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

Optimal. Leaf size=106 \[ \frac{1}{4 x \left (x^4+1\right )}-\frac{5 \log \left (x^2-\sqrt{2} x+1\right )}{16 \sqrt{2}}+\frac{5 \log \left (x^2+\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{5}{4 x}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} x\right )}{8 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} x+1\right )}{8 \sqrt{2}} \]

[Out]

-5/(4*x) + 1/(4*x*(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])

________________________________________________________________________________________

Rubi [A]  time = 0.0515349, antiderivative size = 106, 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, 290, 325, 297, 1162, 617, 204, 1165, 628} \[ \frac{1}{4 x \left (x^4+1\right )}-\frac{5 \log \left (x^2-\sqrt{2} x+1\right )}{16 \sqrt{2}}+\frac{5 \log \left (x^2+\sqrt{2} x+1\right )}{16 \sqrt{2}}-\frac{5}{4 x}+\frac{5 \tan ^{-1}\left (1-\sqrt{2} x\right )}{8 \sqrt{2}}-\frac{5 \tan ^{-1}\left (\sqrt{2} x+1\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(1 + 2*x^4 + x^8)),x]

[Out]

-5/(4*x) + 1/(4*x*(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 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 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{1}{x^2 \left (1+2 x^4+x^8\right )} \, dx &=\int \frac{1}{x^2 \left (1+x^4\right )^2} \, dx\\ &=\frac{1}{4 x \left (1+x^4\right )}+\frac{5}{4} \int \frac{1}{x^2 \left (1+x^4\right )} \, dx\\ &=-\frac{5}{4 x}+\frac{1}{4 x \left (1+x^4\right )}-\frac{5}{4} \int \frac{x^2}{1+x^4} \, dx\\ &=-\frac{5}{4 x}+\frac{1}{4 x \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}{4 x}+\frac{1}{4 x \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}{4 x}+\frac{1}{4 x \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}{4 x}+\frac{1}{4 x \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 = 0.0699025, size = 98, normalized size = 0.92 \[ \frac{1}{32} \left (-\frac{8 x^3}{x^4+1}-5 \sqrt{2} \log \left (x^2-\sqrt{2} x+1\right )+5 \sqrt{2} \log \left (x^2+\sqrt{2} x+1\right )-\frac{32}{x}+10 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} x\right )-10 \sqrt{2} \tan ^{-1}\left (\sqrt{2} x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32/x - (8*x^3)/(1 + x^4) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] - 5*Sqrt[2]*L
og[1 - Sqrt[2]*x + x^2] + 5*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/32

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 75, normalized size = 0.7 \begin{align*} -{x}^{-1}-{\frac{{x}^{3}}{4\,{x}^{4}+4}}-{\frac{5\,\arctan \left ( 1+x\sqrt{2} \right ) \sqrt{2}}{16}}-{\frac{5\,\arctan \left ( -1+x\sqrt{2} \right ) \sqrt{2}}{16}}-{\frac{5\,\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(1/x^2/(x^8+2*x^4+1),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.47945, size = 119, normalized size = 1.12 \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 ) - \frac{5 \, x^{4} + 4}{4 \,{\left (x^{5} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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) - 1/4*(5*x^4 + 4)/(x^5 + x)

________________________________________________________________________________________

Fricas [A]  time = 1.56656, size = 390, normalized size = 3.68 \begin{align*} -\frac{40 \, x^{4} - 20 \, \sqrt{2}{\left (x^{5} + x\right )} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} + \sqrt{2} x + 1} - 1\right ) - 20 \, \sqrt{2}{\left (x^{5} + x\right )} \arctan \left (-\sqrt{2} x + \sqrt{2} \sqrt{x^{2} - \sqrt{2} x + 1} + 1\right ) - 5 \, \sqrt{2}{\left (x^{5} + x\right )} \log \left (x^{2} + \sqrt{2} x + 1\right ) + 5 \, \sqrt{2}{\left (x^{5} + x\right )} \log \left (x^{2} - \sqrt{2} x + 1\right ) + 32}{32 \,{\left (x^{5} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.213433, size = 95, normalized size = 0.9 \begin{align*} - \frac{5 x^{4} + 4}{4 x^{5} + 4 x} - \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(1/x**2/(x**8+2*x**4+1),x)

[Out]

-(5*x**4 + 4)/(4*x**5 + 4*x) - 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

________________________________________________________________________________________

Giac [A]  time = 1.09023, size = 119, normalized size = 1.12 \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 ) - \frac{5 \, x^{4} + 4}{4 \,{\left (x^{5} + x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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) - 1/4*(5*x^4 + 4)/(x^5 + x)