3.37 \(\int \frac {1}{-1+5 x^4} \, dx\)

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)

________________________________________________________________________________________

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 = {212, 206, 203} \[ -\frac {\tan ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}}-\frac {\tanh ^{-1}\left (\sqrt [4]{5} x\right )}{2 \sqrt [4]{5}} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + 5*x^4)^(-1),x]

[Out]

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

Rule 203

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

Rule 206

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

Rule 212

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] &&
!GtQ[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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 43, normalized size = 1.23 \[ -\frac {-\log \left (1-\sqrt [4]{5} x\right )+\log \left (\sqrt [4]{5} x+1\right )+2 \tan ^{-1}\left (\sqrt [4]{5} x\right )}{4 \sqrt [4]{5}} \]

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)

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 58, normalized size = 1.66 \[ \frac {1}{5} \cdot 5^{\frac {3}{4}} \arctan \left (\frac {1}{5} \cdot 5^{\frac {3}{4}} \sqrt {5 \, x^{2} + \sqrt {5}} - 5^{\frac {1}{4}} x\right ) - \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (5 \, x + 5^{\frac {3}{4}}\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (5 \, x - 5^{\frac {3}{4}}\right ) \]

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

________________________________________________________________________________________

giac [A]  time = 1.08, size = 39, normalized size = 1.11 \[ -\frac {1}{10} \cdot 5^{\frac {3}{4}} \arctan \left (5 \, \left (\frac {1}{5}\right )^{\frac {3}{4}} x\right ) - \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left ({\left | x + \left (\frac {1}{5}\right )^{\frac {1}{4}} \right |}\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left ({\left | x - \left (\frac {1}{5}\right )^{\frac {1}{4}} \right |}\right ) \]

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)))

________________________________________________________________________________________

maple [A]  time = 0.00, size = 36, normalized size = 1.03 \[ -\frac {5^{\frac {3}{4}} \arctan \left (5^{\frac {1}{4}} x \right )}{10}-\frac {5^{\frac {3}{4}} \ln \left (\frac {x +\frac {5^{\frac {3}{4}}}{5}}{x -\frac {5^{\frac {3}{4}}}{5}}\right )}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.98, size = 41, normalized size = 1.17 \[ -\frac {1}{10} \cdot 5^{\frac {3}{4}} \arctan \left (5^{\frac {1}{4}} x\right ) + \frac {1}{20} \cdot 5^{\frac {3}{4}} \log \left (\frac {\sqrt {5} x - 5^{\frac {1}{4}}}{\sqrt {5} x + 5^{\frac {1}{4}}}\right ) \]

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)))

________________________________________________________________________________________

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

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

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 48, normalized size = 1.37 \[ \frac {5^{\frac {3}{4}} \log {\left (x - \frac {5^{\frac {3}{4}}}{5} \right )}}{20} - \frac {5^{\frac {3}{4}} \log {\left (x + \frac {5^{\frac {3}{4}}}{5} \right )}}{20} - \frac {5^{\frac {3}{4}} \operatorname {atan}{\left (\sqrt [4]{5} x \right )}}{10} \]

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

________________________________________________________________________________________