\(\int \frac {1}{x^4 (1+2 x^4+x^8)} \, dx\) [287]

Optimal result

Integrand size = 16, antiderivative size = 106 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]

[Out]

-7/12/x^3+1/4/x^3/(x^4+1)-7/16*arctan(-1+x*2^(1/2))*2^(1/2)-7/16*arctan(1+x*2^(1/2))*2^(1/2)+7/32*ln(1+x^2-x*2
^(1/2))*2^(1/2)-7/32*ln(1+x^2+x*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 296, 331, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {7 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \arctan \left (\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {7}{12 x^3}+\frac {7 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {7 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^3 \left (x^4+1\right )} \]

[In]

Int[1/(x^4*(1 + 2*x^4 + x^8)),x]

[Out]

-7/(12*x^3) + 1/(4*x^3*(1 + x^4)) + (7*ArcTan[1 - Sqrt[2]*x])/(8*Sqrt[2]) - (7*ArcTan[1 + Sqrt[2]*x])/(8*Sqrt[
2]) + (7*Log[1 - Sqrt[2]*x + x^2])/(16*Sqrt[2]) - (7*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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 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^4 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7}{4} \int \frac {1}{x^4 \left (1+x^4\right )} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{4} \int \frac {1}{1+x^4} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{8} \int \frac {1-x^2}{1+x^4} \, dx-\frac {7}{8} \int \frac {1+x^2}{1+x^4} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {7}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {7 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {7 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}} \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}} \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{96} \left (-\frac {32}{x^3}-\frac {24 x}{1+x^4}+42 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-42 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+21 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-21 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]

[In]

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

[Out]

(-32/x^3 - (24*x)/(1 + x^4) + 42*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 42*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] + 21*Sqrt[2]
*Log[1 - Sqrt[2]*x + x^2] - 21*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/96

Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.37

method	result	size
risch	\(\frac {-\frac {7 x^{4}}{12}-\frac {1}{3}}{x^{3} \left (x^{4}+1\right )}+\frac {7 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{16}\)	\(39\)
default	\(-\frac {1}{3 x^{3}}-\frac {x}{4 \left (x^{4}+1\right )}-\frac {7 \sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\)	\(68\)

[In]

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

[Out]

(-7/12*x^4-1/3)/x^3/(x^4+1)+7/16*sum(_R*ln(x-_R),_R=RootOf(_Z^4+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {56 \, x^{4} + 21 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{7} + \left (i + 1\right ) \, x^{3}\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{7} - \left (i - 1\right ) \, x^{3}\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{7} + \left (i - 1\right ) \, x^{3}\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{7} - \left (i + 1\right ) \, x^{3}\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) + 32}{96 \, {\left (x^{7} + x^{3}\right )}} \]

[In]

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

[Out]

-1/96*(56*x^4 + 21*sqrt(2)*((I + 1)*x^7 + (I + 1)*x^3)*log(2*x + (I + 1)*sqrt(2)) + 21*sqrt(2)*(-(I - 1)*x^7 -
(I - 1)*x^3)*log(2*x - (I - 1)*sqrt(2)) + 21*sqrt(2)*((I - 1)*x^7 + (I - 1)*x^3)*log(2*x + (I - 1)*sqrt(2)) +
21*sqrt(2)*(-(I + 1)*x^7 - (I + 1)*x^3)*log(2*x - (I + 1)*sqrt(2)) + 32)/(x^7 + x^3)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 7 x^{4} - 4}{12 x^{7} + 12 x^{3}} + \frac {7 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]

[In]

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

[Out]

(-7*x**4 - 4)/(12*x**7 + 12*x**3) + 7*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 7*sqrt(2)*log(x**2 + sqrt(2)*x +
1)/32 - 7*sqrt(2)*atan(sqrt(2)*x - 1)/16 - 7*sqrt(2)*atan(sqrt(2)*x + 1)/16

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {7 \, x^{4} + 4}{12 \, {\left (x^{7} + x^{3}\right )}} \]

[In]

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

[Out]

-7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 7/32*sq
rt(2)*log(x^2 + sqrt(2)*x + 1) + 7/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/12*(7*x^4 + 4)/(x^7 + x^3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{3 \, x^{3}} \]

[In]

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

[Out]

-7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 7/32*sq
rt(2)*log(x^2 + sqrt(2)*x + 1) + 7/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*x/(x^4 + 1) - 1/3/x^3

Mupad [B] (verification not implemented)

Time = 8.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {\frac {7\,x^4}{12}+\frac {1}{3}}{x^7+x^3}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}-\frac {7}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}+\frac {7}{16}{}\mathrm {i}\right ) \]

[In]

int(1/(x^4*(2*x^4 + x^8 + 1)),x)

[Out]

- 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(7/16 + 7i/16) - 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(7/16 - 7i/16) -
((7*x^4)/12 + 1/3)/(x^3 + x^7)