3.2.7 \(\int \frac {x^4}{\text {ArcCos}(a x)^{5/2}} \, dx\) [107]

Optimal. Leaf size=235 \[ \frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\text {ArcCos}(a x)}}+\frac {20 x^5}{3 \sqrt {\text {ArcCos}(a x)}}+\frac {25 \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{3 a^5}-\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{a^5}+\frac {25 \sqrt {\frac {\pi }{6}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{2 a^5}-\frac {4 \sqrt {\frac {2 \pi }{3}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{a^5}+\frac {5 \sqrt {\frac {5 \pi }{2}} S\left (\sqrt {\frac {10}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{6 a^5} \]

[Out]

3/4*FresnelS(6^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*6^(1/2)*Pi^(1/2)/a^5+1/6*FresnelS(2^(1/2)/Pi^(1/2)*arccos(a*x
)^(1/2))*2^(1/2)*Pi^(1/2)/a^5+5/12*FresnelS(10^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*10^(1/2)*Pi^(1/2)/a^5+2/3*x^4
*(-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(3/2)-16/3*x^3/a^2/arccos(a*x)^(1/2)+20/3*x^5/arccos(a*x)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4730, 4808, 4732, 4491, 3386, 3432} \begin {gather*} -\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{a^5}+\frac {25 \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{3 a^5}-\frac {4 \sqrt {\frac {2 \pi }{3}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{a^5}+\frac {25 \sqrt {\frac {\pi }{6}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{2 a^5}+\frac {5 \sqrt {\frac {5 \pi }{2}} S\left (\sqrt {\frac {10}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{6 a^5}-\frac {16 x^3}{3 a^2 \sqrt {\text {ArcCos}(a x)}}+\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \text {ArcCos}(a x)^{3/2}}+\frac {20 x^5}{3 \sqrt {\text {ArcCos}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCos[a*x]^(5/2),x]

[Out]

(2*x^4*Sqrt[1 - a^2*x^2])/(3*a*ArcCos[a*x]^(3/2)) - (16*x^3)/(3*a^2*Sqrt[ArcCos[a*x]]) + (20*x^5)/(3*Sqrt[ArcC
os[a*x]]) + (25*Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Sqrt[ArcCos[a*x]]])/(3*a^5) - (4*Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]
*Sqrt[ArcCos[a*x]]])/a^5 + (25*Sqrt[Pi/6]*FresnelS[Sqrt[6/Pi]*Sqrt[ArcCos[a*x]]])/(2*a^5) - (4*Sqrt[(2*Pi)/3]*
FresnelS[Sqrt[6/Pi]*Sqrt[ArcCos[a*x]]])/a^5 + (5*Sqrt[(5*Pi)/2]*FresnelS[Sqrt[10/Pi]*Sqrt[ArcCos[a*x]]])/(6*a^
5)

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

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^4}{\cos ^{-1}(a x)^{5/2}} \, dx &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {8 \int \frac {x^3}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}} \, dx}{3 a}+\frac {1}{3} (10 a) \int \frac {x^5}{\sqrt {1-a^2 x^2} \cos ^{-1}(a x)^{3/2}} \, dx\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}-\frac {100}{3} \int \frac {x^4}{\sqrt {\cos ^{-1}(a x)}} \, dx+\frac {16 \int \frac {x^2}{\sqrt {\cos ^{-1}(a x)}} \, dx}{a^2}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}-\frac {16 \text {Subst}\left (\int \frac {\cos ^2(x) \sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{a^5}+\frac {100 \text {Subst}\left (\int \frac {\cos ^4(x) \sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{3 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}-\frac {16 \text {Subst}\left (\int \left (\frac {\sin (x)}{4 \sqrt {x}}+\frac {\sin (3 x)}{4 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^5}+\frac {100 \text {Subst}\left (\int \left (\frac {\sin (x)}{8 \sqrt {x}}+\frac {3 \sin (3 x)}{16 \sqrt {x}}+\frac {\sin (5 x)}{16 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{3 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {25 \text {Subst}\left (\int \frac {\sin (5 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{12 a^5}-\frac {4 \text {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{a^5}-\frac {4 \text {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{a^5}+\frac {25 \text {Subst}\left (\int \frac {\sin (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{6 a^5}+\frac {25 \text {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{4 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {25 \text {Subst}\left (\int \sin \left (5 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{6 a^5}-\frac {8 \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{a^5}-\frac {8 \text {Subst}\left (\int \sin \left (3 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{a^5}+\frac {25 \text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{3 a^5}+\frac {25 \text {Subst}\left (\int \sin \left (3 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{3 a \cos ^{-1}(a x)^{3/2}}-\frac {16 x^3}{3 a^2 \sqrt {\cos ^{-1}(a x)}}+\frac {20 x^5}{3 \sqrt {\cos ^{-1}(a x)}}+\frac {25 \sqrt {\frac {\pi }{2}} S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{3 a^5}-\frac {4 \sqrt {2 \pi } S\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{a^5}+\frac {25 \sqrt {\frac {\pi }{6}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}-\frac {4 \sqrt {\frac {2 \pi }{3}} S\left (\sqrt {\frac {6}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{a^5}+\frac {5 \sqrt {\frac {5 \pi }{2}} S\left (\sqrt {\frac {10}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{6 a^5}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.16, size = 322, normalized size = 1.37 \begin {gather*} -\frac {2 \left (-\sqrt {1-a^2 x^2}-e^{-i \text {ArcCos}(a x)} \text {ArcCos}(a x)-e^{i \text {ArcCos}(a x)} \text {ArcCos}(a x)+\sqrt {-i \text {ArcCos}(a x)} \text {ArcCos}(a x) \text {Gamma}\left (\frac {1}{2},-i \text {ArcCos}(a x)\right )+\sqrt {i \text {ArcCos}(a x)} \text {ArcCos}(a x) \text {Gamma}\left (\frac {1}{2},i \text {ArcCos}(a x)\right )\right )-5 \text {ArcCos}(a x) \left (e^{-5 i \text {ArcCos}(a x)}+e^{5 i \text {ArcCos}(a x)}-\sqrt {5} \sqrt {-i \text {ArcCos}(a x)} \text {Gamma}\left (\frac {1}{2},-5 i \text {ArcCos}(a x)\right )-\sqrt {5} \sqrt {i \text {ArcCos}(a x)} \text {Gamma}\left (\frac {1}{2},5 i \text {ArcCos}(a x)\right )\right )-3 \left (3 \text {ArcCos}(a x) \left (e^{-3 i \text {ArcCos}(a x)}+e^{3 i \text {ArcCos}(a x)}-\sqrt {3} \sqrt {-i \text {ArcCos}(a x)} \text {Gamma}\left (\frac {1}{2},-3 i \text {ArcCos}(a x)\right )-\sqrt {3} \sqrt {i \text {ArcCos}(a x)} \text {Gamma}\left (\frac {1}{2},3 i \text {ArcCos}(a x)\right )\right )+\sin (3 \text {ArcCos}(a x))\right )-\sin (5 \text {ArcCos}(a x))}{24 a^5 \text {ArcCos}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcCos[a*x]^(5/2),x]

[Out]

-1/24*(2*(-Sqrt[1 - a^2*x^2] - ArcCos[a*x]/E^(I*ArcCos[a*x]) - E^(I*ArcCos[a*x])*ArcCos[a*x] + Sqrt[(-I)*ArcCo
s[a*x]]*ArcCos[a*x]*Gamma[1/2, (-I)*ArcCos[a*x]] + Sqrt[I*ArcCos[a*x]]*ArcCos[a*x]*Gamma[1/2, I*ArcCos[a*x]])
- 5*ArcCos[a*x]*(E^((-5*I)*ArcCos[a*x]) + E^((5*I)*ArcCos[a*x]) - Sqrt[5]*Sqrt[(-I)*ArcCos[a*x]]*Gamma[1/2, (-
5*I)*ArcCos[a*x]] - Sqrt[5]*Sqrt[I*ArcCos[a*x]]*Gamma[1/2, (5*I)*ArcCos[a*x]]) - 3*(3*ArcCos[a*x]*(E^((-3*I)*A
rcCos[a*x]) + E^((3*I)*ArcCos[a*x]) - Sqrt[3]*Sqrt[(-I)*ArcCos[a*x]]*Gamma[1/2, (-3*I)*ArcCos[a*x]] - Sqrt[3]*
Sqrt[I*ArcCos[a*x]]*Gamma[1/2, (3*I)*ArcCos[a*x]]) + Sin[3*ArcCos[a*x]]) - Sin[5*ArcCos[a*x]])/(a^5*ArcCos[a*x
]^(3/2))

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Maple [A]
time = 0.25, size = 173, normalized size = 0.74

method	result	size
default	\(\frac {10 \sqrt {2}\, \sqrt {\pi }\, \sqrt {5}\, \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {3}{2}}+18 \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {3}{2}}+4 \sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right ) \arccos \left (a x \right )^{\frac {3}{2}}+4 a x \arccos \left (a x \right )+18 \arccos \left (a x \right ) \cos \left (3 \arccos \left (a x \right )\right )+10 \arccos \left (a x \right ) \cos \left (5 \arccos \left (a x \right )\right )+2 \sqrt {-a^{2} x^{2}+1}+3 \sin \left (3 \arccos \left (a x \right )\right )+\sin \left (5 \arccos \left (a x \right )\right )}{24 a^{5} \arccos \left (a x \right )^{\frac {3}{2}}}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arccos(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/24/a^5*(10*2^(1/2)*Pi^(1/2)*5^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*5^(1/2)*arccos(a*x)^(1/2))*arccos(a*x)^(3/2)+1
8*2^(1/2)*Pi^(1/2)*3^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)*arccos(a*x)^(1/2))*arccos(a*x)^(3/2)+4*2^(1/2)*Pi
^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*arccos(a*x)^(3/2)+4*a*x*arccos(a*x)+18*arccos(a*x)*cos(3*a
rccos(a*x))+10*arccos(a*x)*cos(5*arccos(a*x))+2*(-a^2*x^2+1)^(1/2)+3*sin(3*arccos(a*x))+sin(5*arccos(a*x)))/ar
ccos(a*x)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\operatorname {acos}^{\frac {5}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/acos(a*x)**(5/2),x)

[Out]

Integral(x**4/acos(a*x)**(5/2), x)

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Giac [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)^(5/2),x, algorithm="giac")

[Out]

integrate(x^4/arccos(a*x)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{{\mathrm {acos}\left (a\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/acos(a*x)^(5/2),x)

[Out]

int(x^4/acos(a*x)^(5/2), x)

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