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

Optimal. Leaf size=171 \[ \frac{\sqrt{\sqrt{p^2+4}-p} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{p^2+4}-p} x \left (\sqrt{p^2+4}+p-2 x^2\right )}{2 \sqrt{2} \sqrt{p x^2-x^4+1}}\right )}{2 \sqrt{2}}-\frac{\sqrt{\sqrt{p^2+4}+p} \tan ^{-1}\left (\frac{\sqrt{\sqrt{p^2+4}+p} x \left (-\sqrt{p^2+4}+p-2 x^2\right )}{2 \sqrt{2} \sqrt{p x^2-x^4+1}}\right )}{2 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0800218, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {2072} \[ \frac{\sqrt{\sqrt{p^2+4}-p} \tanh ^{-1}\left (\frac{\sqrt{\sqrt{p^2+4}-p} x \left (\sqrt{p^2+4}+p-2 x^2\right )}{2 \sqrt{2} \sqrt{p x^2-x^4+1}}\right )}{2 \sqrt{2}}-\frac{\sqrt{\sqrt{p^2+4}+p} \tan ^{-1}\left (\frac{\sqrt{\sqrt{p^2+4}+p} x \left (-\sqrt{p^2+4}+p-2 x^2\right )}{2 \sqrt{2} \sqrt{p x^2-x^4+1}}\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + p*x^2 - x^4]/(1 + x^4),x]

[Out]

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

Rule 2072

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^4), x_Symbol] :> With[{q = Sqrt[b^2 - 4*a*c]},
-Simp[(a*Sqrt[b + q]*ArcTan[(Sqrt[b + q]*x*(b - q + 2*c*x^2))/(2*Sqrt[2]*Rt[-(a*c), 2]*Sqrt[a + b*x^2 + c*x^4
])])/(2*Sqrt[2]*Rt[-(a*c), 2]*d), x] + Simp[(a*Sqrt[-b + q]*ArcTanh[(Sqrt[-b + q]*x*(b + q + 2*c*x^2))/(2*Sqrt
[2]*Rt[-(a*c), 2]*Sqrt[a + b*x^2 + c*x^4])])/(2*Sqrt[2]*Rt[-(a*c), 2]*d), x]] /; FreeQ[{a, b, c, d, e}, x] &&
EqQ[c*d + a*e, 0] && NegQ[a*c]

Rubi steps

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

Mathematica [C]  time = 0.432111, size = 322, normalized size = 1.88 \[ \frac{\sqrt{\frac{4 x^2}{\sqrt{p^2+4}-p}+2} \sqrt{1-\frac{2 x^2}{\sqrt{p^2+4}+p}} \left (2 i \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{1}{\sqrt{p^2+4}-p}} x\right ),\frac{p-\sqrt{p^2+4}}{\sqrt{p^2+4}+p}\right )-(p+2 i) \Pi \left (\frac{1}{2} i \left (p-\sqrt{p^2+4}\right );i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{1}{\sqrt{p^2+4}-p}} x\right )|\frac{p-\sqrt{p^2+4}}{p+\sqrt{p^2+4}}\right )+(p-2 i) \Pi \left (\frac{1}{2} i \left (\sqrt{p^2+4}-p\right );i \sinh ^{-1}\left (\sqrt{2} \sqrt{\frac{1}{\sqrt{p^2+4}-p}} x\right )|\frac{p-\sqrt{p^2+4}}{p+\sqrt{p^2+4}}\right )\right )}{4 \sqrt{\frac{1}{\sqrt{p^2+4}-p}} \sqrt{p x^2-x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + p*x^2 - x^4]/(1 + x^4),x]

[Out]

(Sqrt[2 + (4*x^2)/(-p + Sqrt[4 + p^2])]*Sqrt[1 - (2*x^2)/(p + Sqrt[4 + p^2])]*((2*I)*EllipticF[I*ArcSinh[Sqrt[
2]*Sqrt[(-p + Sqrt[4 + p^2])^(-1)]*x], (p - Sqrt[4 + p^2])/(p + Sqrt[4 + p^2])] - (2*I + p)*EllipticPi[(I/2)*(
p - Sqrt[4 + p^2]), I*ArcSinh[Sqrt[2]*Sqrt[(-p + Sqrt[4 + p^2])^(-1)]*x], (p - Sqrt[4 + p^2])/(p + Sqrt[4 + p^
2])] + (-2*I + p)*EllipticPi[(I/2)*(-p + Sqrt[4 + p^2]), I*ArcSinh[Sqrt[2]*Sqrt[(-p + Sqrt[4 + p^2])^(-1)]*x],
(p - Sqrt[4 + p^2])/(p + Sqrt[4 + p^2])]))/(4*Sqrt[(-p + Sqrt[4 + p^2])^(-1)]*Sqrt[1 + p*x^2 - x^4])

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Maple [B]  time = 0.072, size = 456, normalized size = 2.7 \begin{align*}{\frac{\sqrt{2}}{32}\sqrt{p+\sqrt{{p}^{2}+4}}\sqrt{{p}^{2}+4}\ln \left ({\frac{-{x}^{4}+p{x}^{2}+1}{{x}^{2}}}+{\frac{\sqrt{2}}{x}\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{p+\sqrt{{p}^{2}+4}}}+\sqrt{{p}^{2}+4} \right ) }-{\frac{\sqrt{2}}{4}\arctan \left ({\frac{1}{2} \left ( 2\,{\frac{\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{2}}{x}}+2\,\sqrt{p+\sqrt{{p}^{2}+4}} \right ){\frac{1}{\sqrt{-p+\sqrt{{p}^{2}+4}}}}} \right ){\frac{1}{\sqrt{-p+\sqrt{{p}^{2}+4}}}}}-{\frac{\sqrt{2}p}{32}\sqrt{p+\sqrt{{p}^{2}+4}}\ln \left ({\frac{-{x}^{4}+p{x}^{2}+1}{{x}^{2}}}+{\frac{\sqrt{2}}{x}\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{p+\sqrt{{p}^{2}+4}}}+\sqrt{{p}^{2}+4} \right ) }-{\frac{\sqrt{2}}{32}\sqrt{p+\sqrt{{p}^{2}+4}}\sqrt{{p}^{2}+4}\ln \left ({\frac{\sqrt{2}}{x}\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{p+\sqrt{{p}^{2}+4}}}-{\frac{-{x}^{4}+p{x}^{2}+1}{{x}^{2}}}-\sqrt{{p}^{2}+4} \right ) }+{\frac{\sqrt{2}}{4}\arctan \left ({\frac{1}{2} \left ( 2\,\sqrt{p+\sqrt{{p}^{2}+4}}-2\,{\frac{\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{2}}{x}} \right ){\frac{1}{\sqrt{-p+\sqrt{{p}^{2}+4}}}}} \right ){\frac{1}{\sqrt{-p+\sqrt{{p}^{2}+4}}}}}+{\frac{\sqrt{2}p}{32}\sqrt{p+\sqrt{{p}^{2}+4}}\ln \left ({\frac{\sqrt{2}}{x}\sqrt{-{x}^{4}+p{x}^{2}+1}\sqrt{p+\sqrt{{p}^{2}+4}}}-{\frac{-{x}^{4}+p{x}^{2}+1}{{x}^{2}}}-\sqrt{{p}^{2}+4} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*2^(1/2)*(p+(p^2+4)^(1/2))^(1/2)*(p^2+4)^(1/2)*ln((-x^4+p*x^2+1)/x^2+(-x^4+p*x^2+1)^(1/2)*2^(1/2)/x*(p+(p^
2+4)^(1/2))^(1/2)+(p^2+4)^(1/2))-1/4*2^(1/2)/(-p+(p^2+4)^(1/2))^(1/2)*arctan(1/2*(2*(-x^4+p*x^2+1)^(1/2)*2^(1/
2)/x+2*(p+(p^2+4)^(1/2))^(1/2))/(-p+(p^2+4)^(1/2))^(1/2))-1/32*2^(1/2)*(p+(p^2+4)^(1/2))^(1/2)*p*ln((-x^4+p*x^
2+1)/x^2+(-x^4+p*x^2+1)^(1/2)*2^(1/2)/x*(p+(p^2+4)^(1/2))^(1/2)+(p^2+4)^(1/2))-1/32*2^(1/2)*(p+(p^2+4)^(1/2))^
(1/2)*(p^2+4)^(1/2)*ln((-x^4+p*x^2+1)^(1/2)*2^(1/2)/x*(p+(p^2+4)^(1/2))^(1/2)-(-x^4+p*x^2+1)/x^2-(p^2+4)^(1/2)
)+1/4*2^(1/2)/(-p+(p^2+4)^(1/2))^(1/2)*arctan(1/2*(2*(p+(p^2+4)^(1/2))^(1/2)-2*(-x^4+p*x^2+1)^(1/2)*2^(1/2)/x)
/(-p+(p^2+4)^(1/2))^(1/2))+1/32*2^(1/2)*(p+(p^2+4)^(1/2))^(1/2)*p*ln((-x^4+p*x^2+1)^(1/2)*2^(1/2)/x*(p+(p^2+4)
^(1/2))^(1/2)-(-x^4+p*x^2+1)/x^2-(p^2+4)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 25.3474, size = 6283, normalized size = 36.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(8*sqrt(2)*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(3/4)*arctan(1/4*(2*(p^3 + 4*p)*x^12 - 2*(p^4 - 2*p
^2 - 24)*x^10 - 20*(p^3 + 4*p)*x^8 + 2*(3*p^4 + 4*p^2 - 32)*x^6 + 10*(p^3 + 4*p)*x^4 + 4*(p^2 + 4)*x^2 - 2*((p
^2 + 4)*x^12 - (p^3 + 4*p)*x^10 - (p^3 + 4*p)*x^6 - (p^2 + 4)*x^4 + (p*x^12 - (p^2 - 6)*x^10 - 10*p*x^8 + (3*p
^2 - 8)*x^6 + 5*p*x^4 + 2*x^2)*sqrt(p^2 + 4))*sqrt(p^2 + 4) + 2*((p^2 + 4)*x^12 - (p^3 + 4*p)*x^10 - (p^3 + 4*
p)*x^6 - (p^2 + 4)*x^4)*sqrt(p^2 + 4) + sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(2*(sqrt(2)*(x^9 - p*x^7 - x^5)*sqrt(-
x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + sqrt(2)*(x^11 - 2*p*x^9 + (p^2 - 2)*x^7 + 2*p*x^5 + x^3)*sqrt(-x^4 + p*x^2 +
1))*(p^2 + 4)^(3/4) - (sqrt(2)*(p*x^9 + 8*x^7 - 6*p*x^5 + 2*p^2*x^3 + p*x)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4
) + sqrt(2)*((p^2 + 4)*x^9 + 4*(p^2 + 4)*x^5 - 2*(p^3 + 4*p)*x^3 - (p^2 + 4)*x)*sqrt(-x^4 + p*x^2 + 1))*(p^2 +
4)^(1/4)) - (2*((p^3 + 4*p)*x^8 + 4*(p^2 + 4)*x^6 - (p^3 + 4*p)*x^4)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + 2
*((p^4 + 6*p^2 + 8)*x^8 + 4*(p^3 + 4*p)*x^6 - (p^4 - 4*p^2 - 32)*x^4 - 4*(p^3 + 4*p)*x^2 - 2*p^2 - 8)*sqrt(-x^
4 + p*x^2 + 1) - 2*((p*x^10 - (p^2 - 4)*x^8 - 6*p*x^6 + (p^2 - 4)*x^4 + p*x^2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2
+ 4) + ((p^2 + 4)*x^10 - (p^3 + 4*p)*x^8 - 2*(p^2 + 4)*x^6 + (p^3 + 4*p)*x^4 + (p^2 + 4)*x^2)*sqrt(-x^4 + p*x
^2 + 1))*sqrt(p^2 + 4) - sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*((sqrt(2)*(x^11 - p*x^9 - p*x^5 - x^3)*sqrt(p^2 + 4)
+ sqrt(2)*(2*x^13 - 5*p*x^11 + (3*p^2 - 8)*x^9 + 10*p*x^7 - (p^2 - 6)*x^5 - p*x^3))*(p^2 + 4)^(3/4) - (sqrt(2)
*(p*x^11 - (p^2 - 6)*x^9 - 10*p*x^7 + (3*p^2 - 8)*x^5 + 5*p*x^3 + 2*x)*sqrt(p^2 + 4) + sqrt(2)*((p^2 + 4)*x^11
- (p^3 + 4*p)*x^9 - (p^3 + 4*p)*x^5 - (p^2 + 4)*x^3))*(p^2 + 4)^(1/4)))*sqrt(-((p^2 + 4)*x^4 - (p^2 + 4)^(3/2
)*x^2 - sqrt(2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(3/4)*x - (p^3 + 4*p)*x^2 - p
^2 - 4)/((p^2 + 4)*x^4 + p^2 + 4)))/((p^2 + 4)*x^12 - 3*(p^3 + 4*p)*x^10 + (2*p^4 + p^2 - 28)*x^8 + 10*(p^3 +
4*p)*x^6 - (2*p^4 + p^2 - 28)*x^4 - 3*(p^3 + 4*p)*x^2 - p^2 - 4)) + 8*sqrt(2)*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*
(p^2 + 4)^(3/4)*arctan(-1/4*(2*(p^3 + 4*p)*x^12 - 2*(p^4 - 2*p^2 - 24)*x^10 - 20*(p^3 + 4*p)*x^8 + 2*(3*p^4 +
4*p^2 - 32)*x^6 + 10*(p^3 + 4*p)*x^4 + 4*(p^2 + 4)*x^2 - 2*((p^2 + 4)*x^12 - (p^3 + 4*p)*x^10 - (p^3 + 4*p)*x^
6 - (p^2 + 4)*x^4 + (p*x^12 - (p^2 - 6)*x^10 - 10*p*x^8 + (3*p^2 - 8)*x^6 + 5*p*x^4 + 2*x^2)*sqrt(p^2 + 4))*sq
rt(p^2 + 4) + 2*((p^2 + 4)*x^12 - (p^3 + 4*p)*x^10 - (p^3 + 4*p)*x^6 - (p^2 + 4)*x^4)*sqrt(p^2 + 4) - sqrt(p^2
+ sqrt(p^2 + 4)*p + 4)*(2*(sqrt(2)*(x^9 - p*x^7 - x^5)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + sqrt(2)*(x^11 -
2*p*x^9 + (p^2 - 2)*x^7 + 2*p*x^5 + x^3)*sqrt(-x^4 + p*x^2 + 1))*(p^2 + 4)^(3/4) - (sqrt(2)*(p*x^9 + 8*x^7 -
6*p*x^5 + 2*p^2*x^3 + p*x)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + sqrt(2)*((p^2 + 4)*x^9 + 4*(p^2 + 4)*x^5 - 2
*(p^3 + 4*p)*x^3 - (p^2 + 4)*x)*sqrt(-x^4 + p*x^2 + 1))*(p^2 + 4)^(1/4)) - (2*((p^3 + 4*p)*x^8 + 4*(p^2 + 4)*x
^6 - (p^3 + 4*p)*x^4)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + 2*((p^4 + 6*p^2 + 8)*x^8 + 4*(p^3 + 4*p)*x^6 - (p
^4 - 4*p^2 - 32)*x^4 - 4*(p^3 + 4*p)*x^2 - 2*p^2 - 8)*sqrt(-x^4 + p*x^2 + 1) - 2*((p*x^10 - (p^2 - 4)*x^8 - 6*
p*x^6 + (p^2 - 4)*x^4 + p*x^2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + 4) + ((p^2 + 4)*x^10 - (p^3 + 4*p)*x^8 - 2*(p
^2 + 4)*x^6 + (p^3 + 4*p)*x^4 + (p^2 + 4)*x^2)*sqrt(-x^4 + p*x^2 + 1))*sqrt(p^2 + 4) + sqrt(p^2 + sqrt(p^2 + 4
)*p + 4)*((sqrt(2)*(x^11 - p*x^9 - p*x^5 - x^3)*sqrt(p^2 + 4) + sqrt(2)*(2*x^13 - 5*p*x^11 + (3*p^2 - 8)*x^9 +
10*p*x^7 - (p^2 - 6)*x^5 - p*x^3))*(p^2 + 4)^(3/4) - (sqrt(2)*(p*x^11 - (p^2 - 6)*x^9 - 10*p*x^7 + (3*p^2 - 8
)*x^5 + 5*p*x^3 + 2*x)*sqrt(p^2 + 4) + sqrt(2)*((p^2 + 4)*x^11 - (p^3 + 4*p)*x^9 - (p^3 + 4*p)*x^5 - (p^2 + 4)
*x^3))*(p^2 + 4)^(1/4)))*sqrt(-((p^2 + 4)*x^4 - (p^2 + 4)^(3/2)*x^2 + sqrt(2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2
+ sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(3/4)*x - (p^3 + 4*p)*x^2 - p^2 - 4)/((p^2 + 4)*x^4 + p^2 + 4)))/((p^2 + 4)*x
^12 - 3*(p^3 + 4*p)*x^10 + (2*p^4 + p^2 - 28)*x^8 + 10*(p^3 + 4*p)*x^6 - (2*p^4 + p^2 - 28)*x^4 - 3*(p^3 + 4*p
)*x^2 - p^2 - 4)) - (sqrt(2)*sqrt(p^2 + 4)*p - sqrt(2)*(p^2 + 4))*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(1
/4)*log(-((p^2 + 4)*x^4 - (p^2 + 4)^(3/2)*x^2 + sqrt(2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)
*(p^2 + 4)^(3/4)*x - (p^3 + 4*p)*x^2 - p^2 - 4)/((p^2 + 4)*x^4 + p^2 + 4)) + (sqrt(2)*sqrt(p^2 + 4)*p - sqrt(2
)*(p^2 + 4))*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(1/4)*log(-((p^2 + 4)*x^4 - (p^2 + 4)^(3/2)*x^2 - sqrt(
2)*sqrt(-x^4 + p*x^2 + 1)*sqrt(p^2 + sqrt(p^2 + 4)*p + 4)*(p^2 + 4)^(3/4)*x - (p^3 + 4*p)*x^2 - p^2 - 4)/((p^2
+ 4)*x^4 + p^2 + 4)))/(p^2 + 4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(p*x**2 - x**4 + 1)/(x**4 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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