\(\int \frac {1}{x^7 (1-3 x^4+x^8)} \, dx\) [395]

Optimal result

Integrand size = 16, antiderivative size = 97 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}-\frac {3}{2 x^2}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

[Out]

-1/6/x^6-3/2/x^2-1/2*arctanh(x^2*2^(1/2)/(3+5^(1/2))^(1/2))*(5/2-11/10*5^(1/2))+1/2*arctanh(x^2*(1/2+1/2*5^(1/
2)))*(5/2+11/10*5^(1/2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1373, 1137, 1295, 1180, 213} \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {1}{6 x^6}-\frac {3}{2 x^2} \]

[In]

Int[1/(x^7*(1 - 3*x^4 + x^8)),x]

[Out]

-1/6*1/x^6 - 3/(2*x^2) - (Sqrt[(123 - 55*Sqrt[5])/10]*ArcTanh[Sqrt[2/(3 + Sqrt[5])]*x^2])/2 + (Sqrt[(123 + 55*
Sqrt[5])/10]*ArcTanh[Sqrt[(3 + Sqrt[5])/2]*x^2])/2

Rule 213

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

Rule 1137

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 1180

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

Rule 1295

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 1373

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 \left (1-3 x^2+x^4\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {1}{6} \text {Subst}\left (\int \frac {9-3 x^2}{x^2 \left (1-3 x^2+x^4\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}-\frac {3}{2 x^2}-\frac {1}{6} \text {Subst}\left (\int \frac {-24+9 x^2}{1-3 x^2+x^4} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}-\frac {3}{2 x^2}-\frac {1}{20} \left (15-7 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{-\frac {3}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )-\frac {1}{20} \left (15+7 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{-\frac {3}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}-\frac {3}{2 x^2}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \tanh ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{20} \sqrt {1230+550 \sqrt {5}} \tanh ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=\frac {1}{120} \left (-\frac {20}{x^6}-\frac {180}{x^2}-3 \left (25+11 \sqrt {5}\right ) \log \left (-1+\sqrt {5}-2 x^2\right )+3 \left (25-11 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 x^2\right )+3 \left (25+11 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 x^2\right )+3 \left (-25+11 \sqrt {5}\right ) \log \left (1+\sqrt {5}+2 x^2\right )\right ) \]

[In]

Integrate[1/(x^7*(1 - 3*x^4 + x^8)),x]

[Out]

(-20/x^6 - 180/x^2 - 3*(25 + 11*Sqrt[5])*Log[-1 + Sqrt[5] - 2*x^2] + 3*(25 - 11*Sqrt[5])*Log[1 + Sqrt[5] - 2*x
^2] + 3*(25 + 11*Sqrt[5])*Log[-1 + Sqrt[5] + 2*x^2] + 3*(-25 + 11*Sqrt[5])*Log[1 + Sqrt[5] + 2*x^2])/120

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {1}{6 x^{6}}-\frac {3}{2 x^{2}}+\frac {5 \ln \left (x^{4}-x^{2}-1\right )}{8}+\frac {11 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x^{2}-1\right ) \sqrt {5}}{5}\right )}{20}-\frac {5 \ln \left (x^{4}+x^{2}-1\right )}{8}+\frac {11 \,\operatorname {arctanh}\left (\frac {\left (2 x^{2}+1\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{20}\)	\(72\)
risch	\(\frac {-\frac {3 x^{4}}{2}-\frac {1}{6}}{x^{6}}+\frac {5 \ln \left (11 x^{2}-\frac {11}{2}+\frac {11 \sqrt {5}}{2}\right )}{8}+\frac {11 \ln \left (11 x^{2}-\frac {11}{2}+\frac {11 \sqrt {5}}{2}\right ) \sqrt {5}}{40}+\frac {5 \ln \left (11 x^{2}-\frac {11}{2}-\frac {11 \sqrt {5}}{2}\right )}{8}-\frac {11 \ln \left (11 x^{2}-\frac {11}{2}-\frac {11 \sqrt {5}}{2}\right ) \sqrt {5}}{40}-\frac {5 \ln \left (11 x^{2}+\frac {11}{2}+\frac {11 \sqrt {5}}{2}\right )}{8}+\frac {11 \ln \left (11 x^{2}+\frac {11}{2}+\frac {11 \sqrt {5}}{2}\right ) \sqrt {5}}{40}-\frac {5 \ln \left (11 x^{2}+\frac {11}{2}-\frac {11 \sqrt {5}}{2}\right )}{8}-\frac {11 \ln \left (11 x^{2}+\frac {11}{2}-\frac {11 \sqrt {5}}{2}\right ) \sqrt {5}}{40}\)	\(145\)

[In]

int(1/x^7/(x^8-3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/6/x^6-3/2/x^2+5/8*ln(x^4-x^2-1)+11/20*5^(1/2)*arctanh(1/5*(2*x^2-1)*5^(1/2))-5/8*ln(x^4+x^2-1)+11/20*arctan
h(1/5*(2*x^2+1)*5^(1/2))*5^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (55) = 110\).

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=\frac {33 \, \sqrt {5} x^{6} \log \left (\frac {2 \, x^{4} + 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} + 1\right )} + 3}{x^{4} + x^{2} - 1}\right ) + 33 \, \sqrt {5} x^{6} \log \left (\frac {2 \, x^{4} - 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} - 1\right )} + 3}{x^{4} - x^{2} - 1}\right ) - 75 \, x^{6} \log \left (x^{4} + x^{2} - 1\right ) + 75 \, x^{6} \log \left (x^{4} - x^{2} - 1\right ) - 180 \, x^{4} - 20}{120 \, x^{6}} \]

[In]

integrate(1/x^7/(x^8-3*x^4+1),x, algorithm="fricas")

[Out]

1/120*(33*sqrt(5)*x^6*log((2*x^4 + 2*x^2 + sqrt(5)*(2*x^2 + 1) + 3)/(x^4 + x^2 - 1)) + 33*sqrt(5)*x^6*log((2*x
^4 - 2*x^2 + sqrt(5)*(2*x^2 - 1) + 3)/(x^4 - x^2 - 1)) - 75*x^6*log(x^4 + x^2 - 1) + 75*x^6*log(x^4 - x^2 - 1)
- 180*x^4 - 20)/x^6

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (75) = 150\).

Time = 0.24 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.05 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=\left (\frac {11 \sqrt {5}}{40} + \frac {5}{8}\right ) \log {\left (x^{2} - \frac {2207}{22} - \frac {2207 \sqrt {5}}{50} + \frac {1152 \left (\frac {11 \sqrt {5}}{40} + \frac {5}{8}\right )^{3}}{11} \right )} + \left (\frac {5}{8} - \frac {11 \sqrt {5}}{40}\right ) \log {\left (x^{2} - \frac {2207}{22} + \frac {1152 \left (\frac {5}{8} - \frac {11 \sqrt {5}}{40}\right )^{3}}{11} + \frac {2207 \sqrt {5}}{50} \right )} + \left (- \frac {5}{8} + \frac {11 \sqrt {5}}{40}\right ) \log {\left (x^{2} - \frac {2207 \sqrt {5}}{50} + \frac {1152 \left (- \frac {5}{8} + \frac {11 \sqrt {5}}{40}\right )^{3}}{11} + \frac {2207}{22} \right )} + \left (- \frac {5}{8} - \frac {11 \sqrt {5}}{40}\right ) \log {\left (x^{2} + \frac {1152 \left (- \frac {5}{8} - \frac {11 \sqrt {5}}{40}\right )^{3}}{11} + \frac {2207 \sqrt {5}}{50} + \frac {2207}{22} \right )} + \frac {- 9 x^{4} - 1}{6 x^{6}} \]

[In]

integrate(1/x**7/(x**8-3*x**4+1),x)

[Out]

(11*sqrt(5)/40 + 5/8)*log(x**2 - 2207/22 - 2207*sqrt(5)/50 + 1152*(11*sqrt(5)/40 + 5/8)**3/11) + (5/8 - 11*sqr
t(5)/40)*log(x**2 - 2207/22 + 1152*(5/8 - 11*sqrt(5)/40)**3/11 + 2207*sqrt(5)/50) + (-5/8 + 11*sqrt(5)/40)*log
(x**2 - 2207*sqrt(5)/50 + 1152*(-5/8 + 11*sqrt(5)/40)**3/11 + 2207/22) + (-5/8 - 11*sqrt(5)/40)*log(x**2 + 115
2*(-5/8 - 11*sqrt(5)/40)**3/11 + 2207*sqrt(5)/50 + 2207/22) + (-9*x**4 - 1)/(6*x**6)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=-\frac {11}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} + 1}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {11}{40} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} - 1}{2 \, x^{2} + \sqrt {5} - 1}\right ) - \frac {9 \, x^{4} + 1}{6 \, x^{6}} - \frac {5}{8} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {5}{8} \, \log \left (x^{4} - x^{2} - 1\right ) \]

[In]

integrate(1/x^7/(x^8-3*x^4+1),x, algorithm="maxima")

[Out]

-11/40*sqrt(5)*log((2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 11/40*sqrt(5)*log((2*x^2 - sqrt(5) - 1)/(2*x
^2 + sqrt(5) - 1)) - 1/6*(9*x^4 + 1)/x^6 - 5/8*log(x^4 + x^2 - 1) + 5/8*log(x^4 - x^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=-\frac {11}{40} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} + 1 \right |}}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {11}{40} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} - 1 \right |}}{{\left | 2 \, x^{2} + \sqrt {5} - 1 \right |}}\right ) - \frac {9 \, x^{4} + 1}{6 \, x^{6}} - \frac {5}{8} \, \log \left ({\left | x^{4} + x^{2} - 1 \right |}\right ) + \frac {5}{8} \, \log \left ({\left | x^{4} - x^{2} - 1 \right |}\right ) \]

[In]

integrate(1/x^7/(x^8-3*x^4+1),x, algorithm="giac")

[Out]

-11/40*sqrt(5)*log(abs(2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 11/40*sqrt(5)*log(abs(2*x^2 - sqrt(5) - 1
)/abs(2*x^2 + sqrt(5) - 1)) - 1/6*(9*x^4 + 1)/x^6 - 5/8*log(abs(x^4 + x^2 - 1)) + 5/8*log(abs(x^4 - x^2 - 1))

Mupad [B] (verification not implemented)

Time = 8.59 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^7 \left (1-3 x^4+x^8\right )} \, dx=\mathrm {atanh}\left (\frac {4126100\,x^2}{1140425\,\sqrt {5}-2550075}-\frac {1845250\,\sqrt {5}\,x^2}{1140425\,\sqrt {5}-2550075}\right )\,\left (\frac {11\,\sqrt {5}}{20}-\frac {5}{4}\right )+\mathrm {atanh}\left (\frac {4126100\,x^2}{1140425\,\sqrt {5}+2550075}+\frac {1845250\,\sqrt {5}\,x^2}{1140425\,\sqrt {5}+2550075}\right )\,\left (\frac {11\,\sqrt {5}}{20}+\frac {5}{4}\right )-\frac {\frac {3\,x^4}{2}+\frac {1}{6}}{x^6} \]

[In]

int(1/(x^7*(x^8 - 3*x^4 + 1)),x)

[Out]

atanh((4126100*x^2)/(1140425*5^(1/2) - 2550075) - (1845250*5^(1/2)*x^2)/(1140425*5^(1/2) - 2550075))*((11*5^(1
/2))/20 - 5/4) + atanh((4126100*x^2)/(1140425*5^(1/2) + 2550075) + (1845250*5^(1/2)*x^2)/(1140425*5^(1/2) + 25
50075))*((11*5^(1/2))/20 + 5/4) - ((3*x^4)/2 + 1/6)/x^6