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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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 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 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 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 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]

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 = 0.11, size = 142, normalized size = 1.60 \[ \frac {1}{24} \left (-\frac {4}{x^6}+\frac {12}{x^2}+\sqrt {3} \left (\sqrt {3}-i\right ) \log \left (x^2-\frac {i \sqrt {3}}{2}-\frac {1}{2}\right )+\sqrt {3} \left (\sqrt {3}+i\right ) \log \left (x^2+\frac {1}{2} i \left (\sqrt {3}+i\right )\right )-3 \log \left (x^2-x+1\right )-3 \log \left (x^2+x+1\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )\right ) \]

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

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 84, normalized size = 0.94 \[ \frac {2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + 2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - 3 \, x^{6} \log \left (x^{4} + x^{2} + 1\right ) + 3 \, x^{6} \log \left (x^{4} - x^{2} + 1\right ) + 12 \, x^{4} - 4}{24 \, x^{6}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.27, size = 73, normalized size = 0.82 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 95, normalized size = 1.07 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{2}-x +1\right )}{8}-\frac {\ln \left (x^{2}+x +1\right )}{8}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {1}{2 x^{2}}-\frac {1}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)*arcta
n(1/3*(2*x^2-1)*3^(1/2))-1/8*ln(x^2-x+1)+1/12*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

________________________________________________________________________________________

maxima [A]  time = 2.37, size = 73, normalized size = 0.82 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 62, normalized size = 0.70 \[ \mathrm {atanh}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atanh}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\frac {\frac {x^4}{2}-\frac {1}{6}}{x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^4 + x^8 + 1)),x)

[Out]

atanh((2*x^2)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/12 + 1/4) + atanh((2*x^2)/(3^(1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1
/4) + (x^4/2 - 1/6)/x^6

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 88, normalized size = 0.99 \[ \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} + \frac {3 x^{4} - 1}{6 x^{6}} \]

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)

________________________________________________________________________________________