3.60 \(\int \frac {x^4}{\cos ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=98 \[ \frac {\text {Si}\left (\cos ^{-1}(a x)\right )}{16 a^5}+\frac {27 \text {Si}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac {25 \text {Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5}-\frac {2 x^3}{a^2 \cos ^{-1}(a x)}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}+\frac {5 x^5}{2 \cos ^{-1}(a x)} \]

[Out]

-2*x^3/a^2/arccos(a*x)+5/2*x^5/arccos(a*x)+1/16*Si(arccos(a*x))/a^5+27/32*Si(3*arccos(a*x))/a^5+25/32*Si(5*arc
cos(a*x))/a^5+1/2*x^4*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^2

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Rubi [A]  time = 0.34, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4634, 4720, 4636, 4406, 3299} \[ \frac {\text {Si}\left (\cos ^{-1}(a x)\right )}{16 a^5}+\frac {27 \text {Si}\left (3 \cos ^{-1}(a x)\right )}{32 a^5}+\frac {25 \text {Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5}+\frac {x^4 \sqrt {1-a^2 x^2}}{2 a \cos ^{-1}(a x)^2}-\frac {2 x^3}{a^2 \cos ^{-1}(a x)}+\frac {5 x^5}{2 \cos ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[1 - a^2*x^2])/(2*a*ArcCos[a*x]^2) - (2*x^3)/(a^2*ArcCos[a*x]) + (5*x^5)/(2*ArcCos[a*x]) + SinIntegra
l[ArcCos[a*x]]/(16*a^5) + (27*SinIntegral[3*ArcCos[a*x]])/(32*a^5) + (25*SinIntegral[5*ArcCos[a*x]])/(32*a^5)

Rule 3299

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 4406

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 4634

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcCo
s[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 4636

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

Rule 4720

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp
[((f*x)^m*(a + b*ArcCos[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] + Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 103, normalized size = 1.05 \[ \frac {80 a^5 x^5 \cos ^{-1}(a x)-64 a^3 x^3 \cos ^{-1}(a x)+16 a^4 x^4 \sqrt {1-a^2 x^2}+2 \cos ^{-1}(a x)^2 \text {Si}\left (\cos ^{-1}(a x)\right )+27 \cos ^{-1}(a x)^2 \text {Si}\left (3 \cos ^{-1}(a x)\right )+25 \cos ^{-1}(a x)^2 \text {Si}\left (5 \cos ^{-1}(a x)\right )}{32 a^5 \cos ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*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] + 2*ArcCos[a*x]^2*SinIntegral[
ArcCos[a*x]] + 27*ArcCos[a*x]^2*SinIntegral[3*ArcCos[a*x]] + 25*ArcCos[a*x]^2*SinIntegral[5*ArcCos[a*x]])/(32*
a^5*ArcCos[a*x]^2)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{4}}{\arccos \left (a x\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4/arccos(a*x)^3, x)

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giac [A]  time = 0.25, size = 86, normalized size = 0.88 \[ \frac {5 \, x^{5}}{2 \, \arccos \left (a x\right )} + \frac {\sqrt {-a^{2} x^{2} + 1} x^{4}}{2 \, a \arccos \left (a x\right )^{2}} - \frac {2 \, x^{3}}{a^{2} \arccos \left (a x\right )} + \frac {25 \, \operatorname {Si}\left (5 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {27 \, \operatorname {Si}\left (3 \, \arccos \left (a x\right )\right )}{32 \, a^{5}} + \frac {\operatorname {Si}\left (\arccos \left (a x\right )\right )}{16 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*x^5/arccos(a*x) + 1/2*sqrt(-a^2*x^2 + 1)*x^4/(a*arccos(a*x)^2) - 2*x^3/(a^2*arccos(a*x)) + 25/32*sin_integ
ral(5*arccos(a*x))/a^5 + 27/32*sin_integral(3*arccos(a*x))/a^5 + 1/16*sin_integral(arccos(a*x))/a^5

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maple [A]  time = 0.13, size = 121, normalized size = 1.23 \[ \frac {\frac {3 \sin \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {9 \cos \left (3 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {27 \Si \left (3 \arccos \left (a x \right )\right )}{32}+\frac {\sin \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )^{2}}+\frac {5 \cos \left (5 \arccos \left (a x \right )\right )}{32 \arccos \left (a x \right )}+\frac {25 \Si \left (5 \arccos \left (a x \right )\right )}{32}+\frac {\sqrt {-a^{2} x^{2}+1}}{16 \arccos \left (a x \right )^{2}}+\frac {a x}{16 \arccos \left (a x \right )}+\frac {\Si \left (\arccos \left (a x \right )\right )}{16}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(3/32/arccos(a*x)^2*sin(3*arccos(a*x))+9/32/arccos(a*x)*cos(3*arccos(a*x))+27/32*Si(3*arccos(a*x))+1/32/
arccos(a*x)^2*sin(5*arccos(a*x))+5/32/arccos(a*x)*cos(5*arccos(a*x))+25/32*Si(5*arccos(a*x))+1/16/arccos(a*x)^
2*(-a^2*x^2+1)^(1/2)+1/16*a*x/arccos(a*x)+1/16*Si(arccos(a*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\sqrt {a x + 1} \sqrt {-a x + 1} a x^{4} - \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{2} \int \frac {{\left (25 \, a^{2} x^{2} - 12\right )} x^{2}}{\arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )}\,{d x} + {\left (5 \, a^{2} x^{5} - 4 \, x^{3}\right )} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )}{2 \, a^{2} \arctan \left (\sqrt {a x + 1} \sqrt {-a x + 1}, a x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(a*x + 1)*sqrt(-a*x + 1)*a*x^4 - arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x)^2*integrate((25*a^2*x^4 -
12*x^2)/arctan2(sqrt(a*x + 1)*sqrt(-a*x + 1), a*x), x) + (5*a^2*x^5 - 4*x^3)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\mathrm {acos}\left (a\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4/acos(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/acos(a*x)**3, x)

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