\(\int \frac {1}{x^6 (1+2 x^4+x^8)} \, dx\) [288]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=-\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]

[Out]

-9/20/x^5+9/4/x+1/4/x^5/(x^4+1)+9/16*arctan(-1+x*2^(1/2))*2^(1/2)+9/16*arctan(1+x*2^(1/2))*2^(1/2)+9/32*ln(1+x
^2-x*2^(1/2))*2^(1/2)-9/32*ln(1+x^2+x*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 296, 331, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=-\frac {9 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \arctan \left (\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {9}{20 x^5}+\frac {9 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {9 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^5 \left (x^4+1\right )}+\frac {9}{4 x} \]

[In]

Int[1/(x^6*(1 + 2*x^4 + x^8)),x]

[Out]

-9/(20*x^5) + 9/(4*x) + 1/(4*x^5*(1 + x^4)) - (9*ArcTan[1 - Sqrt[2]*x])/(8*Sqrt[2]) + (9*ArcTan[1 + Sqrt[2]*x]
)/(8*Sqrt[2]) + (9*Log[1 - Sqrt[2]*x + x^2])/(16*Sqrt[2]) - (9*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 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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

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 331

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 631

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 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 1176

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 1179

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^6 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{4} \int \frac {1}{x^6 \left (1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9}{4} \int \frac {1}{x^2 \left (1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{4} \int \frac {x^2}{1+x^4} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9}{8} \int \frac {1-x^2}{1+x^4} \, dx+\frac {9}{8} \int \frac {1+x^2}{1+x^4} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {9}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {9 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {9 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}} \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {9 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {9 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}} \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{160} \left (-\frac {32}{x^5}+\frac {320}{x}+\frac {40 x^3}{1+x^4}-90 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )+90 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+45 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-45 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]

[In]

Integrate[1/(x^6*(1 + 2*x^4 + x^8)),x]

[Out]

(-32/x^5 + 320/x + (40*x^3)/(1 + x^4) - 90*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 90*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] +
45*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 45*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/160

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.39

method	result	size
risch	\(\frac {\frac {9}{4} x^{8}+\frac {9}{5} x^{4}-\frac {1}{5}}{x^{5} \left (x^{4}+1\right )}+\frac {9 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{3}+x \right )\right )}{16}\)	\(44\)
default	\(-\frac {1}{5 x^{5}}+\frac {2}{x}+\frac {x^{3}}{4 x^{4}+4}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\)	\(75\)

[In]

int(1/x^6/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

(9/4*x^8+9/5*x^4-1/5)/x^5/(x^4+1)+9/16*sum(_R*ln(_R^3+x),_R=RootOf(_Z^4+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {360 \, x^{8} + 288 \, x^{4} - 45 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{9} - \left (i - 1\right ) \, x^{5}\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{9} + \left (i + 1\right ) \, x^{5}\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{9} - \left (i + 1\right ) \, x^{5}\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{9} + \left (i - 1\right ) \, x^{5}\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) - 32}{160 \, {\left (x^{9} + x^{5}\right )}} \]

[In]

integrate(1/x^6/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

1/160*(360*x^8 + 288*x^4 - 45*sqrt(2)*(-(I - 1)*x^9 - (I - 1)*x^5)*log(2*x + (I + 1)*sqrt(2)) - 45*sqrt(2)*((I
+ 1)*x^9 + (I + 1)*x^5)*log(2*x - (I - 1)*sqrt(2)) - 45*sqrt(2)*(-(I + 1)*x^9 - (I + 1)*x^5)*log(2*x + (I - 1
)*sqrt(2)) - 45*sqrt(2)*((I - 1)*x^9 + (I - 1)*x^5)*log(2*x - (I + 1)*sqrt(2)) - 32)/(x^9 + x^5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {9 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {9 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} + \frac {45 x^{8} + 36 x^{4} - 4}{20 x^{9} + 20 x^{5}} \]

[In]

integrate(1/x**6/(x**8+2*x**4+1),x)

[Out]

9*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 9*sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 + 9*sqrt(2)*atan(sqrt(2)*x - 1
)/16 + 9*sqrt(2)*atan(sqrt(2)*x + 1)/16 + (45*x**8 + 36*x**4 - 4)/(20*x**9 + 20*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {9}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {9}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {45 \, x^{8} + 36 \, x^{4} - 4}{20 \, {\left (x^{9} + x^{5}\right )}} \]

[In]

integrate(1/x^6/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

9/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 9/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 9/32*sqr
t(2)*log(x^2 + sqrt(2)*x + 1) + 9/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/20*(45*x^8 + 36*x^4 - 4)/(x^9 + x^5)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {9}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {9}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {10 \, x^{4} - 1}{5 \, x^{5}} \]

[In]

integrate(1/x^6/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

1/4*x^3/(x^4 + 1) + 9/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 9/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x -
sqrt(2))) - 9/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 9/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) + 1/5*(10*x^4 - 1)/x
^5

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {\frac {9\,x^8}{4}+\frac {9\,x^4}{5}-\frac {1}{5}}{x^9+x^5}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {9}{16}-\frac {9}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {9}{16}+\frac {9}{16}{}\mathrm {i}\right ) \]

[In]

int(1/(x^6*(2*x^4 + x^8 + 1)),x)

[Out]

2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(9/16 - 9i/16) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(9/16 + 9i/16) + ((
9*x^4)/5 + (9*x^8)/4 - 1/5)/(x^5 + x^9)