3.1.37 \(\int \frac {1}{-1+5 x^4} \, dx\) [37]

Optimal. Leaf size=35 \[ -\frac {\tan ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}}-\frac {\tanh ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}} \]

[Out]

-1/10*arctan(5^(1/4)*x)*5^(3/4)-1/10*arctanh(5^(1/4)*x)*5^(3/4)

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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 = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}}-\frac {\tanh ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + 5*x^4)^(-1),x]

[Out]

-1/2*ArcTan[5^(1/4)*x]/5^(1/4) - ArcTanh[5^(1/4)*x]/(2*5^(1/4))

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}{-1+5 x^4} \, dx &=-\left (\frac {1}{2} \int \frac {1}{1-\sqrt {5} x^2} \, dx\right )-\frac {1}{2} \int \frac {1}{1+\sqrt {5} x^2} \, dx\\ &=-\frac {\tan ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}}-\frac {\tanh ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 1.23 \begin {gather*} -\frac {2 \tan ^{-1}\left (\sqrt [4]{5} x\right )-\log \left (1-\sqrt [4]{5} x\right )+\log \left (1+\sqrt [4]{5} x\right )}{4 \sqrt [4]{5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 5*x^4)^(-1),x]

[Out]

-1/4*(2*ArcTan[5^(1/4)*x] - Log[1 - 5^(1/4)*x] + Log[1 + 5^(1/4)*x])/5^(1/4)

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Maple [A]
time = 0.03, size = 33, normalized size = 0.94

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\RootOf \left (5 \textit {\_Z}^{4}-1\right )}{\sum }\frac {\ln \left (-\textit {\_R} +x \right )}{\textit {\_R}^{3}}\right )}{20}\)	\(24\)
default	\(-\frac {5^{\frac {3}{4}} \left (\ln \left (\frac {x +\frac {5^{\frac {3}{4}}}{5}}{x -\frac {5^{\frac {3}{4}}}{5}}\right )+2 \arctan \left (5^{\frac {1}{4}} x \right )\right )}{20}\)	\(33\)
meijerg	\(\frac {5^{\frac {3}{4}} x \left (\ln \left (1-5^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}\right )-\ln \left (1+5^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}\right )-2 \arctan \left (5^{\frac {1}{4}} \left (x^{4}\right )^{\frac {1}{4}}\right )\right )}{20 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^4-1),x,method=_RETURNVERBOSE)

[Out]

-1/20*5^(3/4)*(ln((x+1/5*5^(3/4))/(x-1/5*5^(3/4)))+2*arctan(5^(1/4)*x))

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Maxima [A]
time = 4.55, size = 41, normalized size = 1.17 \begin {gather*} -\frac {1}{10} \cdot 5^{\frac {3}{4}} \arctan \left (5^{\frac {1}{4}} x\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (\frac {\sqrt {5} x - 5^{\frac {1}{4}}}{\sqrt {5} x + 5^{\frac {1}{4}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^4-1),x, algorithm="maxima")

[Out]

-1/10*5^(3/4)*arctan(5^(1/4)*x) + 1/20*5^(3/4)*log((sqrt(5)*x - 5^(1/4))/(sqrt(5)*x + 5^(1/4)))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (23) = 46\).
time = 0.95, size = 60, normalized size = 1.71 \begin {gather*} \frac {1}{5} \cdot 5^{\frac {3}{4}} \arctan \left (-5^{\frac {1}{4}} x + \frac {1}{5} \cdot 5^{\frac {1}{4}} \sqrt {25 \, x^{2} + 5 \, \sqrt {5}}\right ) - \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (5 \, x + 5^{\frac {3}{4}}\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (5 \, x - 5^{\frac {3}{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^4-1),x, algorithm="fricas")

[Out]

1/5*5^(3/4)*arctan(-5^(1/4)*x + 1/5*5^(1/4)*sqrt(25*x^2 + 5*sqrt(5))) - 1/20*5^(3/4)*log(5*x + 5^(3/4)) + 1/20
*5^(3/4)*log(5*x - 5^(3/4))

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Sympy [A]
time = 0.15, size = 48, normalized size = 1.37 \begin {gather*} \frac {5^{\frac {3}{4}} \log {\left (x - \frac {5^{\frac {3}{4}}}{5} \right )}}{20} - \frac {5^{\frac {3}{4}} \log {\left (x + \frac {5^{\frac {3}{4}}}{5} \right )}}{20} - \frac {5^{\frac {3}{4}} \operatorname {atan}{\left (\sqrt [4]{5} x \right )}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x**4-1),x)

[Out]

5**(3/4)*log(x - 5**(3/4)/5)/20 - 5**(3/4)*log(x + 5**(3/4)/5)/20 - 5**(3/4)*atan(5**(1/4)*x)/10

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Giac [A]
time = 0.45, size = 39, normalized size = 1.11 \begin {gather*} -\frac {1}{10} \cdot 5^{\frac {3}{4}} \arctan \left (5 \, \left (\frac {1}{5}\right )^{\frac {3}{4}} x\right ) - \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left ({\left | x + \left (\frac {1}{5}\right )^{\frac {1}{4}} \right |}\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left ({\left | x - \left (\frac {1}{5}\right )^{\frac {1}{4}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(5*x^4-1),x, algorithm="giac")

[Out]

-1/10*5^(3/4)*arctan(5*(1/5)^(3/4)*x) - 1/20*5^(3/4)*log(abs(x + (1/5)^(1/4))) + 1/20*5^(3/4)*log(abs(x - (1/5
)^(1/4)))

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Mupad [B]
time = 0.18, size = 18, normalized size = 0.51 \begin {gather*} -\frac {5^{3/4}\,\left (\mathrm {atan}\left (5^{1/4}\,x\right )+\mathrm {atanh}\left (5^{1/4}\,x\right )\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*x^4 - 1),x)

[Out]

-(5^(3/4)*(atan(5^(1/4)*x) + atanh(5^(1/4)*x)))/10

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