3.289 \(\int \frac {1}{x^8 (1+2 x^4+x^8)} \, dx\)

Optimal. Leaf size=113 \[ -\frac {11}{28 x^7}+\frac {11}{12 x^3}-\frac {11 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {11 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^7 \left (x^4+1\right )}-\frac {11 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {11 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

[Out]

-11/28/x^7+11/12/x^3+1/4/x^7/(x^4+1)+11/16*arctan(-1+x*2^(1/2))*2^(1/2)+11/16*arctan(1+x*2^(1/2))*2^(1/2)-11/3
2*ln(1+x^2-x*2^(1/2))*2^(1/2)+11/32*ln(1+x^2+x*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 290, 325, 211, 1165, 628, 1162, 617, 204} \[ \frac {1}{4 x^7 \left (x^4+1\right )}+\frac {11}{12 x^3}-\frac {11}{28 x^7}-\frac {11 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {11 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {11 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {11 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(1 + 2*x^4 + x^8)),x]

[Out]

-11/(28*x^7) + 11/(12*x^3) + 1/(4*x^7*(1 + x^4)) - (11*ArcTan[1 - Sqrt[2]*x])/(8*Sqrt[2]) + (11*ArcTan[1 + Sqr
t[2]*x])/(8*Sqrt[2]) - (11*Log[1 - Sqrt[2]*x + x^2])/(16*Sqrt[2]) + (11*Log[1 + Sqrt[2]*x + x^2])/(16*Sqrt[2])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {1}{x^8 \left (1+2 x^4+x^8\right )} \, dx &=\int \frac {1}{x^8 \left (1+x^4\right )^2} \, dx\\ &=\frac {1}{4 x^7 \left (1+x^4\right )}+\frac {11}{4} \int \frac {1}{x^8 \left (1+x^4\right )} \, dx\\ &=-\frac {11}{28 x^7}+\frac {1}{4 x^7 \left (1+x^4\right )}-\frac {11}{4} \int \frac {1}{x^4 \left (1+x^4\right )} \, dx\\ &=-\frac {11}{28 x^7}+\frac {11}{12 x^3}+\frac {1}{4 x^7 \left (1+x^4\right )}+\frac {11}{4} \int \frac {1}{1+x^4} \, dx\\ &=-\frac {11}{28 x^7}+\frac {11}{12 x^3}+\frac {1}{4 x^7 \left (1+x^4\right )}+\frac {11}{8} \int \frac {1-x^2}{1+x^4} \, dx+\frac {11}{8} \int \frac {1+x^2}{1+x^4} \, dx\\ &=-\frac {11}{28 x^7}+\frac {11}{12 x^3}+\frac {1}{4 x^7 \left (1+x^4\right )}+\frac {11}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {11}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {11 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}-\frac {11 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}\\ &=-\frac {11}{28 x^7}+\frac {11}{12 x^3}+\frac {1}{4 x^7 \left (1+x^4\right )}-\frac {11 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {11 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {11 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {11 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}}\\ &=-\frac {11}{28 x^7}+\frac {11}{12 x^3}+\frac {1}{4 x^7 \left (1+x^4\right )}-\frac {11 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {11 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {11 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {11 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 101, normalized size = 0.89 \[ \frac {1}{672} \left (-\frac {96}{x^7}+\frac {168 x}{x^4+1}+\frac {448}{x^3}-231 \sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )+231 \sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )-462 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )+462 \sqrt {2} \tan ^{-1}\left (\sqrt {2} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(1 + 2*x^4 + x^8)),x]

[Out]

(-96/x^7 + 448/x^3 + (168*x)/(1 + x^4) - 462*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 462*Sqrt[2]*ArcTan[1 + Sqrt[2]*x]
- 231*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] + 231*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/672

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fricas [A]  time = 0.95, size = 145, normalized size = 1.28 \[ \frac {616 \, x^{8} + 352 \, x^{4} - 924 \, \sqrt {2} {\left (x^{11} + x^{7}\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) - 924 \, \sqrt {2} {\left (x^{11} + x^{7}\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) + 231 \, \sqrt {2} {\left (x^{11} + x^{7}\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - 231 \, \sqrt {2} {\left (x^{11} + x^{7}\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) - 96}{672 \, {\left (x^{11} + x^{7}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

1/672*(616*x^8 + 352*x^4 - 924*sqrt(2)*(x^11 + x^7)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 + sqrt(2)*x + 1) - 1)
- 924*sqrt(2)*(x^11 + x^7)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 - sqrt(2)*x + 1) + 1) + 231*sqrt(2)*(x^11 + x
^7)*log(x^2 + sqrt(2)*x + 1) - 231*sqrt(2)*(x^11 + x^7)*log(x^2 - sqrt(2)*x + 1) - 96)/(x^11 + x^7)

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giac [A]  time = 0.36, size = 94, normalized size = 0.83 \[ \frac {11}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {11}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {11}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {11}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {x}{4 \, {\left (x^{4} + 1\right )}} + \frac {14 \, x^{4} - 3}{21 \, x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

11/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 11/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 11/32*
sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 11/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/4*x/(x^4 + 1) + 1/21*(14*x^4 - 3
)/x^7

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maple [A]  time = 0.01, size = 78, normalized size = 0.69 \[ \frac {x}{4 x^{4}+4}+\frac {11 \sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{16}+\frac {11 \sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{16}+\frac {11 \sqrt {2}\, \ln \left (\frac {x^{2}+\sqrt {2}\, x +1}{x^{2}-\sqrt {2}\, x +1}\right )}{32}+\frac {2}{3 x^{3}}-\frac {1}{7 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^8/(x^8+2*x^4+1),x)

[Out]

-1/7/x^7+2/3/x^3+1/4/(x^4+1)*x+11/16*2^(1/2)*arctan(2^(1/2)*x-1)+11/32*2^(1/2)*ln((x^2+2^(1/2)*x+1)/(x^2-2^(1/
2)*x+1))+11/16*2^(1/2)*arctan(2^(1/2)*x+1)

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maxima [A]  time = 1.93, size = 95, normalized size = 0.84 \[ \frac {11}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {11}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {11}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {11}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {77 \, x^{8} + 44 \, x^{4} - 12}{84 \, {\left (x^{11} + x^{7}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

11/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 11/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) + 11/32*
sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 11/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/84*(77*x^8 + 44*x^4 - 12)/(x^11
+ x^7)

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mupad [B]  time = 0.10, size = 55, normalized size = 0.49 \[ \frac {\frac {11\,x^8}{12}+\frac {11\,x^4}{21}-\frac {1}{7}}{x^{11}+x^7}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {11}{16}+\frac {11}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {11}{16}-\frac {11}{16}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8*(2*x^4 + x^8 + 1)),x)

[Out]

2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(11/16 + 11i/16) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(11/16 - 11i/16)
+ ((11*x^4)/21 + (11*x^8)/12 - 1/7)/(x^7 + x^11)

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sympy [A]  time = 0.23, size = 102, normalized size = 0.90 \[ - \frac {11 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} + \frac {11 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {11 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {11 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} + \frac {77 x^{8} + 44 x^{4} - 12}{84 x^{11} + 84 x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(x**8+2*x**4+1),x)

[Out]

-11*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 + 11*sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 + 11*sqrt(2)*atan(sqrt(2)*x
- 1)/16 + 11*sqrt(2)*atan(sqrt(2)*x + 1)/16 + (77*x**8 + 44*x**4 - 12)/(84*x**11 + 84*x**7)

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