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

Optimal. Leaf size=104 \[ \frac {5 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {5 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {x^5}{4 \left (x^4+1\right )}+\frac {5 x}{4}+\frac {5 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

[Out]

5/4*x-1/4*x^5/(x^4+1)-5/16*arctan(-1+x*2^(1/2))*2^(1/2)-5/16*arctan(1+x*2^(1/2))*2^(1/2)+5/32*ln(1+x^2-x*2^(1/
2))*2^(1/2)-5/32*ln(1+x^2+x*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 288, 321, 211, 1165, 628, 1162, 617, 204} \[ -\frac {x^5}{4 \left (x^4+1\right )}+\frac {5 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {5 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {5 x}{4}+\frac {5 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(1 + 2*x^4 + x^8),x]

[Out]

(5*x)/4 - x^5/(4*(1 + x^4)) + (5*ArcTan[1 - Sqrt[2]*x])/(8*Sqrt[2]) - (5*ArcTan[1 + Sqrt[2]*x])/(8*Sqrt[2]) +
(5*Log[1 - Sqrt[2]*x + x^2])/(16*Sqrt[2]) - (5*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 288

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {x^8}{1+2 x^4+x^8} \, dx &=\int \frac {x^8}{\left (1+x^4\right )^2} \, dx\\ &=-\frac {x^5}{4 \left (1+x^4\right )}+\frac {5}{4} \int \frac {x^4}{1+x^4} \, dx\\ &=\frac {5 x}{4}-\frac {x^5}{4 \left (1+x^4\right )}-\frac {5}{4} \int \frac {1}{1+x^4} \, dx\\ &=\frac {5 x}{4}-\frac {x^5}{4 \left (1+x^4\right )}-\frac {5}{8} \int \frac {1-x^2}{1+x^4} \, dx-\frac {5}{8} \int \frac {1+x^2}{1+x^4} \, dx\\ &=\frac {5 x}{4}-\frac {x^5}{4 \left (1+x^4\right )}-\frac {5}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {5}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {5 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {5 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}\\ &=\frac {5 x}{4}-\frac {x^5}{4 \left (1+x^4\right )}+\frac {5 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {5 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}}\\ &=\frac {5 x}{4}-\frac {x^5}{4 \left (1+x^4\right )}+\frac {5 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {5 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {5 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 94, normalized size = 0.90 \[ \frac {1}{32} \left (\frac {8 x}{x^4+1}+5 \sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )-5 \sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )+32 x+10 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )-10 \sqrt {2} \tan ^{-1}\left (\sqrt {2} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(1 + 2*x^4 + x^8),x]

[Out]

(32*x + (8*x)/(1 + x^4) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] + 5*Sqrt[2]*Log[
1 - Sqrt[2]*x + x^2] - 5*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/32

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fricas [A]  time = 0.80, size = 132, normalized size = 1.27 \[ \frac {32 \, x^{5} + 20 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + 20 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) - 5 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) + 5 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + 40 \, x}{32 \, {\left (x^{4} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(32*x^5 + 20*sqrt(2)*(x^4 + 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 + sqrt(2)*x + 1) - 1) + 20*sqrt(2)*(x
^4 + 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 - sqrt(2)*x + 1) + 1) - 5*sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x +
1) + 5*sqrt(2)*(x^4 + 1)*log(x^2 - sqrt(2)*x + 1) + 40*x)/(x^4 + 1)

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giac [A]  time = 0.34, size = 83, normalized size = 0.80 \[ -\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + x + \frac {x}{4 \, {\left (x^{4} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 5/32*sq
rt(2)*log(x^2 + sqrt(2)*x + 1) + 5/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + x + 1/4*x/(x^4 + 1)

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maple [A]  time = 0.01, size = 69, normalized size = 0.66 \[ x +\frac {x}{4 x^{4}+4}-\frac {5 \sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{16}-\frac {5 \sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{16}-\frac {5 \sqrt {2}\, \ln \left (\frac {x^{2}+\sqrt {2}\, x +1}{x^{2}-\sqrt {2}\, x +1}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/4*x/(x^4+1)-5/32*2^(1/2)*ln((1+x^2+2^(1/2)*x)/(1+x^2-2^(1/2)*x))-5/16*arctan(-1+2^(1/2)*x)*2^(1/2)-5/16*ar
ctan(1+2^(1/2)*x)*2^(1/2)

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maxima [A]  time = 2.02, size = 83, normalized size = 0.80 \[ -\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + x + \frac {x}{4 \, {\left (x^{4} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 5/32*sq
rt(2)*log(x^2 + sqrt(2)*x + 1) + 5/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + x + 1/4*x/(x^4 + 1)

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mupad [B]  time = 1.37, size = 45, normalized size = 0.43 \[ x+\frac {x}{4\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}-\frac {5}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}+\frac {5}{16}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(2*x^4 + x^8 + 1),x)

[Out]

x - 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(5/16 + 5i/16) - 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(5/16 - 5i/16)
+ x/(4*(x^4 + 1))

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sympy [A]  time = 0.18, size = 90, normalized size = 0.87 \[ x + \frac {x}{4 x^{4} + 4} + \frac {5 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + x/(4*x**4 + 4) + 5*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 5*sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 - 5*sqrt(
2)*atan(sqrt(2)*x - 1)/16 - 5*sqrt(2)*atan(sqrt(2)*x + 1)/16

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