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

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

[Out]

(-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))

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Rubi [A]  time = 0.0284355, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, 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 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]

Rule 12

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

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 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]

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*}

Mathematica [C]  time = 0.155284, size = 63, normalized size = 1.15 \[ \frac{1}{9} \left (\frac{2 i \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-a} x\right ),-1\right )}{(-a)^{3/2}}-\frac{2 x \sqrt{1-a^2 x^4}}{a}+3 x^3 \cos ^{-1}\left (a x^2\right )\right ) \]

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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Maple [A]  time = 0.007, size = 79, normalized size = 1.4 \begin{align*}{\frac{{x}^{3}\arccos \left ( a{x}^{2} \right ) }{3}}+{\frac{2\,a}{3} \left ( -{\frac{x}{3\,{a}^{2}}\sqrt{-{a}^{2}{x}^{4}+1}}+{\frac{1}{3}\sqrt{-a{x}^{2}+1}\sqrt{a{x}^{2}+1}{\it EllipticF} \left ( x\sqrt{a},i \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{-{a}^{2}{x}^{4}+1}}}} \right ) } \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \arccos \left (a x^{2}\right ), x\right ) \end{align*}

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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Sympy [A]  time = 1.42681, size = 48, normalized size = 0.87 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int x^{2} \arccos \left (a x^{2}\right )\,{d x} \end{align*}

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)