\(\int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx\) [929]

Optimal result

Integrand size = 17, antiderivative size = 70 \[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {x}{2 \sqrt {x^2+\sqrt {1+x^4}}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )}{\sqrt {2}} \]

[Out]

1/2*x/(x^2+(x^4+1)^(1/2))^(1/2)+1/2*arctanh(2^(1/2)*x*(x^2+(x^4+1)^(1/2))^(1/2)/(1+x^2+(x^4+1)^(1/2)))*2^(1/2)

Rubi [F]

\[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx \]

[In]

Int[1/Sqrt[x^2 + Sqrt[1 + x^4]],x]

[Out]

Defer[Int][1/Sqrt[x^2 + Sqrt[1 + x^4]], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {x}{2 \sqrt {x^2+\sqrt {1+x^4}}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )}{\sqrt {2}} \]

[In]

Integrate[1/Sqrt[x^2 + Sqrt[1 + x^4]],x]

[Out]

x/(2*Sqrt[x^2 + Sqrt[1 + x^4]]) + ArcTanh[(Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^4]])/(1 + x^2 + Sqrt[1 + x^4])]/Sqr
t[2]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.73

method	result	size
meijerg	\(-\frac {-\frac {\sqrt {\pi }\, \sqrt {2}\, \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}, \frac {7}{4}\right ], \left [2, 2, \frac {5}{2}\right ], -\frac {1}{x^{4}}\right )}{4 x^{4}}+2 \left (-4 \ln \left (2\right )-2-4 \ln \left (x \right )\right ) \sqrt {\pi }\, \sqrt {2}}{16 \sqrt {\pi }}\)	\(51\)

[In]

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

[Out]

-1/16/Pi^(1/2)*(-1/4*Pi^(1/2)*2^(1/2)/x^4*hypergeom([1,1,5/4,7/4],[2,2,5/2],-1/x^4)+2*(-4*ln(2)-2-4*ln(x))*Pi^
(1/2)*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.29 \[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=-\frac {1}{2} \, {\left (x^{3} - \sqrt {x^{4} + 1} x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + \frac {1}{8} \, \sqrt {2} \log \left (4 \, x^{4} + 4 \, \sqrt {x^{4} + 1} x^{2} + 2 \, {\left (\sqrt {2} x^{3} + \sqrt {2} \sqrt {x^{4} + 1} x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + 1\right ) \]

[In]

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

[Out]

-1/2*(x^3 - sqrt(x^4 + 1)*x)*sqrt(x^2 + sqrt(x^4 + 1)) + 1/8*sqrt(2)*log(4*x^4 + 4*sqrt(x^4 + 1)*x^2 + 2*(sqrt
(2)*x^3 + sqrt(2)*sqrt(x^4 + 1)*x)*sqrt(x^2 + sqrt(x^4 + 1)) + 1)

Sympy [A] (verification not implemented)

Time = 1.75 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} 1, 1 & \frac {3}{2} \\\frac {1}{4}, \frac {3}{4} & 0 \end {matrix} \middle | {x^{4}} \right )}}{16 \sqrt {\pi }} \]

[In]

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

[Out]

meijerg(((1, 1), (3/2,)), ((1/4, 3/4), (0,)), x**4)/(16*sqrt(pi))

Maxima [F]

\[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int { \frac {1}{\sqrt {x^{2} + \sqrt {x^{4} + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int { \frac {1}{\sqrt {x^{2} + \sqrt {x^{4} + 1}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\int \frac {1}{\sqrt {\sqrt {x^4+1}+x^2}} \,d x \]

[In]

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

[Out]

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