3.4.45 \(\int \frac {1}{x^6 (1+x^4+x^8)} \, dx\) [345]

Optimal. Leaf size=98 \[ -\frac {1}{5 x^5}+\frac {1}{x}-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \]

[Out]

-1/5/x^5+1/x-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/12*ln(1+x^2-x*3
^(1/2))*3^(1/2)-1/12*ln(1+x^2+x*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {1382, 1518, 12, 1386, 1178, 642, 1175, 632, 210} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{5 x^5}+\frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}}+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(1 + x^4 + x^8)),x]

[Out]

-1/5*1/x^5 + x^(-1) - ArcTan[(1 - 2*x)/Sqrt[3]]/(2*Sqrt[3]) + ArcTan[(1 + 2*x)/Sqrt[3]]/(2*Sqrt[3]) + Log[1 -
Sqrt[3]*x + x^2]/(4*Sqrt[3]) - Log[1 + Sqrt[3]*x + x^2]/(4*Sqrt[3])

Rule 12

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 1175

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 1178

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 1382

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

Rule 1386

Int[(x_)^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[
2*q - b/c, 2]}, -Dist[1/(2*c*r), Int[x^(m - 3*(n/2))*((q - r*x^(n/2))/(q - r*x^(n/2) + x^n)), x], x] + Dist[1/
(2*c*r), Int[x^(m - 3*(n/2))*((q + r*x^(n/2))/(q + r*x^(n/2) + x^n)), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n
2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 0] && IGtQ[m, 0] && GeQ[m, 3*(n/2)] && LtQ[m, 2*n] && NegQ[b^2 - 4
*a*c]

Rule 1518

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
Simp[d*(f*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*f*(m + 1))), x] + Dist[1/(a*f^n*(m + 1)), Int[(f*x)^
(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(p + 1) + 1)*x^n,
x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -
1] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (1+x^4+x^8\right )} \, dx &=-\frac {1}{5 x^5}+\frac {1}{5} \int \frac {-5-5 x^4}{x^2 \left (1+x^4+x^8\right )} \, dx\\ &=-\frac {1}{5 x^5}+\frac {1}{x}-\frac {1}{5} \int -\frac {5 x^6}{1+x^4+x^8} \, dx\\ &=-\frac {1}{5 x^5}+\frac {1}{x}+\int \frac {x^6}{1+x^4+x^8} \, dx\\ &=-\frac {1}{5 x^5}+\frac {1}{x}-\frac {1}{2} \int \frac {1-x^2}{1-x^2+x^4} \, dx+\frac {1}{2} \int \frac {1+x^2}{1+x^2+x^4} \, dx\\ &=-\frac {1}{5 x^5}+\frac {1}{x}+\frac {1}{4} \int \frac {1}{1-x+x^2} \, dx+\frac {1}{4} \int \frac {1}{1+x+x^2} \, dx+\frac {\int \frac {\sqrt {3}+2 x}{-1-\sqrt {3} x-x^2} \, dx}{4 \sqrt {3}}+\frac {\int \frac {\sqrt {3}-2 x}{-1+\sqrt {3} x-x^2} \, dx}{4 \sqrt {3}}\\ &=-\frac {1}{5 x^5}+\frac {1}{x}+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac {1}{5 x^5}+\frac {1}{x}-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 95, normalized size = 0.97 \begin {gather*} \frac {1}{60} \left (-\frac {12}{x^5}+\frac {60}{x}+10 \sqrt {3} \tan ^{-1}\left (\frac {-1+2 x}{\sqrt {3}}\right )+10 \sqrt {3} \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )+5 \sqrt {3} \log \left (-1+\sqrt {3} x-x^2\right )-5 \sqrt {3} \log \left (1+\sqrt {3} x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(1 + x^4 + x^8)),x]

[Out]

(-12/x^5 + 60/x + 10*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 10*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] + 5*Sqrt[3]*Log
[-1 + Sqrt[3]*x - x^2] - 5*Sqrt[3]*Log[1 + Sqrt[3]*x + x^2])/60

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Maple [A]
time = 0.04, size = 75, normalized size = 0.77

method	result	size
default	\(\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {1}{5 x^{5}}+\frac {1}{x}+\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{12}-\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\)	\(75\)
risch	\(\frac {x^{4}-\frac {1}{5}}{x^{5}}+\frac {\arctan \left (\frac {x \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {x^{3} \sqrt {3}}{3}+\frac {2 x \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{12}-\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{12}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/6*arctan(1/3*(2*x+1)*3^(1/2))*3^(1/2)-1/5/x^5+1/x+1/12*ln(1+x^2-x*3^(1/2))*3^(1/2)-1/12*ln(1+x^2+x*3^(1/2))*
3^(1/2)+1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/5*(5*x^4 - 1)/x^5 +
1/2*integrate((x^2 - 1)/(x^4 - x^2 + 1), x)

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Fricas [A]
time = 0.35, size = 90, normalized size = 0.92 \begin {gather*} \frac {10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) + 10 \, \sqrt {3} x^{5} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + 5 \, \sqrt {3} x^{5} \log \left (\frac {x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) + 60 \, x^{4} - 12}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/60*(10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*(x^3 + 2*x)) + 10*sqrt(3)*x^5*arctan(1/3*sqrt(3)*x) + 5*sqrt(3)*x^5*lo
g((x^4 + 5*x^2 - 2*sqrt(3)*(x^3 + x) + 1)/(x^4 - x^2 + 1)) + 60*x^4 - 12)/x^5

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Sympy [A]
time = 0.09, size = 94, normalized size = 0.96 \begin {gather*} \frac {\sqrt {3} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{12} + \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} + \frac {5 x^{4} - 1}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(x**8+x**4+1),x)

[Out]

sqrt(3)*(2*atan(sqrt(3)*x/3) + 2*atan(sqrt(3)*x**3/3 + 2*sqrt(3)*x/3))/12 + sqrt(3)*log(x**2 - sqrt(3)*x + 1)/
12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + (5*x**4 - 1)/(5*x**5)

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Giac [A]
time = 4.01, size = 100, normalized size = 1.02 \begin {gather*} \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {5 \, x^{4} - 1}{5 \, x^{5}} + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/24*sqrt(3)*log(x^2 +
sqrt(3)*x + 1) + 1/24*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/5*(5*x^4 - 1)/x^5 + 1/4*arctan(2*x + sqrt(3)) + 1/
4*arctan(2*x - sqrt(3))

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Mupad [B]
time = 0.04, size = 52, normalized size = 0.53 \begin {gather*} \frac {x^4-\frac {1}{5}}{x^5}-\frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )}{6}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}-\frac {2}{3}\right )}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(x^4 + x^8 + 1)),x)

[Out]

(x^4 - 1/5)/x^5 - (3^(1/2)*atanh((2*3^(1/2)*x)/(3*((2*x^2)/3 + 2/3))))/6 - (3^(1/2)*atan((2*3^(1/2)*x)/(3*((2*
x^2)/3 - 2/3))))/6

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