3.48 \(\int x^2 \cos ^{-1}(a x^2) \, dx\)

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

[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 = 0.03, 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 = {4843, 12, 321, 221} \[ -\frac {2 x \sqrt {1-a^2 x^4}}{9 a}+\frac {2 F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{9 a^{3/2}}+\frac {1}{3} x^3 \cos ^{-1}\left (a x^2\right ) \]

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 221

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 321

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 4843

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]  time = 0.16, size = 63, normalized size = 1.15 \[ \frac {1}{9} \left (-\frac {2 x \sqrt {1-a^2 x^4}}{a}+3 x^3 \cos ^{-1}\left (a x^2\right )+\frac {2 i F\left (\left .i \sinh ^{-1}\left (\sqrt {-a} x\right )\right |-1\right )}{(-a)^{3/2}}\right ) \]

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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \arccos \left (a x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*arccos(a*x^2), x)

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

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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maple [A]  time = 0.01, size = 79, normalized size = 1.44 \[ \frac {x^{3} \arccos \left (a \,x^{2}\right )}{3}+\frac {2 a \left (-\frac {x \sqrt {-a^{2} x^{4}+1}}{3 a^{2}}+\frac {\sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \EllipticF \left (x \sqrt {a}, i\right )}{3 a^{\frac {5}{2}} \sqrt {-a^{2} x^{4}+1}}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \[ \frac {1}{3} \, x^{3} \arctan \left (\sqrt {a x^{2} + 1} \sqrt {-a x^{2} + 1}, a x^{2}\right ) - 2 \, a \int \frac {x^{4} e^{\left (\frac {1}{2} \, \log \left (a x^{2} + 1\right ) + \frac {1}{2} \, \log \left (-a x^{2} + 1\right )\right )}}{3 \, {\left (a^{4} x^{8} - a^{2} x^{4} + {\left (a^{2} x^{4} - 1\right )} e^{\left (\log \left (a x^{2} + 1\right ) + \log \left (-a x^{2} + 1\right )\right )}\right )}}\,{d x} \]

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

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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sympy [A]  time = 1.23, size = 48, normalized size = 0.87 \[ \frac {a x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {a^{2} x^{4} e^{2 i \pi }} \right )}}{6 \Gamma \left (\frac {9}{4}\right )} + \frac {x^{3} \operatorname {acos}{\left (a x^{2} \right )}}{3} \]

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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