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

Optimal. Leaf size=17 \[ \tanh ^{-1}\left (\frac {x \left (x^2-1\right )}{\sqrt {x^6-1}}\right ) \]

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Rubi [F]  time = 0.39, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx \end {gather*}

Verification is not applicable to the result.

[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*} \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx &=\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}} F\left (\cos ^{-1}\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} \operatorname {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}} F\left (\cos ^{-1}\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} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^3}{\sqrt {-1+x^6}}\right )\\ &=\frac {1}{3} \tanh ^{-1}\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}} F\left (\cos ^{-1}\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*}

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Mathematica [F]  time = 0.14, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+2 x^2+2 x^4}{\left (1+2 x^2\right ) \sqrt {-1+x^6}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 8.53, size = 17, normalized size = 1.00 \begin {gather*} \tanh ^{-1}\left (\frac {x \left (-1+x^2\right )}{\sqrt {-1+x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-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]]

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fricas [B]  time = 0.53, size = 72, normalized size = 4.24 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} + 2 \, x^{2} - 1}{\sqrt {x^{6} - 1} {\left (2 \, x^{2} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.24, size = 32, normalized size = 1.88

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{4} + 2 \, x^{2} - 1}{\sqrt {x^{6} - 1} {\left (2 \, x^{2} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {2\,x^4+2\,x^2-1}{\sqrt {x^6-1}\,\left (2\,x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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