\(\int \frac {-1+2 x^2+2 x^4}{(1+2 x^2) \sqrt {-1+x^6}} \, dx\) [119]

Optimal result

Integrand size = 31, antiderivative size = 17 \[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\text {arctanh}\left (\frac {x \left (-1+x^2\right )}{\sqrt {-1+x^6}}\right ) \]

[Out]

arctanh(x*(x^2-1)/(x^6-1)^(1/2))

Rubi [F]

\[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx \]

[In]

Int[(-1 + 2*x^2 + 2*x^4)/((1 + 2*x^2)*Sqrt[-1 + x^6]),x]

[Out]

ArcTanh[x^3/Sqrt[-1 + x^6]]/3 + (x*(1 - x^2)*Sqrt[(1 + x^2 + x^4)/(1 - (1 + Sqrt[3])*x^2)^2]*EllipticF[ArcCos[
(1 - (1 - Sqrt[3])*x^2)/(1 - (1 + Sqrt[3])*x^2)], (2 + Sqrt[3])/4])/(4*3^(1/4)*Sqrt[-((x^2*(1 - x^2))/(1 - (1
+ Sqrt[3])*x^2)^2)]*Sqrt[-1 + x^6]) - ((3*I)/4)*Defer[Int][1/((I - Sqrt[2]*x)*Sqrt[-1 + x^6]), x] - ((3*I)/4)*
Defer[Int][1/((I + Sqrt[2]*x)*Sqrt[-1 + x^6]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2 \sqrt {-1+x^6}}+\frac {x^2}{\sqrt {-1+x^6}}-\frac {3}{2 \left (1+2 x^2\right ) \sqrt {-1+x^6}}\right ) \, dx \\ & = \frac {1}{2} \int \frac {1}{\sqrt {-1+x^6}} \, dx-\frac {3}{2} \int \frac {1}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx+\int \frac {x^2}{\sqrt {-1+x^6}} \, dx \\ & = \frac {x \left (1-x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x^2}{1-\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {-\frac {x^2 \left (1-x^2\right )}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {-1+x^6}}+\frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,x^3\right )-\frac {3}{2} \int \left (\frac {i}{2 \left (i-\sqrt {2} x\right ) \sqrt {-1+x^6}}+\frac {i}{2 \left (i+\sqrt {2} x\right ) \sqrt {-1+x^6}}\right ) \, dx \\ & = \frac {x \left (1-x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x^2}{1-\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {-\frac {x^2 \left (1-x^2\right )}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {-1+x^6}}-\frac {3}{4} i \int \frac {1}{\left (i-\sqrt {2} x\right ) \sqrt {-1+x^6}} \, dx-\frac {3}{4} i \int \frac {1}{\left (i+\sqrt {2} x\right ) \sqrt {-1+x^6}} \, dx+\frac {1}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right ) \\ & = \frac {1}{3} \text {arctanh}\left (\frac {x^3}{\sqrt {-1+x^6}}\right )+\frac {x \left (1-x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1-\left (1-\sqrt {3}\right ) x^2}{1-\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 \sqrt [4]{3} \sqrt {-\frac {x^2 \left (1-x^2\right )}{\left (1-\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {-1+x^6}}-\frac {3}{4} i \int \frac {1}{\left (i-\sqrt {2} x\right ) \sqrt {-1+x^6}} \, dx-\frac {3}{4} i \int \frac {1}{\left (i+\sqrt {2} x\right ) \sqrt {-1+x^6}} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 7.35 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\text {arctanh}\left (\frac {x \left (-1+x^2\right )}{\sqrt {-1+x^6}}\right ) \]

[In]

Integrate[(-1 + 2*x^2 + 2*x^4)/((1 + 2*x^2)*Sqrt[-1 + x^6]),x]

[Out]

ArcTanh[(x*(-1 + x^2))/Sqrt[-1 + x^6]]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(31\) vs. \(2(15)=30\).

Time = 1.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.88

method	result	size
trager	\(-\frac {\ln \left (-\frac {-2 x^{4}+2 x \sqrt {x^{6}-1}-1}{2 x^{2}+1}\right )}{2}\)	\(32\)

[In]

int((2*x^4+2*x^2-1)/(2*x^2+1)/(x^6-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(-(-2*x^4+2*x*(x^6-1)^(1/2)-1)/(2*x^2+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 4.24 \[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\frac {1}{3} \, \log \left (x^{3} + \sqrt {x^{6} - 1}\right ) + \frac {1}{6} \, \log \left (-\frac {10 \, x^{6} + 6 \, x^{4} + 12 \, x^{2} + 6 \, \sqrt {x^{6} - 1} {\left (x^{3} - x\right )} - 1}{8 \, x^{6} + 12 \, x^{4} + 6 \, x^{2} + 1}\right ) \]

[In]

integrate((2*x^4+2*x^2-1)/(2*x^2+1)/(x^6-1)^(1/2),x, algorithm="fricas")

[Out]

1/3*log(x^3 + sqrt(x^6 - 1)) + 1/6*log(-(10*x^6 + 6*x^4 + 12*x^2 + 6*sqrt(x^6 - 1)*(x^3 - x) - 1)/(8*x^6 + 12*
x^4 + 6*x^2 + 1))

Sympy [F]

\[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\int \frac {2 x^{4} + 2 x^{2} - 1}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{2} + 1\right )}\, dx \]

[In]

integrate((2*x**4+2*x**2-1)/(2*x**2+1)/(x**6-1)**(1/2),x)

[Out]

Integral((2*x**4 + 2*x**2 - 1)/(sqrt((x - 1)*(x + 1)*(x**2 - x + 1)*(x**2 + x + 1))*(2*x**2 + 1)), x)

Maxima [F]

\[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\int { \frac {2 \, x^{4} + 2 \, x^{2} - 1}{\sqrt {x^{6} - 1} {\left (2 \, x^{2} + 1\right )}} \,d x } \]

[In]

integrate((2*x^4+2*x^2-1)/(2*x^2+1)/(x^6-1)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*x^4 + 2*x^2 - 1)/(sqrt(x^6 - 1)*(2*x^2 + 1)), x)

Giac [F]

\[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\int { \frac {2 \, x^{4} + 2 \, x^{2} - 1}{\sqrt {x^{6} - 1} {\left (2 \, x^{2} + 1\right )}} \,d x } \]

[In]

integrate((2*x^4+2*x^2-1)/(2*x^2+1)/(x^6-1)^(1/2),x, algorithm="giac")

[Out]

integrate((2*x^4 + 2*x^2 - 1)/(sqrt(x^6 - 1)*(2*x^2 + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx=\int \frac {2\,x^4+2\,x^2-1}{\sqrt {x^6-1}\,\left (2\,x^2+1\right )} \,d x \]

[In]

int((2*x^2 + 2*x^4 - 1)/((x^6 - 1)^(1/2)*(2*x^2 + 1)),x)

[Out]

int((2*x^2 + 2*x^4 - 1)/((x^6 - 1)^(1/2)*(2*x^2 + 1)), x)