\(\int \frac {x^2}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx\) [1058]

Optimal result

Integrand size = 21, antiderivative size = 79 \[ \int \frac {x^2}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=-\frac {x}{8 \left (x^2+\sqrt {1+x^4}\right )^{3/2}}+\frac {1}{4} x \sqrt {x^2+\sqrt {1+x^4}}-\frac {\arctan \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}\right )}{8 \sqrt {2}} \]

[Out]

-1/8*x/(x^2+(x^4+1)^(1/2))^(3/2)+1/4*x*(x^2+(x^4+1)^(1/2))^(1/2)-1/16*arctan(2^(1/2)*x*(x^2+(x^4+1)^(1/2))^(1/
2))*2^(1/2)

Rubi [F]

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

[In]

Int[x^2/Sqrt[x^2 + Sqrt[1 + x^4]],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=-\frac {x}{8 \left (x^2+\sqrt {1+x^4}\right )^{3/2}}+\frac {1}{4} x \sqrt {x^2+\sqrt {1+x^4}}-\frac {\arctan \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}\right )}{8 \sqrt {2}} \]

[In]

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

[Out]

-1/8*x/(x^2 + Sqrt[1 + x^4])^(3/2) + (x*Sqrt[x^2 + Sqrt[1 + x^4]])/4 - ArcTan[Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^
4]]]/(8*Sqrt[2])

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28

method	result	size
meijerg	\(\frac {\sqrt {2}\, x^{2} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {3}{2}\right ], -\frac {1}{x^{4}}\right )}{4}\)	\(22\)

[In]

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

[Out]

1/4*2^(1/2)*x^2*hypergeom([-1/2,1/4,3/4],[1/2,3/2],-1/x^4)

Fricas [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03 \[ \int \frac {x^2}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=-\frac {1}{8} \, {\left (2 \, x^{5} - 2 \, \sqrt {x^{4} + 1} x^{3} - x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {{\left (\sqrt {2} x^{2} - \sqrt {2} \sqrt {x^{4} + 1}\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}}}{2 \, x}\right ) \]

[In]

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

[Out]

-1/8*(2*x^5 - 2*sqrt(x^4 + 1)*x^3 - x)*sqrt(x^2 + sqrt(x^4 + 1)) + 1/16*sqrt(2)*arctan(-1/2*(sqrt(2)*x^2 - sqr
t(2)*sqrt(x^4 + 1))*sqrt(x^2 + sqrt(x^4 + 1))/x)

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.19 \[ \int \frac {x^2}{\sqrt {x^2+\sqrt {1+x^4}}} \, dx=\frac {{G_{3, 3}^{2, 2}\left (\begin {matrix} \frac {3}{2}, 1 & 2 \\\frac {3}{4}, \frac {5}{4} & 0 \end {matrix} \middle | {x^{4}} \right )}}{16 \sqrt {\pi }} \]

[In]

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

[Out]

meijerg(((3/2, 1), (2,)), ((3/4, 5/4), (0,)), x**4)/(16*sqrt(pi))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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