3.8.34 \(\int \sqrt {1-x^2-y^4} \, dx\)

Optimal. Leaf size=56 \[ \frac {1}{2} x \sqrt {-x^2-y^4+1}-\frac {1}{2} i \left (y^4-1\right ) \log \left (\sqrt {-x^2-y^4+1}-i x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 52, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {195, 217, 203} \begin {gather*} \frac {1}{2} x \sqrt {-x^2-y^4+1}+\frac {1}{2} \left (1-y^4\right ) \tan ^{-1}\left (\frac {x}{\sqrt {-x^2-y^4+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^2 - y^4],x]

[Out]

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

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 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])

Rule 217

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.88 \begin {gather*} \frac {1}{2} \left (x \sqrt {-x^2-y^4+1}-\left (y^4-1\right ) \tan ^{-1}\left (\frac {x}{\sqrt {-x^2-y^4+1}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - x^2 - y^4],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 56, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \sqrt {1-x^2-y^4}-\frac {1}{2} i \left (-1+y^4\right ) \log \left (-i x+\sqrt {1-x^2-y^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[1 - x^2 - y^4],x]

[Out]

(x*Sqrt[1 - x^2 - y^4])/2 - (I/2)*(-1 + y^4)*Log[(-I)*x + Sqrt[1 - x^2 - y^4]]

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fricas [A]  time = 0.48, size = 44, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, {\left (y^{4} - 1\right )} \arctan \left (\frac {\sqrt {-y^{4} - x^{2} + 1}}{x}\right ) + \frac {1}{2} \, \sqrt {-y^{4} - x^{2} + 1} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 37, normalized size = 0.66 \begin {gather*} -\frac {1}{2} \, {\left (y^{4} - 1\right )} \arcsin \left (\frac {x}{\sqrt {-y^{4} + 1}}\right ) + \frac {1}{2} \, \sqrt {-y^{4} - x^{2} + 1} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 45, normalized size = 0.80

method	result	size
default	\(\frac {x \sqrt {-y^{4}-x^{2}+1}}{2}-\frac {\left (4 y^{4}-4\right ) \arctan \left (\frac {x}{\sqrt {-y^{4}-x^{2}+1}}\right )}{8}\)	\(45\)
risch	\(-\frac {x \left (y^{4}+x^{2}-1\right )}{2 \sqrt {-y^{4}-x^{2}+1}}-\frac {\arctan \left (\frac {x}{\sqrt {-y^{4}-x^{2}+1}}\right ) y^{4}}{2}+\frac {\arctan \left (\frac {x}{\sqrt {-y^{4}-x^{2}+1}}\right )}{2}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-y^4-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*x*(-y^4-x^2+1)^(1/2)-1/8*(4*y^4-4)*arctan(1/(-y^4-x^2+1)^(1/2)*x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((y-1)*(y+1)>0)', see `assume?`
for more details)Is (y-1)*(y+1) positive or negative?

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mupad [B]  time = 0.10, size = 48, normalized size = 0.86 \begin {gather*} \frac {x\,\sqrt {-x^2-y^4+1}}{2}+\ln \left (\sqrt {-x^2-y^4+1}+x\,1{}\mathrm {i}\right )\,\left (\frac {y^4\,1{}\mathrm {i}}{2}-\frac {1}{2}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - y^4 - x^2)^(1/2),x)

[Out]

(x*(1 - y^4 - x^2)^(1/2))/2 + log(x*1i + (1 - y^4 - x^2)^(1/2))*((y^4*1i)/2 - 1i/2)

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sympy [B]  time = 1.65, size = 745, normalized size = 13.30

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*I*x**3*sqrt(polar_lift(1 - y**4))/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) - 4*sqrt(x**2/pola
r_lift(1 - y**4) - 1)) - 2*I*x*y**4*sqrt(polar_lift(1 - y**4))/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) - 4
*sqrt(x**2/polar_lift(1 - y**4) - 1)) + 2*I*x*sqrt(polar_lift(1 - y**4))/(4*y**4*sqrt(x**2/polar_lift(1 - y**4
) - 1) - 4*sqrt(x**2/polar_lift(1 - y**4) - 1)) - 2*I*y**4*(1 - y**4)*sqrt(x**2/polar_lift(1 - y**4) - 1)*acos
h(x/sqrt(polar_lift(1 - y**4)))/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) - 4*sqrt(x**2/polar_lift(1 - y**4)
- 1)) + pi*y**4*(1 - y**4)*sqrt(x**2/polar_lift(1 - y**4) - 1)/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) -
4*sqrt(x**2/polar_lift(1 - y**4) - 1)) + 2*I*(1 - y**4)*sqrt(x**2/polar_lift(1 - y**4) - 1)*acosh(x/sqrt(polar
_lift(1 - y**4)))/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) - 4*sqrt(x**2/polar_lift(1 - y**4) - 1)) - pi*(1
- y**4)*sqrt(x**2/polar_lift(1 - y**4) - 1)/(4*y**4*sqrt(x**2/polar_lift(1 - y**4) - 1) - 4*sqrt(x**2/polar_l
ift(1 - y**4) - 1)), Abs(x**2/(y**4 - 1)) > 1), (x**3*sqrt(polar_lift(1 - y**4))/(2*y**4*sqrt(-x**2/polar_lift
(1 - y**4) + 1) - 2*sqrt(-x**2/polar_lift(1 - y**4) + 1)) + x*y**4*sqrt(polar_lift(1 - y**4))/(2*y**4*sqrt(-x*
*2/polar_lift(1 - y**4) + 1) - 2*sqrt(-x**2/polar_lift(1 - y**4) + 1)) - x*sqrt(polar_lift(1 - y**4))/(2*y**4*
sqrt(-x**2/polar_lift(1 - y**4) + 1) - 2*sqrt(-x**2/polar_lift(1 - y**4) + 1)) + y**4*(1 - y**4)*sqrt(-x**2/po
lar_lift(1 - y**4) + 1)*asin(x/sqrt(polar_lift(1 - y**4)))/(2*y**4*sqrt(-x**2/polar_lift(1 - y**4) + 1) - 2*sq
rt(-x**2/polar_lift(1 - y**4) + 1)) - (1 - y**4)*sqrt(-x**2/polar_lift(1 - y**4) + 1)*asin(x/sqrt(polar_lift(1
- y**4)))/(2*y**4*sqrt(-x**2/polar_lift(1 - y**4) + 1) - 2*sqrt(-x**2/polar_lift(1 - y**4) + 1)), True))

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