3.15 \(\int \frac{\sqrt{-x^2+\sqrt{1+x^4}}}{\sqrt{1+x^4}} \, dx\)

Optimal. Leaf size=33 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{\sqrt{x^4+1}-x^2}}\right )}{\sqrt{2}} \]

[Out]

ArcTan[(Sqrt[2]*x)/Sqrt[-x^2 + Sqrt[1 + x^4]]]/Sqrt[2]

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Rubi [A]  time = 0.0626, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2132, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{\sqrt{x^4+1}-x^2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-x^2 + Sqrt[1 + x^4]]/Sqrt[1 + x^4],x]

[Out]

ArcTan[(Sqrt[2]*x)/Sqrt[-x^2 + Sqrt[1 + x^4]]]/Sqrt[2]

Rule 2132

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{-x^2+\sqrt{1+x^4}}}{\sqrt{1+x^4}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{1+2 x^2} \, dx,x,\frac{x}{\sqrt{-x^2+\sqrt{1+x^4}}}\right )\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{-x^2+\sqrt{1+x^4}}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0113053, size = 33, normalized size = 1. \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{\sqrt{x^4+1}-x^2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-x^2 + Sqrt[1 + x^4]]/Sqrt[1 + x^4],x]

[Out]

ArcTan[(Sqrt[2]*x)/Sqrt[-x^2 + Sqrt[1 + x^4]]]/Sqrt[2]

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Maple [C]  time = 0.033, size = 22, normalized size = 0.7 \begin{align*} -{\frac{\sqrt{2}}{4\,{x}^{2}}{\mbox{$_3$F$_2$}({\frac{1}{2}},{\frac{3}{4}},{\frac{5}{4}};\,{\frac{3}{2}},{\frac{3}{2}};\,-{x}^{-4})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+(x^4+1)^(1/2))^(1/2)/(x^4+1)^(1/2),x)

[Out]

-1/4*2^(1/2)/x^2*hypergeom([1/2,3/4,5/4],[3/2,3/2],-1/x^4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-x^{2} + \sqrt{x^{4} + 1}}}{\sqrt{x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+(x^4+1)^(1/2))^(1/2)/(x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-x^2 + sqrt(x^4 + 1))/sqrt(x^4 + 1), x)

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Fricas [A]  time = 4.07396, size = 85, normalized size = 2.58 \begin{align*} -\frac{1}{2} \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-x^{2} + \sqrt{x^{4} + 1}}}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+(x^4+1)^(1/2))^(1/2)/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-x^2 + sqrt(x^4 + 1))/x)

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Sympy [A]  time = 0.705757, size = 15, normalized size = 0.45 \begin{align*} \frac{{G_{3, 3}^{2, 2}\left (\begin{matrix} \frac{1}{2}, 1 & 1 \\\frac{1}{4}, \frac{3}{4} & 0 \end{matrix} \middle |{x^{4}} \right )}}{4 \sqrt{\pi }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+(x**4+1)**(1/2))**(1/2)/(x**4+1)**(1/2),x)

[Out]

meijerg(((1/2, 1), (1,)), ((1/4, 3/4), (0,)), x**4)/(4*sqrt(pi))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-x^{2} + \sqrt{x^{4} + 1}}}{\sqrt{x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+(x^4+1)^(1/2))^(1/2)/(x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-x^2 + sqrt(x^4 + 1))/sqrt(x^4 + 1), x)