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

Optimal. Leaf size=99 \[ \frac{x^3}{4 \left (x^4+1\right )}+\frac{\log \left (x^2-\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{\log \left (x^2+\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x+1\right )}{8 \sqrt{2}} \]

[Out]

x^3/(4*(1 + x^4)) - ArcTan[1 - Sqrt[2]*x]/(8*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(8*Sqrt[2]) + Log[1 - Sqrt[2]*x
+ x^2]/(16*Sqrt[2]) - Log[1 + Sqrt[2]*x + x^2]/(16*Sqrt[2])

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Rubi [A]  time = 0.050228, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {28, 290, 297, 1162, 617, 204, 1165, 628} \[ \frac{x^3}{4 \left (x^4+1\right )}+\frac{\log \left (x^2-\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{\log \left (x^2+\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x+1\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + 2*x^4 + x^8),x]

[Out]

x^3/(4*(1 + x^4)) - ArcTan[1 - Sqrt[2]*x]/(8*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(8*Sqrt[2]) + Log[1 - Sqrt[2]*x
+ x^2]/(16*Sqrt[2]) - 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 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 297

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

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.0495853, size = 92, normalized size = 0.93 \[ \frac{1}{32} \left (\frac{8 x^3}{x^4+1}+\sqrt{2} \log \left (x^2-\sqrt{2} x+1\right )-\sqrt{2} \log \left (x^2+\sqrt{2} x+1\right )-2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} x\right )+2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + 2*x^4 + x^8),x]

[Out]

((8*x^3)/(1 + x^4) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] + Sqrt[2]*Log[1 - Sqrt[
2]*x + x^2] - Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/32

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Maple [A]  time = 0.005, size = 70, normalized size = 0.7 \begin{align*}{\frac{{x}^{3}}{4\,{x}^{4}+4}}+{\frac{\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{16}}+{\frac{\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{16}}+{\frac{\sqrt{2}}{32}\ln \left ({\frac{1+{x}^{2}-x\sqrt{2}}{1+{x}^{2}+x\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.4797, size = 113, normalized size = 1.14 \begin{align*} \frac{x^{3}}{4 \,{\left (x^{4} + 1\right )}} + \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) - \frac{1}{32} \, \sqrt{2} \log \left (x^{2} + \sqrt{2} x + 1\right ) + \frac{1}{32} \, \sqrt{2} \log \left (x^{2} - \sqrt{2} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.65132, size = 373, normalized size = 3.77 \begin{align*} \frac{8 \, x^{3} - 4 \, \sqrt{2}{\left (x^{4} + 1\right )} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} + \sqrt{2} x + 1} - 1\right ) - 4 \, \sqrt{2}{\left (x^{4} + 1\right )} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} - \sqrt{2} x + 1} + 1\right ) - \sqrt{2}{\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt{2} x + 1\right ) + \sqrt{2}{\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt{2} x + 1\right )}{32 \,{\left (x^{4} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*x^3 - 4*sqrt(2)*(x^4 + 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 + sqrt(2)*x + 1) - 1) - 4*sqrt(2)*(x^4
+ 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 - sqrt(2)*x + 1) + 1) - sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x + 1) +
sqrt(2)*(x^4 + 1)*log(x^2 - sqrt(2)*x + 1))/(x^4 + 1)

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Sympy [A]  time = 0.178476, size = 83, normalized size = 0.84 \begin{align*} \frac{x^{3}}{4 x^{4} + 4} + \frac{\sqrt{2} \log{\left (x^{2} - \sqrt{2} x + 1 \right )}}{32} - \frac{\sqrt{2} \log{\left (x^{2} + \sqrt{2} x + 1 \right )}}{32} + \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x - 1 \right )}}{16} + \frac{\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**2/(x**8+2*x**4+1),x)

[Out]

x**3/(4*x**4 + 4) + sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 + sqrt(2)*atan
(sqrt(2)*x - 1)/16 + sqrt(2)*atan(sqrt(2)*x + 1)/16

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Giac [A]  time = 1.10743, size = 113, normalized size = 1.14 \begin{align*} \frac{x^{3}}{4 \,{\left (x^{4} + 1\right )}} + \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x + \sqrt{2}\right )}\right ) + \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x - \sqrt{2}\right )}\right ) - \frac{1}{32} \, \sqrt{2} \log \left (x^{2} + \sqrt{2} x + 1\right ) + \frac{1}{32} \, \sqrt{2} \log \left (x^{2} - \sqrt{2} x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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