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

Optimal. Leaf size=26 \[ -\tan ^{-1}\left (\frac {x \sqrt {x^6+1}}{x^4-x^2+1}\right ) \]

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Rubi [F]  time = 0.89, 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^4\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^4)*Sqrt[1 + x^6]),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx &=\int \left (\frac {1}{\sqrt {1+x^6}}-\frac {2 \left (1+x^2\right )}{\left (1+2 x^4\right ) \sqrt {1+x^6}}\right ) \, dx\\ &=-\left (2 \int \frac {1+x^2}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx\right )+\int \frac {1}{\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 )}{2 \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}}-2 \int \left (\frac {i \left (i+\sqrt {2}\right )}{2 \sqrt {2} \left (i-\sqrt {2} x^2\right ) \sqrt {1+x^6}}-\frac {i \left (i-\sqrt {2}\right )}{2 \sqrt {2} \left (i+\sqrt {2} x^2\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 )}{2 \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}{2} \left (2 i-\sqrt {2}\right ) \int \frac {1}{\left (i-\sqrt {2} x^2\right ) \sqrt {1+x^6}} \, dx-\frac {1}{2} \left (2 i+\sqrt {2}\right ) \int \frac {1}{\left (i+\sqrt {2} x^2\right ) \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 )}{2 \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}{2} \left (2 i-\sqrt {2}\right ) \int \left (-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}}\right ) \, dx-\frac {1}{2} \left (2 i+\sqrt {2}\right ) \int \left (-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}+\sqrt [4]{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 )}{2 \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}}+\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+i \sqrt {2}\right )\right ) \int \frac {1}{\left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx+\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+i \sqrt {2}\right )\right ) \int \frac {1}{\left (-(-1)^{3/4}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i+\sqrt {2}\right )\right ) \int \frac {1}{\left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i+\sqrt {2}\right )\right ) \int \frac {1}{\left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\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^4)*Sqrt[1 + x^6]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 12.13, size = 26, normalized size = 1.00 \begin {gather*} -\tan ^{-1}\left (\frac {x \sqrt {1+x^6}}{1-x^2+x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {x^{6} + 1} x}{2 \, 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^4+1)/(x^6+1)^(1/2),x, algorithm="fricas")

[Out]

1/2*arctan(2*sqrt(x^6 + 1)*x/(2*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^{4} + 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^4+1)/(x^6+1)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.29, size = 51, normalized size = 1.96

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \sqrt {x^{6}+1}\, x -\RootOf \left (\textit {\_Z}^{2}+1\right )}{2 x^{4}+1}\right )}{2}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*RootOf(_Z^2+1)*ln(-(2*RootOf(_Z^2+1)*x^2-2*(x^6+1)^(1/2)*x-RootOf(_Z^2+1))/(2*x^4+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^{4} + 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^4+1)/(x^6+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {-2\,x^4+2\,x^2+1}{\sqrt {x^6+1}\,\left (2\,x^4+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^4 + 1)),x)

[Out]

int(-(2*x^2 - 2*x^4 + 1)/((x^6 + 1)^(1/2)*(2*x^4 + 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^{2} + 1\right ) \left (x^{4} - x^{2} + 1\right )} \left (2 x^{4} + 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**4+1)/(x**6+1)**(1/2),x)

[Out]

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

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