3.1.75 \(\int x^3 \sqrt {\text {ArcCos}(a x)} \, dx\) [75]

Optimal. Leaf size=95 \[ -\frac {3 \sqrt {\text {ArcCos}(a x)}}{32 a^4}+\frac {1}{4} x^4 \sqrt {\text {ArcCos}(a x)}-\frac {\sqrt {\frac {\pi }{2}} \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{64 a^4}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{16 a^4} \]

[Out]

-1/128*FresnelC(2*2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-1/16*FresnelC(2*arccos(a*x)^(1/2)/P
i^(1/2))*Pi^(1/2)/a^4-3/32*arccos(a*x)^(1/2)/a^4+1/4*x^4*arccos(a*x)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4726, 4810, 3393, 3385, 3433} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \text {FresnelC}\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\text {ArcCos}(a x)}\right )}{64 a^4}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcCos}(a x)}}{\sqrt {\pi }}\right )}{16 a^4}-\frac {3 \sqrt {\text {ArcCos}(a x)}}{32 a^4}+\frac {1}{4} x^4 \sqrt {\text {ArcCos}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[ArcCos[a*x]],x]

[Out]

(-3*Sqrt[ArcCos[a*x]])/(32*a^4) + (x^4*Sqrt[ArcCos[a*x]])/4 - (Sqrt[Pi/2]*FresnelC[2*Sqrt[2/Pi]*Sqrt[ArcCos[a*
x]]])/(64*a^4) - (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcCos[a*x]])/Sqrt[Pi]])/(16*a^4)

Rule 3385

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3433

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

Rule 4726

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCos[c*x])^n/(m
+ 1)), x] + Dist[b*c*(n/(m + 1)), Int[x^(m + 1)*((a + b*ArcCos[c*x])^(n - 1)/Sqrt[1 - c^2*x^2]), x], x] /; Fre
eQ[{a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]

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 x^3 \sqrt {\cos ^{-1}(a x)} \, dx &=\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}+\frac {1}{8} a \int \frac {x^4}{\sqrt {1-a^2 x^2} \sqrt {\cos ^{-1}(a x)}} \, dx\\ &=\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cos ^4(x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{8 a^4}\\ &=\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \left (\frac {3}{8 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}+\frac {\cos (4 x)}{8 \sqrt {x}}\right ) \, dx,x,\cos ^{-1}(a x)\right )}{8 a^4}\\ &=-\frac {3 \sqrt {\cos ^{-1}(a x)}}{32 a^4}+\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{64 a^4}-\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\cos ^{-1}(a x)\right )}{16 a^4}\\ &=-\frac {3 \sqrt {\cos ^{-1}(a x)}}{32 a^4}+\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}-\frac {\text {Subst}\left (\int \cos \left (4 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{32 a^4}-\frac {\text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\cos ^{-1}(a x)}\right )}{8 a^4}\\ &=-\frac {3 \sqrt {\cos ^{-1}(a x)}}{32 a^4}+\frac {1}{4} x^4 \sqrt {\cos ^{-1}(a x)}-\frac {\sqrt {\frac {\pi }{2}} C\left (2 \sqrt {\frac {2}{\pi }} \sqrt {\cos ^{-1}(a x)}\right )}{64 a^4}-\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\cos ^{-1}(a x)}}{\sqrt {\pi }}\right )}{16 a^4}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.09, size = 125, normalized size = 1.32 \begin {gather*} -\frac {\sqrt {\text {ArcCos}(a x)} \left (4 i \sqrt {2} \text {ArcCos}(a x) \text {Gamma}\left (\frac {3}{2},-2 i \text {ArcCos}(a x)\right )+4 \sqrt {2} \sqrt {\text {ArcCos}(a x)^2} \text {Gamma}\left (\frac {3}{2},2 i \text {ArcCos}(a x)\right )+i \text {ArcCos}(a x) \text {Gamma}\left (\frac {3}{2},-4 i \text {ArcCos}(a x)\right )+\sqrt {\text {ArcCos}(a x)^2} \text {Gamma}\left (\frac {3}{2},4 i \text {ArcCos}(a x)\right )\right )}{128 a^4 (-i \text {ArcCos}(a x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[ArcCos[a*x]],x]

[Out]

-1/128*(Sqrt[ArcCos[a*x]]*((4*I)*Sqrt[2]*ArcCos[a*x]*Gamma[3/2, (-2*I)*ArcCos[a*x]] + 4*Sqrt[2]*Sqrt[ArcCos[a*
x]^2]*Gamma[3/2, (2*I)*ArcCos[a*x]] + I*ArcCos[a*x]*Gamma[3/2, (-4*I)*ArcCos[a*x]] + Sqrt[ArcCos[a*x]^2]*Gamma
[3/2, (4*I)*ArcCos[a*x]]))/(a^4*((-I)*ArcCos[a*x])^(3/2))

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Maple [A]
time = 0.29, size = 91, normalized size = 0.96

method	result	size
default	\(\frac {-\sqrt {2}\, \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )+16 \arccos \left (a x \right ) \cos \left (2 \arccos \left (a x \right )\right )+4 \arccos \left (a x \right ) \cos \left (4 \arccos \left (a x \right )\right )-8 \sqrt {\arccos \left (a x \right )}\, \sqrt {\pi }\, \FresnelC \left (\frac {2 \sqrt {\arccos \left (a x \right )}}{\sqrt {\pi }}\right )}{128 a^{4} \sqrt {\arccos \left (a x \right )}}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccos(a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/128/a^4/arccos(a*x)^(1/2)*(-2^(1/2)*arccos(a*x)^(1/2)*Pi^(1/2)*FresnelC(2*2^(1/2)/Pi^(1/2)*arccos(a*x)^(1/2)
)+16*arccos(a*x)*cos(2*arccos(a*x))+4*arccos(a*x)*cos(4*arccos(a*x))-8*arccos(a*x)^(1/2)*Pi^(1/2)*FresnelC(2*a
rccos(a*x)^(1/2)/Pi^(1/2)))

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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^3*arccos(a*x)^(1/2),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^3*arccos(a*x)^(1/2),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 x^{3} \sqrt {\operatorname {acos}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acos(a*x)**(1/2),x)

[Out]

Integral(x**3*sqrt(acos(a*x)), x)

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Giac [C] Result contains complex when optimal does not.
time = 0.48, size = 153, normalized size = 1.61 \begin {gather*} \frac {\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{4}} - \frac {\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\arccos \left (a x\right )}\right )}{512 \, a^{4}} + \frac {\left (i + 1\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{64 \, a^{4}} - \frac {\left (i - 1\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, \sqrt {\arccos \left (a x\right )}\right )}{64 \, a^{4}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (4 i \, \arccos \left (a x\right )\right )}}{64 \, a^{4}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (2 i \, \arccos \left (a x\right )\right )}}{16 \, a^{4}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (-2 i \, \arccos \left (a x\right )\right )}}{16 \, a^{4}} + \frac {\sqrt {\arccos \left (a x\right )} e^{\left (-4 i \, \arccos \left (a x\right )\right )}}{64 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccos(a*x)^(1/2),x, algorithm="giac")

[Out]

(1/512*I + 1/512)*sqrt(2)*sqrt(pi)*erf((I - 1)*sqrt(2)*sqrt(arccos(a*x)))/a^4 - (1/512*I - 1/512)*sqrt(2)*sqrt
(pi)*erf(-(I + 1)*sqrt(2)*sqrt(arccos(a*x)))/a^4 + (1/64*I + 1/64)*sqrt(pi)*erf((I - 1)*sqrt(arccos(a*x)))/a^4
- (1/64*I - 1/64)*sqrt(pi)*erf(-(I + 1)*sqrt(arccos(a*x)))/a^4 + 1/64*sqrt(arccos(a*x))*e^(4*I*arccos(a*x))/a
^4 + 1/16*sqrt(arccos(a*x))*e^(2*I*arccos(a*x))/a^4 + 1/16*sqrt(arccos(a*x))*e^(-2*I*arccos(a*x))/a^4 + 1/64*s
qrt(arccos(a*x))*e^(-4*I*arccos(a*x))/a^4

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\sqrt {\mathrm {acos}\left (a\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(a*x)^(1/2),x)

[Out]

int(x^3*acos(a*x)^(1/2), x)

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