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

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

[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])

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Rubi [A]  time = 0.054602, antiderivative size = 104, normalized size of antiderivative = 1., 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 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 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.0704203, size = 94, normalized size = 0.9 \[ \frac{1}{32} \left (\frac{8 x}{x^4+1}+5 \sqrt{2} \log \left (x^2-\sqrt{2} x+1\right )-5 \sqrt{2} \log \left (x^2+\sqrt{2} x+1\right )+32 x+10 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} x\right )-10 \sqrt{2} \tan ^{-1}\left (\sqrt{2} x+1\right )\right ) \]

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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Maple [A]  time = 0.007, size = 69, normalized size = 0.7 \begin{align*} x+{\frac{x}{4\,{x}^{4}+4}}-{\frac{5\,\sqrt{2}}{32}\ln \left ({\frac{1+{x}^{2}+x\sqrt{2}}{1+{x}^{2}-x\sqrt{2}}} \right ) }-{\frac{5\,\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{16}}-{\frac{5\,\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{16}} \end{align*}

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+x*2^(1/2))/(1+x^2-x*2^(1/2)))-5/16*arctan(1+x*2^(1/2))*2^(1/2)-5/16*arc
tan(-1+x*2^(1/2))*2^(1/2)

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Maxima [A]  time = 1.49095, size = 112, normalized size = 1.08 \begin{align*} -\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 )}} \end{align*}

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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Fricas [A]  time = 1.51484, size = 392, normalized size = 3.77 \begin{align*} \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 )}} \end{align*}

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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Sympy [A]  time = 0.185903, size = 90, normalized size = 0.87 \begin{align*} 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} \end{align*}

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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Giac [A]  time = 1.10433, size = 112, normalized size = 1.08 \begin{align*} -\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 )}} \end{align*}

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)