3.1.30 \(\int \frac {1-x^4}{1-6 x^4+x^8} \, dx\)

Optimal. Leaf size=125 \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}} \]

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Rubi [A]  time = 0.07, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1419, 1093, 207, 203} \begin {gather*} \frac {\tan ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{4 \sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^4)/(1 - 6*x^4 + x^8),x]

[Out]

ArcTan[x/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[2*(-1 + Sqrt[2])]) + ArcTan[x/Sqrt[1 + Sqrt[2]]]/(4*Sqrt[2*(1 + Sqrt[2])]
) + ArcTanh[x/Sqrt[-1 + Sqrt[2]]]/(4*Sqrt[2*(-1 + Sqrt[2])]) + ArcTanh[x/Sqrt[1 + Sqrt[2]]]/(4*Sqrt[2*(1 + Sqr
t[2])])

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 207

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

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/
2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 1419

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[(2*d)/e -
b/c, 2]}, Dist[e/(2*c), Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x^(n/2
) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2,
0] && IGtQ[n/2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !LtQ[(2*d)/e - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))

Rubi steps

\begin {align*} \int \frac {1-x^4}{1-6 x^4+x^8} \, dx &=-\left (\frac {1}{2} \int \frac {1}{-1-2 x^2+x^4} \, dx\right )-\frac {1}{2} \int \frac {1}{-1+2 x^2+x^4} \, dx\\ &=-\frac {\int \frac {1}{-1-\sqrt {2}+x^2} \, dx}{4 \sqrt {2}}-\frac {\int \frac {1}{1-\sqrt {2}+x^2} \, dx}{4 \sqrt {2}}+\frac {\int \frac {1}{-1+\sqrt {2}+x^2} \, dx}{4 \sqrt {2}}+\frac {\int \frac {1}{1+\sqrt {2}+x^2} \, dx}{4 \sqrt {2}}\\ &=\frac {\tan ^{-1}\left (\frac {x}{\sqrt {-1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (-1+\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {-1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (-1+\sqrt {2}\right )}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2 \left (1+\sqrt {2}\right )}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 114, normalized size = 0.91 \begin {gather*} \frac {\sqrt {1+\sqrt {2}} \tan ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )+\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )+\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )+\sqrt {\sqrt {2}-1} \tanh ^{-1}\left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x^4)/(1 - 6*x^4 + x^8),x]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[x/Sqrt[-1 + Sqrt[2]]] + Sqrt[-1 + Sqrt[2]]*ArcTan[x/Sqrt[1 + Sqrt[2]]] + Sqrt[1 + Sq
rt[2]]*ArcTanh[x/Sqrt[-1 + Sqrt[2]]] + Sqrt[-1 + Sqrt[2]]*ArcTanh[x/Sqrt[1 + Sqrt[2]]])/(4*Sqrt[2])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-x^4}{1-6 x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(1 - x^4)/(1 - 6*x^4 + x^8),x]

[Out]

IntegrateAlgebraic[(1 - x^4)/(1 - 6*x^4 + x^8), x]

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fricas [B]  time = 1.40, size = 199, normalized size = 1.59 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {2} + 1} \arctan \left (-x \sqrt {\sqrt {2} + 1} + \sqrt {x^{2} + \sqrt {2} - 1} \sqrt {\sqrt {2} + 1}\right ) - \frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {2} - 1} \arctan \left (-x \sqrt {\sqrt {2} - 1} + \sqrt {x^{2} + \sqrt {2} + 1} \sqrt {\sqrt {2} - 1}\right ) + \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} - 1} \log \left ({\left (\sqrt {2} + 1\right )} \sqrt {\sqrt {2} - 1} + x\right ) - \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} - 1} \log \left (-{\left (\sqrt {2} + 1\right )} \sqrt {\sqrt {2} - 1} + x\right ) + \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} + 1} \log \left (\sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + x\right ) - \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} + 1} \log \left (-\sqrt {\sqrt {2} + 1} {\left (\sqrt {2} - 1\right )} + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8-6*x^4+1),x, algorithm="fricas")

[Out]

-1/4*sqrt(2)*sqrt(sqrt(2) + 1)*arctan(-x*sqrt(sqrt(2) + 1) + sqrt(x^2 + sqrt(2) - 1)*sqrt(sqrt(2) + 1)) - 1/4*
sqrt(2)*sqrt(sqrt(2) - 1)*arctan(-x*sqrt(sqrt(2) - 1) + sqrt(x^2 + sqrt(2) + 1)*sqrt(sqrt(2) - 1)) + 1/16*sqrt
(2)*sqrt(sqrt(2) - 1)*log((sqrt(2) + 1)*sqrt(sqrt(2) - 1) + x) - 1/16*sqrt(2)*sqrt(sqrt(2) - 1)*log(-(sqrt(2)
+ 1)*sqrt(sqrt(2) - 1) + x) + 1/16*sqrt(2)*sqrt(sqrt(2) + 1)*log(sqrt(sqrt(2) + 1)*(sqrt(2) - 1) + x) - 1/16*s
qrt(2)*sqrt(sqrt(2) + 1)*log(-sqrt(sqrt(2) + 1)*(sqrt(2) - 1) + x)

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giac [A]  time = 0.63, size = 135, normalized size = 1.08 \begin {gather*} \frac {1}{8} \, \sqrt {2 \, \sqrt {2} - 2} \arctan \left (\frac {x}{\sqrt {\sqrt {2} + 1}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 2} \arctan \left (\frac {x}{\sqrt {\sqrt {2} - 1}}\right ) + \frac {1}{16} \, \sqrt {2 \, \sqrt {2} - 2} \log \left ({\left | x + \sqrt {\sqrt {2} + 1} \right |}\right ) - \frac {1}{16} \, \sqrt {2 \, \sqrt {2} - 2} \log \left ({\left | x - \sqrt {\sqrt {2} + 1} \right |}\right ) + \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 2} \log \left ({\left | x + \sqrt {\sqrt {2} - 1} \right |}\right ) - \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 2} \log \left ({\left | x - \sqrt {\sqrt {2} - 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8-6*x^4+1),x, algorithm="giac")

[Out]

1/8*sqrt(2*sqrt(2) - 2)*arctan(x/sqrt(sqrt(2) + 1)) + 1/8*sqrt(2*sqrt(2) + 2)*arctan(x/sqrt(sqrt(2) - 1)) + 1/
16*sqrt(2*sqrt(2) - 2)*log(abs(x + sqrt(sqrt(2) + 1))) - 1/16*sqrt(2*sqrt(2) - 2)*log(abs(x - sqrt(sqrt(2) + 1
))) + 1/16*sqrt(2*sqrt(2) + 2)*log(abs(x + sqrt(sqrt(2) - 1))) - 1/16*sqrt(2*sqrt(2) + 2)*log(abs(x - sqrt(sqr
t(2) - 1)))

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maple [A]  time = 0.03, size = 90, normalized size = 0.72 \begin {gather*} \frac {\sqrt {2}\, \arctanh \left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {2}\, \arctanh \left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{8 \sqrt {\sqrt {2}-1}}+\frac {\sqrt {2}\, \arctan \left (\frac {x}{\sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {2}\, \arctan \left (\frac {x}{\sqrt {\sqrt {2}-1}}\right )}{8 \sqrt {\sqrt {2}-1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^4+1)/(x^8-6*x^4+1),x)

[Out]

1/8*2^(1/2)/(2^(1/2)-1)^(1/2)*arctan(1/(2^(1/2)-1)^(1/2)*x)+1/8*2^(1/2)/(1+2^(1/2))^(1/2)*arctanh(1/(1+2^(1/2)
)^(1/2)*x)+1/8*2^(1/2)/(1+2^(1/2))^(1/2)*arctan(1/(1+2^(1/2))^(1/2)*x)+1/8*2^(1/2)/(2^(1/2)-1)^(1/2)*arctanh(1
/(2^(1/2)-1)^(1/2)*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {x^{4} - 1}{x^{8} - 6 \, x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^4+1)/(x^8-6*x^4+1),x, algorithm="maxima")

[Out]

-integrate((x^4 - 1)/(x^8 - 6*x^4 + 1), x)

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mupad [B]  time = 0.20, size = 245, normalized size = 1.96 \begin {gather*} -\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {x\,\sqrt {1-\sqrt {2}}\,4352{}\mathrm {i}}{3072\,\sqrt {2}-4352}-\frac {\sqrt {2}\,x\,\sqrt {1-\sqrt {2}}\,3072{}\mathrm {i}}{3072\,\sqrt {2}-4352}\right )\,\sqrt {1-\sqrt {2}}\,1{}\mathrm {i}}{8}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {x\,\sqrt {-\sqrt {2}-1}\,4352{}\mathrm {i}}{3072\,\sqrt {2}+4352}+\frac {\sqrt {2}\,x\,\sqrt {-\sqrt {2}-1}\,3072{}\mathrm {i}}{3072\,\sqrt {2}+4352}\right )\,\sqrt {-\sqrt {2}-1}\,1{}\mathrm {i}}{8}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-1}\,4352{}\mathrm {i}}{3072\,\sqrt {2}-4352}-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}-1}\,3072{}\mathrm {i}}{3072\,\sqrt {2}-4352}\right )\,\sqrt {\sqrt {2}-1}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}+1}\,4352{}\mathrm {i}}{3072\,\sqrt {2}+4352}+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+1}\,3072{}\mathrm {i}}{3072\,\sqrt {2}+4352}\right )\,\sqrt {\sqrt {2}+1}\,1{}\mathrm {i}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4 - 1)/(x^8 - 6*x^4 + 1),x)

[Out]

(2^(1/2)*atan((x*(- 2^(1/2) - 1)^(1/2)*4352i)/(3072*2^(1/2) + 4352) + (2^(1/2)*x*(- 2^(1/2) - 1)^(1/2)*3072i)/
(3072*2^(1/2) + 4352))*(- 2^(1/2) - 1)^(1/2)*1i)/8 - (2^(1/2)*atan((x*(1 - 2^(1/2))^(1/2)*4352i)/(3072*2^(1/2)
- 4352) - (2^(1/2)*x*(1 - 2^(1/2))^(1/2)*3072i)/(3072*2^(1/2) - 4352))*(1 - 2^(1/2))^(1/2)*1i)/8 + (2^(1/2)*a
tan((x*(2^(1/2) - 1)^(1/2)*4352i)/(3072*2^(1/2) - 4352) - (2^(1/2)*x*(2^(1/2) - 1)^(1/2)*3072i)/(3072*2^(1/2)
- 4352))*(2^(1/2) - 1)^(1/2)*1i)/8 - (2^(1/2)*atan((x*(2^(1/2) + 1)^(1/2)*4352i)/(3072*2^(1/2) + 4352) + (2^(1
/2)*x*(2^(1/2) + 1)^(1/2)*3072i)/(3072*2^(1/2) + 4352))*(2^(1/2) + 1)^(1/2)*1i)/8

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sympy [A]  time = 1.16, size = 51, normalized size = 0.41 \begin {gather*} - \operatorname {RootSum} {\left (16384 t^{4} - 256 t^{2} - 1, \left (t \mapsto t \log {\left (65536 t^{5} - 28 t + x \right )} \right )\right )} - \operatorname {RootSum} {\left (16384 t^{4} + 256 t^{2} - 1, \left (t \mapsto t \log {\left (65536 t^{5} - 28 t + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**4+1)/(x**8-6*x**4+1),x)

[Out]

-RootSum(16384*_t**4 - 256*_t**2 - 1, Lambda(_t, _t*log(65536*_t**5 - 28*_t + x))) - RootSum(16384*_t**4 + 256
*_t**2 - 1, Lambda(_t, _t*log(65536*_t**5 - 28*_t + x)))

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