3.1.36 \(\int \frac {1}{-2+x^4} \, dx\) [36]

Optimal. Leaf size=35 \[ -\frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}} \]

[Out]

-1/4*arctan(1/2*x*2^(3/4))*2^(1/4)-1/4*arctanh(1/2*x*2^(3/4))*2^(1/4)

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Rubi [A]
time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {x}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )}{2\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + x^4)^(-1),x]

[Out]

-1/2*ArcTan[x/2^(1/4)]/2^(3/4) - ArcTanh[x/2^(1/4)]/(2*2^(3/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 43, normalized size = 1.23 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {x}{\sqrt [4]{2}}\right )-\log \left (2-2^{3/4} x\right )+\log \left (2+2^{3/4} x\right )}{4\ 2^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x^4)^(-1),x]

[Out]

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

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Maple [A]
time = 0.03, size = 32, normalized size = 0.91

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}-2\right )}{\sum }\frac {\ln \left (-\textit {\_R} +x \right )}{\textit {\_R}^{3}}\right )}{4}\)	\(22\)
default	\(-\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {x +2^{\frac {1}{4}}}{x -2^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x 2^{\frac {3}{4}}}{2}\right )\right )}{8}\)	\(32\)
meijerg	\(\frac {2^{\frac {1}{4}} x \left (\ln \left (1-\frac {2^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{2}\right )-\ln \left (1+\frac {2^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{2}\right )-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}\right )^{\frac {1}{4}}}{2}\right )\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 5.05, size = 34, normalized size = 0.97 \begin {gather*} -\frac {1}{4} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} x\right ) + \frac {1}{8} \cdot 2^{\frac {1}{4}} \log \left (\frac {x - 2^{\frac {1}{4}}}{x + 2^{\frac {1}{4}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
time = 0.71, size = 56, normalized size = 1.60 \begin {gather*} \frac {1}{8} \cdot 8^{\frac {3}{4}} \arctan \left (-\frac {1}{2} \cdot 8^{\frac {1}{4}} x + \frac {1}{2} \cdot 8^{\frac {1}{4}} \sqrt {x^{2} + \sqrt {2}}\right ) - \frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (4 \, x + 8^{\frac {3}{4}}\right ) + \frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (4 \, x - 8^{\frac {3}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*8^(3/4)*arctan(-1/2*8^(1/4)*x + 1/2*8^(1/4)*sqrt(x^2 + sqrt(2))) - 1/32*8^(3/4)*log(4*x + 8^(3/4)) + 1/32*
8^(3/4)*log(4*x - 8^(3/4))

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Sympy [A]
time = 0.15, size = 46, normalized size = 1.31 \begin {gather*} \frac {\sqrt [4]{2} \log {\left (x - \sqrt [4]{2} \right )}}{8} - \frac {\sqrt [4]{2} \log {\left (x + \sqrt [4]{2} \right )}}{8} - \frac {\sqrt [4]{2} \operatorname {atan}{\left (\frac {2^{\frac {3}{4}} x}{2} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2**(1/4)*log(x - 2**(1/4))/8 - 2**(1/4)*log(x + 2**(1/4))/8 - 2**(1/4)*atan(2**(3/4)*x/2)/4

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Giac [A]
time = 0.45, size = 39, normalized size = 1.11 \begin {gather*} -\frac {1}{4} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} x\right ) - \frac {1}{8} \cdot 2^{\frac {1}{4}} \log \left ({\left | x + 2^{\frac {1}{4}} \right |}\right ) + \frac {1}{8} \cdot 2^{\frac {1}{4}} \log \left ({\left | x - 2^{\frac {1}{4}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.16, size = 20, normalized size = 0.57 \begin {gather*} -\frac {2^{1/4}\,\left (\mathrm {atan}\left (\frac {2^{3/4}\,x}{2}\right )+\mathrm {atanh}\left (\frac {2^{3/4}\,x}{2}\right )\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4 - 2),x)

[Out]

-(2^(1/4)*(atan((2^(3/4)*x)/2) + atanh((2^(3/4)*x)/2)))/4

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