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

Optimal. Leaf size=23 \[ \frac{1}{2} \sqrt{1-x^2} x+\frac{1}{2} \sin ^{-1}(x) \]

[Out]

(x*Sqrt[1 - x^2])/2 + ArcSin[x]/2

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Rubi [A]  time = 0.0030979, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {26, 195, 216} \[ \frac{1}{2} \sqrt{1-x^2} x+\frac{1}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^4]/Sqrt[1 + x^2],x]

[Out]

(x*Sqrt[1 - x^2])/2 + ArcSin[x]/2

Rule 26

Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(m_.)*((c_) + (d_.)*(x_)^(j_))^(p_.), x_Symbol] :> Dist[(-(b^2/d))^m, Int[
u/(a - b*x^n)^m, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[j, 2*n] && EqQ[p, -m] && EqQ[b^2*c + a^2*d,
0] && GtQ[a, 0] && LtQ[d, 0]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-x^4}}{\sqrt{1+x^2}} \, dx &=\int \sqrt{1-x^2} \, dx\\ &=\frac{1}{2} x \sqrt{1-x^2}+\frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx\\ &=\frac{1}{2} x \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1}(x)\\ \end{align*}

Mathematica [B]  time = 0.0411808, size = 50, normalized size = 2.17 \[ \frac{1}{2} \left (\frac{\sqrt{1-x^4} x}{\sqrt{x^2+1}}+\tan ^{-1}\left (\frac{x \sqrt{x^2+1}}{\sqrt{1-x^4}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.008, size = 42, normalized size = 1.8 \begin{align*}{\frac{1}{2}\sqrt{-{x}^{4}+1} \left ( x\sqrt{-{x}^{2}+1}+\arcsin \left ( x \right ) \right ){\frac{1}{\sqrt{{x}^{2}+1}}}{\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-x^4+1)^(1/2)/(x^2+1)^(1/2)*(x*(-x^2+1)^(1/2)+arcsin(x))/(-x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.45123, size = 144, normalized size = 6.26 \begin{align*} \frac{\sqrt{-x^{4} + 1} \sqrt{x^{2} + 1} x -{\left (x^{2} + 1\right )} \arctan \left (\frac{\sqrt{-x^{4} + 1} \sqrt{x^{2} + 1}}{x^{3} + x}\right )}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(-x^4 + 1)*sqrt(x^2 + 1)*x - (x^2 + 1)*arctan(sqrt(-x^4 + 1)*sqrt(x^2 + 1)/(x^3 + x)))/(x^2 + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}{\sqrt{x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(x - 1)*(x + 1)*(x**2 + 1))/sqrt(x**2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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