3.1.48 \(\int x^2 \text {ArcCos}(a x^2) \, dx\) [48]

Optimal. Leaf size=55 \[ -\frac {2 x \sqrt {1-a^2 x^4}}{9 a}+\frac {1}{3} x^3 \text {ArcCos}\left (a x^2\right )+\frac {2 F\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{9 a^{3/2}} \]

[Out]

1/3*x^3*arccos(a*x^2)+2/9*EllipticF(x*a^(1/2),I)/a^(3/2)-2/9*x*(-a^2*x^4+1)^(1/2)/a

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Rubi [A]
time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4927, 12, 327, 227} \begin {gather*} \frac {2 F\left (\left .\text {ArcSin}\left (\sqrt {a} x\right )\right |-1\right )}{9 a^{3/2}}-\frac {2 x \sqrt {1-a^2 x^4}}{9 a}+\frac {1}{3} x^3 \text {ArcCos}\left (a x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCos[a*x^2],x]

[Out]

(-2*x*Sqrt[1 - a^2*x^4])/(9*a) + (x^3*ArcCos[a*x^2])/3 + (2*EllipticF[ArcSin[Sqrt[a]*x], -1])/(9*a^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4927

Int[((a_.) + ArcCos[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcCos[
u])/(d*(m + 1))), x] + Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 - u^2]), x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int x^2 \cos ^{-1}\left (a x^2\right ) \, dx &=\frac {1}{3} x^3 \cos ^{-1}\left (a x^2\right )+\frac {1}{3} \int \frac {2 a x^4}{\sqrt {1-a^2 x^4}} \, dx\\ &=\frac {1}{3} x^3 \cos ^{-1}\left (a x^2\right )+\frac {1}{3} (2 a) \int \frac {x^4}{\sqrt {1-a^2 x^4}} \, dx\\ &=-\frac {2 x \sqrt {1-a^2 x^4}}{9 a}+\frac {1}{3} x^3 \cos ^{-1}\left (a x^2\right )+\frac {2 \int \frac {1}{\sqrt {1-a^2 x^4}} \, dx}{9 a}\\ &=-\frac {2 x \sqrt {1-a^2 x^4}}{9 a}+\frac {1}{3} x^3 \cos ^{-1}\left (a x^2\right )+\frac {2 F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{9 a^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.11, size = 63, normalized size = 1.15 \begin {gather*} \frac {1}{9} \left (-\frac {2 x \sqrt {1-a^2 x^4}}{a}+3 x^3 \text {ArcCos}\left (a x^2\right )+\frac {2 i F\left (\left .i \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )}{(-a)^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCos[a*x^2],x]

[Out]

((-2*x*Sqrt[1 - a^2*x^4])/a + 3*x^3*ArcCos[a*x^2] + ((2*I)*EllipticF[I*ArcSinh[Sqrt[-a]*x], -1])/(-a)^(3/2))/9

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Maple [A]
time = 0.01, size = 79, normalized size = 1.44

method	result	size
default	\(\frac {x^{3} \arccos \left (a \,x^{2}\right )}{3}+\frac {2 a \left (-\frac {x \sqrt {-a^{2} x^{4}+1}}{3 a^{2}}+\frac {\sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \EllipticF \left (x \sqrt {a}, i\right )}{3 a^{\frac {5}{2}} \sqrt {-a^{2} x^{4}+1}}\right )}{3}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccos(a*x^2)+2/3*a*(-1/3/a^2*x*(-a^2*x^4+1)^(1/2)+1/3/a^(5/2)*(-a*x^2+1)^(1/2)*(a*x^2+1)^(1/2)/(-a^2*
x^4+1)^(1/2)*EllipticF(x*a^(1/2),I))

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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^2*arccos(a*x^2),x, algorithm="maxima")

[Out]

1/3*x^3*arctan2(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1), a*x^2) - 2*a*integrate(1/3*x^4*e^(1/2*log(a*x^2 + 1) + 1/2*l
og(-a*x^2 + 1))/(a^4*x^8 - a^2*x^4 + (a^2*x^4 - 1)*e^(log(a*x^2 + 1) + log(-a*x^2 + 1))), x)

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Fricas [A]
time = 0.29, size = 33, normalized size = 0.60 \begin {gather*} \frac {3 \, a x^{3} \arccos \left (a x^{2}\right ) - 2 \, \sqrt {-a^{2} x^{4} + 1} x}{9 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*x^3*arccos(a*x^2) - 2*sqrt(-a^2*x^4 + 1)*x)/a

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Sympy [A]
time = 0.75, size = 48, normalized size = 0.87 \begin {gather*} \frac {a x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {a^{2} x^{4} e^{2 i \pi }} \right )}}{6 \Gamma \left (\frac {9}{4}\right )} + \frac {x^{3} \operatorname {acos}{\left (a x^{2} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acos(a*x**2),x)

[Out]

a*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), a**2*x**4*exp_polar(2*I*pi))/(6*gamma(9/4)) + x**3*acos(a*x**2)/3

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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^2*arccos(a*x^2),x, algorithm="giac")

[Out]

integrate(x^2*arccos(a*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,\mathrm {acos}\left (a\,x^2\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(a*x^2),x)

[Out]

int(x^2*acos(a*x^2), x)

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