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]

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

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Rubi [A]  time = 0.08, antiderivative size = 171, normalized size of antiderivative = 1.00, 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*}

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Mathematica [C]  time = 0.42, size = 322, normalized size = 1.88 \[ \frac {\sqrt {\frac {4 x^2}{\sqrt {p^2+4}-p}+2} \sqrt {1-\frac {2 x^2}{\sqrt {p^2+4}+p}} \left (2 i \operatorname {EllipticF}\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {1}{\sqrt {p^2+4}-p}} x\right ),\frac {p-\sqrt {p^2+4}}{\sqrt {p^2+4}+p}\right )-(p+2 i) \Pi \left (\frac {1}{2} i \left (p-\sqrt {p^2+4}\right );i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {1}{\sqrt {p^2+4}-p}} x\right )|\frac {p-\sqrt {p^2+4}}{p+\sqrt {p^2+4}}\right )+(p-2 i) \Pi \left (\frac {1}{2} i \left (\sqrt {p^2+4}-p\right );i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {1}{\sqrt {p^2+4}-p}} x\right )|\frac {p-\sqrt {p^2+4}}{p+\sqrt {p^2+4}}\right )\right )}{4 \sqrt {\frac {1}{\sqrt {p^2+4}-p}} \sqrt {p x^2-x^4+1}} \]

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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fricas [B]  time = 5.83, size = 2667, normalized size = 15.60 \[ \text {result too large to display} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{4} + p x^{2} + 1}}{x^{4} + 1}\,{d x} \]

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)

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maple [B]  time = 0.11, size = 456, normalized size = 2.67 \[ -\frac {\sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}\, p \ln \left (\frac {\sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}}{x}+\frac {-x^{4}+p \,x^{2}+1}{x^{2}}+\sqrt {p^{2}+4}\right )}{32}+\frac {\sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}\, p \ln \left (\frac {\sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}}{x}-\frac {-x^{4}+p \,x^{2}+1}{x^{2}}-\sqrt {p^{2}+4}\right )}{32}+\frac {\sqrt {2}\, \arctan \left (\frac {-\frac {2 \sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}}{x}+2 \sqrt {p +\sqrt {p^{2}+4}}}{2 \sqrt {-p +\sqrt {p^{2}+4}}}\right )}{4 \sqrt {-p +\sqrt {p^{2}+4}}}-\frac {\sqrt {2}\, \arctan \left (\frac {\frac {2 \sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}}{x}+2 \sqrt {p +\sqrt {p^{2}+4}}}{2 \sqrt {-p +\sqrt {p^{2}+4}}}\right )}{4 \sqrt {-p +\sqrt {p^{2}+4}}}+\frac {\sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}\, \sqrt {p^{2}+4}\, \ln \left (\frac {\sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}}{x}+\frac {-x^{4}+p \,x^{2}+1}{x^{2}}+\sqrt {p^{2}+4}\right )}{32}-\frac {\sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}\, \sqrt {p^{2}+4}\, \ln \left (\frac {\sqrt {-x^{4}+p \,x^{2}+1}\, \sqrt {2}\, \sqrt {p +\sqrt {p^{2}+4}}}{x}-\frac {-x^{4}+p \,x^{2}+1}{x^{2}}-\sqrt {p^{2}+4}\right )}{32} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{4} + p x^{2} + 1}}{x^{4} + 1}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {-x^4+p\,x^2+1}}{x^4+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {p x^{2} - x^{4} + 1}}{x^{4} + 1}\, dx \]

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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