\(\int \sqrt {1+x^4} \, dx\) [809]

Optimal result

Integrand size = 9, antiderivative size = 58 \[ \int \sqrt {1+x^4} \, dx=\frac {1}{3} x \sqrt {1+x^4}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{3 \sqrt {1+x^4}} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {201, 226} \[ \int \sqrt {1+x^4} \, dx=\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{3 \sqrt {x^4+1}}+\frac {1}{3} \sqrt {x^4+1} x \]

[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 201

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \sqrt {1+x^4} \, dx=\frac {x+x^5-2 \sqrt [4]{-1} \sqrt {1+x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} x\right ),-1\right )}{3 \sqrt {1+x^4}} \]

[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])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.24

method	result	size
meijerg	\(x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-x^{4}\right )\)	\(14\)
default	\(\frac {x \sqrt {x^{4}+1}}{3}+\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)
risch	\(\frac {x \sqrt {x^{4}+1}}{3}+\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)
elliptic	\(\frac {x \sqrt {x^{4}+1}}{3}+\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, F\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{3 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\)	\(72\)

[In]

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

[Out]

x*hypergeom([-1/2,1/4],[5/4],-x^4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.45 \[ \int \sqrt {1+x^4} \, dx=\frac {1}{3} \, \sqrt {x^{4} + 1} x + \frac {2}{3} i \, \sqrt {i} F(\arcsin \left (\frac {\sqrt {i}}{x}\right )\,|\,-1) \]

[In]

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

[Out]

1/3*sqrt(x^4 + 1)*x + 2/3*I*sqrt(I)*elliptic_f(arcsin(sqrt(I)/x), -1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.50 \[ \int \sqrt {1+x^4} \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[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))

Maxima [F]

\[ \int \sqrt {1+x^4} \, dx=\int { \sqrt {x^{4} + 1} \,d x } \]

[In]

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

[Out]

integrate(sqrt(x^4 + 1), x)

Giac [F]

\[ \int \sqrt {1+x^4} \, dx=\int { \sqrt {x^{4} + 1} \,d x } \]

[In]

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

[Out]

integrate(sqrt(x^4 + 1), x)

Mupad [B] (verification not implemented)

Time = 5.44 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.21 \[ \int \sqrt {1+x^4} \, dx=x\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{4};\ \frac {5}{4};\ -x^4\right ) \]

[In]

int((x^4 + 1)^(1/2),x)

[Out]

x*hypergeom([-1/2, 1/4], 5/4, -x^4)