\(\int \frac {(-1+3 x^4) \sqrt {1+x^2+2 x^4+x^8}}{(1-x+x^4)^2 (1+x+x^4)} \, dx\) [954]

Optimal result

Integrand size = 43, antiderivative size = 72 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=-\frac {\sqrt {1+x^2+2 x^4+x^8}}{2 \left (1-x+x^4\right )}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{1+x+x^4+\sqrt {1+x^2+2 x^4+x^8}}\right )}{\sqrt {2}} \]

[Out]

-(x^8+2*x^4+x^2+1)^(1/2)/(2*x^4-2*x+2)-1/2*arctanh(2^(1/2)*x/(1+x+x^4+(x^8+2*x^4+x^2+1)^(1/2)))*2^(1/2)

Rubi [F]

\[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (-1+4 x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{2 \left (1-x+x^4\right )^2}+\frac {\left (-1-4 x^2+x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{4 \left (1-x+x^4\right )}+\frac {\left (-1+4 x^2-x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{4 \left (1+x+x^4\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {\left (-1-4 x^2+x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{1-x+x^4} \, dx+\frac {1}{4} \int \frac {\left (-1+4 x^2-x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{1+x+x^4} \, dx+\frac {1}{2} \int \frac {\left (-1+4 x^3\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2} \, dx \\ & = \frac {1}{4} \int \left (\frac {\sqrt {1+x^2+2 x^4+x^8}}{-1+x-x^4}-\frac {4 x^2 \sqrt {1+x^2+2 x^4+x^8}}{1-x+x^4}+\frac {x^3 \sqrt {1+x^2+2 x^4+x^8}}{1-x+x^4}\right ) \, dx+\frac {1}{4} \int \left (\frac {\sqrt {1+x^2+2 x^4+x^8}}{-1-x-x^4}+\frac {4 x^2 \sqrt {1+x^2+2 x^4+x^8}}{1+x+x^4}-\frac {x^3 \sqrt {1+x^2+2 x^4+x^8}}{1+x+x^4}\right ) \, dx+\frac {1}{2} \int \left (-\frac {\sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2}+\frac {4 x^3 \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2}\right ) \, dx \\ & = \frac {1}{4} \int \frac {\sqrt {1+x^2+2 x^4+x^8}}{-1-x-x^4} \, dx+\frac {1}{4} \int \frac {\sqrt {1+x^2+2 x^4+x^8}}{-1+x-x^4} \, dx+\frac {1}{4} \int \frac {x^3 \sqrt {1+x^2+2 x^4+x^8}}{1-x+x^4} \, dx-\frac {1}{4} \int \frac {x^3 \sqrt {1+x^2+2 x^4+x^8}}{1+x+x^4} \, dx-\frac {1}{2} \int \frac {\sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2} \, dx+2 \int \frac {x^3 \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2} \, dx-\int \frac {x^2 \sqrt {1+x^2+2 x^4+x^8}}{1-x+x^4} \, dx+\int \frac {x^2 \sqrt {1+x^2+2 x^4+x^8}}{1+x+x^4} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=-\frac {\sqrt {1+x^2+2 x^4+x^8}}{2 \left (1-x+x^4\right )}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{1+x+x^4+\sqrt {1+x^2+2 x^4+x^8}}\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01

method	result	size
pseudoelliptic	\(\frac {\left (x^{4}-x +1\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x^{4}-x +1\right ) \sqrt {2}}{2 \sqrt {x^{8}+2 x^{4}+x^{2}+1}}\right )-2 \sqrt {x^{8}+2 x^{4}+x^{2}+1}}{4 x^{4}-4 x +4}\)	\(73\)
trager	\(-\frac {\sqrt {x^{8}+2 x^{4}+x^{2}+1}}{2 \left (x^{4}-x +1\right )}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )-2 \sqrt {x^{8}+2 x^{4}+x^{2}+1}}{x^{4}+x +1}\right )}{4}\)	\(90\)
risch	\(-\frac {\sqrt {x^{8}+2 x^{4}+x^{2}+1}}{2 \left (x^{4}-x +1\right )}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{4}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )+2 \sqrt {x^{8}+2 x^{4}+x^{2}+1}}{x^{4}+x +1}\right )}{4}\)	\(91\)

[In]

int((3*x^4-1)*(x^8+2*x^4+x^2+1)^(1/2)/(x^4-x+1)^2/(x^4+x+1),x,method=_RETURNVERBOSE)

[Out]

((x^4-x+1)*2^(1/2)*arctanh(1/2*(x^4-x+1)*2^(1/2)/(x^8+2*x^4+x^2+1)^(1/2))-2*(x^8+2*x^4+x^2+1)^(1/2))/(4*x^4-4*
x+4)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.67 \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\frac {\sqrt {2} {\left (x^{4} - x + 1\right )} \log \left (\frac {3 \, x^{8} - 2 \, x^{5} + 6 \, x^{4} + 2 \, \sqrt {2} \sqrt {x^{8} + 2 \, x^{4} + x^{2} + 1} {\left (x^{4} - x + 1\right )} + 3 \, x^{2} - 2 \, x + 3}{x^{8} + 2 \, x^{5} + 2 \, x^{4} + x^{2} + 2 \, x + 1}\right ) - 4 \, \sqrt {x^{8} + 2 \, x^{4} + x^{2} + 1}}{8 \, {\left (x^{4} - x + 1\right )}} \]

[In]

integrate((3*x^4-1)*(x^8+2*x^4+x^2+1)^(1/2)/(x^4-x+1)^2/(x^4+x+1),x, algorithm="fricas")

[Out]

1/8*(sqrt(2)*(x^4 - x + 1)*log((3*x^8 - 2*x^5 + 6*x^4 + 2*sqrt(2)*sqrt(x^8 + 2*x^4 + x^2 + 1)*(x^4 - x + 1) +
3*x^2 - 2*x + 3)/(x^8 + 2*x^5 + 2*x^4 + x^2 + 2*x + 1)) - 4*sqrt(x^8 + 2*x^4 + x^2 + 1))/(x^4 - x + 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate((3*x**4-1)*(x**8+2*x**4+x**2+1)**(1/2)/(x**4-x+1)**2/(x**4+x+1),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\int { \frac {\sqrt {x^{8} + 2 \, x^{4} + x^{2} + 1} {\left (3 \, x^{4} - 1\right )}}{{\left (x^{4} + x + 1\right )} {\left (x^{4} - x + 1\right )}^{2}} \,d x } \]

[In]

integrate((3*x^4-1)*(x^8+2*x^4+x^2+1)^(1/2)/(x^4-x+1)^2/(x^4+x+1),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\int { \frac {\sqrt {x^{8} + 2 \, x^{4} + x^{2} + 1} {\left (3 \, x^{4} - 1\right )}}{{\left (x^{4} + x + 1\right )} {\left (x^{4} - x + 1\right )}^{2}} \,d x } \]

[In]

integrate((3*x^4-1)*(x^8+2*x^4+x^2+1)^(1/2)/(x^4-x+1)^2/(x^4+x+1),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+3 x^4\right ) \sqrt {1+x^2+2 x^4+x^8}}{\left (1-x+x^4\right )^2 \left (1+x+x^4\right )} \, dx=\int \frac {\left (3\,x^4-1\right )\,\sqrt {x^8+2\,x^4+x^2+1}}{{\left (x^4-x+1\right )}^2\,\left (x^4+x+1\right )} \,d x \]

[In]

int(((3*x^4 - 1)*(x^2 + 2*x^4 + x^8 + 1)^(1/2))/((x^4 - x + 1)^2*(x + x^4 + 1)),x)

[Out]

int(((3*x^4 - 1)*(x^2 + 2*x^4 + x^8 + 1)^(1/2))/((x^4 - x + 1)^2*(x + x^4 + 1)), x)