3.809 \(\int \sqrt {1+x^4} \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{3} \sqrt {x^4+1} x+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+1}} \]

[Out]

1/3*x*(x^4+1)^(1/2)+1/3*(x^2+1)*(cos(2*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2*2^(
1/2))*((x^4+1)/(x^2+1)^2)^(1/2)/(x^4+1)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {195, 220} \[ \frac {1}{3} \sqrt {x^4+1} x+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^4],x]

[Out]

(x*Sqrt[1 + x^4])/3 + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(3*Sqrt[1 + x^4])

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin {align*} \int \sqrt {1+x^4} \, dx &=\frac {1}{3} x \sqrt {1+x^4}+\frac {2}{3} \int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=\frac {1}{3} x \sqrt {1+x^4}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{3 \sqrt {1+x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 48, normalized size = 0.83 \[ \frac {x^5-2 \sqrt [4]{-1} \sqrt {x^4+1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+x}{3 \sqrt {x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^4],x]

[Out]

(x + x^5 - 2*(-1)^(1/4)*Sqrt[1 + x^4]*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1])/(3*Sqrt[1 + x^4])

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {x^{4} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(x^4 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 1), x)

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maple [C]  time = 0.01, size = 72, normalized size = 1.24 \[ \frac {\sqrt {x^{4}+1}\, x}{3}+\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) x , i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^4+1)^(1/2)*x+2/3/(1/2*2^(1/2)+1/2*I*2^(1/2))*(-I*x^2+1)^(1/2)*(I*x^2+1)^(1/2)/(x^4+1)^(1/2)*EllipticF((
1/2*2^(1/2)+1/2*I*2^(1/2))*x,I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x^{4} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 1), x)

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mupad [B]  time = 0.04, size = 12, normalized size = 0.21 \[ x\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{4};\ \frac {5}{4};\ -x^4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)^(1/2),x)

[Out]

x*hypergeom([-1/2, 1/4], 5/4, -x^4)

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sympy [C]  time = 1.43, size = 29, normalized size = 0.50 \[ \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*exp_polar(I*pi))/(4*gamma(5/4))

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