3.101 \(\int \frac {x^4}{\cos ^{-1}(a x)^{3/2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {\sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}-\frac {3 \sqrt {\frac {3 \pi }{2}} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}-\frac {\sqrt {\frac {5 \pi }{2}} C\left (\sqrt {\frac {10}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}+\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}} \]

[Out]

-1/4*FresnelC(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^5-3/8*FresnelC(6^(1/2)/Pi^(1/2)*arccos(a*
x)^(1/2))*6^(1/2)*Pi^(1/2)/a^5-1/8*FresnelC(10^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*10^(1/2)*Pi^(1/2)/a^5+2*x^4*(
-a^2*x^2+1)^(1/2)/a/arccos(a*x)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4632, 3304, 3352} \[ -\frac {\sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}-\frac {3 \sqrt {\frac {3 \pi }{2}} \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}-\frac {\sqrt {\frac {5 \pi }{2}} \text {FresnelC}\left (\sqrt {\frac {10}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}+\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCos[a*x]^(3/2),x]

[Out]

(2*x^4*Sqrt[1 - a^2*x^2])/(a*Sqrt[ArcCos[a*x]]) - (Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcCos[a*x]]])/(2*a^5)
- (3*Sqrt[(3*Pi)/2]*FresnelC[Sqrt[6/Pi]*Sqrt[ArcCos[a*x]]])/(4*a^5) - (Sqrt[(5*Pi)/2]*FresnelC[Sqrt[10/Pi]*Sqr
t[ArcCos[a*x]]])/(4*a^5)

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(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 3352

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

Rule 4632

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1
), Cos[x]^(m - 1)*(m - (m + 1)*Cos[x]^2), x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] &&
GeQ[n, -2] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {x^4}{\cos ^{-1}(a x)^{3/2}} \, dx &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}}+\frac {2 \operatorname {Subst}\left (\int \left (-\frac {\cos (x)}{8 \sqrt {x}}-\frac {9 \cos (3 x)}{16 \sqrt {x}}-\frac {5 \cos (5 x)}{16 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}}-\frac {\operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{4 a^5}-\frac {5 \operatorname {Subst}\left (\int \frac {\cos (5 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^5}-\frac {9 \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}}-\frac {\operatorname {Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}-\frac {5 \operatorname {Subst}\left (\int \cos \left (5 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}-\frac {9 \operatorname {Subst}\left (\int \cos \left (3 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}\\ &=\frac {2 x^4 \sqrt {1-a^2 x^2}}{a \sqrt {\cos ^{-1}(a x)}}-\frac {\sqrt {\frac {\pi }{2}} C\left (\sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{2 a^5}-\frac {3 \sqrt {\frac {3 \pi }{2}} C\left (\sqrt {\frac {6}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}-\frac {\sqrt {\frac {5 \pi }{2}} C\left (\sqrt {\frac {10}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{4 a^5}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 233, normalized size = 1.71 \[ \frac {i \left (-4 i \sqrt {1-a^2 x^2}-6 i \sin \left (3 \cos ^{-1}(a x)\right )-2 i \sin \left (5 \cos ^{-1}(a x)\right )+2 \sqrt {-i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},-i \cos ^{-1}(a x)\right )-2 \sqrt {i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},i \cos ^{-1}(a x)\right )+3 \sqrt {3} \sqrt {-i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},-3 i \cos ^{-1}(a x)\right )-3 \sqrt {3} \sqrt {i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},3 i \cos ^{-1}(a x)\right )+\sqrt {5} \sqrt {-i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},-5 i \cos ^{-1}(a x)\right )-\sqrt {5} \sqrt {i \cos ^{-1}(a x)} \Gamma \left (\frac {1}{2},5 i \cos ^{-1}(a x)\right )\right )}{16 a^5 \sqrt {\cos ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/ArcCos[a*x]^(3/2),x]

[Out]

((I/16)*((-4*I)*Sqrt[1 - a^2*x^2] + 2*Sqrt[(-I)*ArcCos[a*x]]*Gamma[1/2, (-I)*ArcCos[a*x]] - 2*Sqrt[I*ArcCos[a*
x]]*Gamma[1/2, I*ArcCos[a*x]] + 3*Sqrt[3]*Sqrt[(-I)*ArcCos[a*x]]*Gamma[1/2, (-3*I)*ArcCos[a*x]] - 3*Sqrt[3]*Sq
rt[I*ArcCos[a*x]]*Gamma[1/2, (3*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]] - (6*I)*Sin[3*ArcCos[a*x]] - (2*I)*Sin[5*ArcCos[
a*x]]))/(a^5*Sqrt[ArcCos[a*x]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\arccos \left (a x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arccos(a*x)^(3/2),x, algorithm="giac")

[Out]

integrate(x^4/arccos(a*x)^(3/2), x)

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maple [A]  time = 0.18, size = 140, normalized size = 1.03 \[ -\frac {\sqrt {5}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (a x \right )}\, \FresnelC \left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )+3 \sqrt {3}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (a x \right )}\, \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )+2 \sqrt {2}\, \sqrt {\pi }\, \sqrt {\arccos \left (a x \right )}\, \FresnelC \left (\frac {\sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )-2 \sqrt {-a^{2} x^{2}+1}-3 \sin \left (3 \arccos \left (a x \right )\right )-\sin \left (5 \arccos \left (a x \right )\right )}{8 a^{5} \sqrt {\arccos \left (a x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arccos(a*x)^(3/2),x)

[Out]

-1/8/a^5*(5^(1/2)*2^(1/2)*Pi^(1/2)*arccos(a*x)^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*5^(1/2)*arccos(a*x)^(1/2))+3*3^
(1/2)*2^(1/2)*Pi^(1/2)*arccos(a*x)^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)*arccos(a*x)^(1/2))+2*2^(1/2)*Pi^(1/
2)*arccos(a*x)^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))-2*(-a^2*x^2+1)^(1/2)-3*sin(3*arccos(a*x))-si
n(5*arccos(a*x)))/arccos(a*x)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/acos(a*x)^(3/2),x)

[Out]

int(x^4/acos(a*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {acos}^{\frac {3}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/acos(a*x)**(3/2),x)

[Out]

Integral(x**4/acos(a*x)**(3/2), x)

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