3.337 \(\int \frac{1}{x^7 (1+x^4+x^8)} \, dx\)

Optimal. Leaf size=89 \[ \frac{1}{2 x^2}-\frac{1}{6 x^6}+\frac{1}{8} \log \left (x^4-x^2+1\right )-\frac{1}{8} \log \left (x^4+x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

[Out]

-1/(6*x^6) + 1/(2*x^2) - ArcTan[(1 - 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + L
og[1 - x^2 + x^4]/8 - Log[1 + x^2 + x^4]/8

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Rubi [A]  time = 0.0962304, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {1359, 1123, 1281, 12, 1127, 1161, 618, 204, 1164, 628} \[ \frac{1}{2 x^2}-\frac{1}{6 x^6}+\frac{1}{8} \log \left (x^4-x^2+1\right )-\frac{1}{8} \log \left (x^4+x^2+1\right )-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + x^4 + x^8)),x]

[Out]

-1/(6*x^6) + 1/(2*x^2) - ArcTan[(1 - 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + L
og[1 - x^2 + x^4]/8 - Log[1 + x^2 + x^4]/8

Rule 1359

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k) + c*x^((2*n)/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 1123

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2 +
c*x^4)^(p + 1))/(a*d*(m + 1)), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[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]

Rubi steps

\begin{align*} \int \frac{1}{x^7 \left (1+x^4+x^8\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+x^2+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{-3-3 x^2}{x^2 \left (1+x^2+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}-\frac{1}{6} \operatorname{Subst}\left (\int -\frac{3 x^2}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^2+x^4} \, dx,x,x^2\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^2+x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1+2 x}{-1-x-x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1-2 x}{-1+x-x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}+\frac{1}{8} \log \left (1-x^2+x^4\right )-\frac{1}{8} \log \left (1+x^2+x^4\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=-\frac{1}{6 x^6}+\frac{1}{2 x^2}-\frac{\tan ^{-1}\left (\frac{1-2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1+2 x^2}{\sqrt{3}}\right )}{4 \sqrt{3}}+\frac{1}{8} \log \left (1-x^2+x^4\right )-\frac{1}{8} \log \left (1+x^2+x^4\right )\\ \end{align*}

Mathematica [C]  time = 0.111959, size = 142, normalized size = 1.6 \[ \frac{1}{24} \left (\frac{12}{x^2}-\frac{4}{x^6}+\sqrt{3} \left (\sqrt{3}-i\right ) \log \left (x^2-\frac{i \sqrt{3}}{2}-\frac{1}{2}\right )+\sqrt{3} \left (\sqrt{3}+i\right ) \log \left (x^2+\frac{1}{2} i \left (\sqrt{3}+i\right )\right )-3 \log \left (x^2-x+1\right )-3 \log \left (x^2+x+1\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + x^4 + x^8)),x]

[Out]

(-4/x^6 + 12/x^2 + 2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + Sqrt[3]*(-I +
Sqrt[3])*Log[-1/2 - (I/2)*Sqrt[3] + x^2] + Sqrt[3]*(I + Sqrt[3])*Log[(I/2)*(I + Sqrt[3]) + x^2] - 3*Log[1 - x
+ x^2] - 3*Log[1 + x + x^2])/24

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Maple [A]  time = 0.008, size = 95, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ({x}^{2}+x+1 \right ) }{8}}-{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{6\,{x}^{6}}}+{\frac{1}{2\,{x}^{2}}}-{\frac{\ln \left ({x}^{2}-x+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\ln \left ({x}^{4}-{x}^{2}+1 \right ) }{8}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 2\,{x}^{2}-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^8+x^4+1),x)

[Out]

-1/8*ln(x^2+x+1)-1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/6/x^6+1/2/x^2-1/8*ln(x^2-x+1)+1/12*3^(1/2)*arctan(
1/3*(2*x-1)*3^(1/2))+1/8*ln(x^4-x^2+1)+1/12*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))

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Maxima [A]  time = 1.51521, size = 99, normalized size = 1.11 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{3 \, x^{4} - 1}{6 \, x^{6}} - \frac{1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + 1/6*(3*x^4 - 1)/
x^6 - 1/8*log(x^4 + x^2 + 1) + 1/8*log(x^4 - x^2 + 1)

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Fricas [A]  time = 1.54108, size = 234, normalized size = 2.63 \begin{align*} \frac{2 \, \sqrt{3} x^{6} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + 2 \, \sqrt{3} x^{6} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) - 3 \, x^{6} \log \left (x^{4} + x^{2} + 1\right ) + 3 \, x^{6} \log \left (x^{4} - x^{2} + 1\right ) + 12 \, x^{4} - 4}{24 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/24*(2*sqrt(3)*x^6*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 2*sqrt(3)*x^6*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 3*x^6*lo
g(x^4 + x^2 + 1) + 3*x^6*log(x^4 - x^2 + 1) + 12*x^4 - 4)/x^6

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Sympy [A]  time = 0.242809, size = 88, normalized size = 0.99 \begin{align*} \frac{\log{\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac{\log{\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} - \frac{\sqrt{3}}{3} \right )}}{12} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x^{2}}{3} + \frac{\sqrt{3}}{3} \right )}}{12} + \frac{3 x^{4} - 1}{6 x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**8+x**4+1),x)

[Out]

log(x**4 - x**2 + 1)/8 - log(x**4 + x**2 + 1)/8 + sqrt(3)*atan(2*sqrt(3)*x**2/3 - sqrt(3)/3)/12 + sqrt(3)*atan
(2*sqrt(3)*x**2/3 + sqrt(3)/3)/12 + (3*x**4 - 1)/(6*x**6)

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Giac [A]  time = 1.09196, size = 99, normalized size = 1.11 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} + 1\right )}\right ) + \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x^{2} - 1\right )}\right ) + \frac{3 \, x^{4} - 1}{6 \, x^{6}} - \frac{1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac{1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + 1/6*(3*x^4 - 1)/
x^6 - 1/8*log(x^4 + x^2 + 1) + 1/8*log(x^4 - x^2 + 1)