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

Optimal. Leaf size=43 \[ x \cos ^{-1}\left (a x^2\right )-\frac {2 F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{\sqrt {a}}+\frac {2 E\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{\sqrt {a}} \]

[Out]

x*arccos(a*x^2)+2*EllipticE(x*a^(1/2),I)/a^(1/2)-2*EllipticF(x*a^(1/2),I)/a^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4841, 12, 307, 221, 1199, 424} \[ x \cos ^{-1}\left (a x^2\right )-\frac {2 F\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{\sqrt {a}}+\frac {2 E\left (\left .\sin ^{-1}\left (\sqrt {a} x\right )\right |-1\right )}{\sqrt {a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 4841

Int[ArcCos[u_], x_Symbol] :> Simp[x*ArcCos[u], x] + Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.79 \[ \frac {2}{3} a x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};a^2 x^4\right )+x \cos ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcCos[a*x^2] + (2*a*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, a^2*x^4])/3

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\arccos \left (a x^{2}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccos(a*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccos(a*x^2), x)

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maple [A]  time = 0.01, size = 65, normalized size = 1.51 \[ x \arccos \left (a \,x^{2}\right )-\frac {2 \sqrt {-a \,x^{2}+1}\, \sqrt {a \,x^{2}+1}\, \left (\EllipticF \left (x \sqrt {a}, i\right )-\EllipticE \left (x \sqrt {a}, i\right )\right )}{\sqrt {a}\, \sqrt {-a^{2} x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccos(a*x^2)-2/a^(1/2)*(-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)-Elliptic
E(x*a^(1/2),I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (\sqrt {a x^{2} + 1} \sqrt {-a x^{2} + 1}, a x^{2}\right ) - 2 \, a \int \frac {x^{2} e^{\left (\frac {1}{2} \, \log \left (a x^{2} + 1\right ) + \frac {1}{2} \, \log \left (-a x^{2} + 1\right )\right )}}{a^{4} x^{8} - a^{2} x^{4} + {\left (a^{2} x^{4} - 1\right )} e^{\left (\log \left (a x^{2} + 1\right ) + \log \left (-a x^{2} + 1\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan2(sqrt(a*x^2 + 1)*sqrt(-a*x^2 + 1), a*x^2) - 2*a*integrate(x^2*e^(1/2*log(a*x^2 + 1) + 1/2*log(-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 \[ \int \mathrm {acos}\left (a\,x^2\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acos(a*x^2), x)

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sympy [A]  time = 1.00, size = 44, normalized size = 1.02 \[ \frac {a x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {a^{2} x^{4} e^{2 i \pi }} \right )}}{2 \Gamma \left (\frac {7}{4}\right )} + x \operatorname {acos}{\left (a x^{2} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), a**2*x**4*exp_polar(2*I*pi))/(2*gamma(7/4)) + x*acos(a*x**2)

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