosineIntegral \left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{48 \arccos \left (a x \right )^{3}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{96 \arccos \left (a x \right )^{2}}-\frac {25 \sin \left (5 \arccos \left (a x \right )\right )}{96 \arccos \left (a x \right )}+\frac {125 \cosineIntegral \left (5 \arccos \left (a x \right )\right )}{96}}{a^{5}}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(1/24/arccos(a*x)^3*(-a^2*x^2+1)^(1/2)+1/48*a*x/arccos(a*x)^2-1/48/arccos(a*x)*(-a^2*x^2+1)^(1/2)+1/48*C
i(arccos(a*x))+1/16/arccos(a*x)^3*sin(3*arccos(a*x))+3/32/arccos(a*x)^2*cos(3*arccos(a*x))-9/32/arccos(a*x)*si
n(3*arccos(a*x))+27/32*Ci(3*arccos(a*x))+1/48/arccos(a*x)^3*sin(5*arccos(a*x))+5/96/arccos(a*x)^2*cos(5*arccos
(a*x))-25/96/arccos(a*x)*sin(5*arccos(a*x))+125/96*Ci(5*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^4/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*(125*a^4*x^5 - 136*a^2*x^3 + 24*x)*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^4 - (25*a^
2*x^4 - 12*x^2)*arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2)*sqrt(a*x + 1)*sqrt(-a*x + 1) + (5*a^3*x^5 - 4*a*
x^3)*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^4/arccos(a*x)^4,x, algorithm="fricas")

[Out]

integral(x^4/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^{4}}{\operatorname {acos}^{4}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/acos(a*x)**4, x)

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Giac [A]
time = 0.44, size = 138, normalized size = 0.87 \begin {gather*} \frac {5 \, x^{5}}{6 \, \arccos \left (a x\right )^{2}} - \frac {25 \, \sqrt {-a^{2} x^{2} + 1} x^{4}}{6 \, a \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{3 \, a \arccos \left (a x\right )^{3}} - \frac {2 \, x^{3}}{3 \, a^{2} \arccos \left (a x\right )^{2}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{3} \arccos \left (a x\right )} + \frac {125 \, \operatorname {Ci}\left (5 \, \arccos \left (a x\right )\right )}{96 \, a^{5}} + \frac {27 \, \operatorname {Ci}\left (3 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {\operatorname {Ci}\left (\arccos \left (a x\right )\right )}{48 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/6*x^5/arccos(a*x)^2 - 25/6*sqrt(-a^2*x^2 + 1)*x^4/(a*arccos(a*x)) + 1/3*sqrt(-a^2*x^2 + 1)*x^4/(a*arccos(a*x
)^3) - 2/3*x^3/(a^2*arccos(a*x)^2) + 2*sqrt(-a^2*x^2 + 1)*x^2/(a^3*arccos(a*x)) + 125/96*cos_integral(5*arccos
(a*x))/a^5 + 27/32*cos_integral(3*arccos(a*x))/a^5 + 1/48*cos_integral(arccos(a*x))/a^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4/acos(a*x)^4, x)

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