\(\int \frac {1}{x^2 (1+2 x^4+x^8)} \, dx\) [37]

Optimal result

Integrand size = 16, antiderivative size = 83 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{x}-\frac {x^3}{4 \left (1+x^4\right )}+\frac {5 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {5 \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}} \] Output:

-1/x-x^3/(4*x^4+4)-5/16*arctan(-1+x*2^(1/2))*2^(1/2)-5/16*arctan(1+x*2^(1/ 
2))*2^(1/2)+5/16*arctanh(2^(1/2)*x/(x^2+1))*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{32} \left (-\frac {32}{x}-\frac {8 x^3}{1+x^4}+10 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-10 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )-5 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )+5 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:

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

Output:

(-32/x - (8*x^3)/(1 + x^4) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 10*Sqrt[2] 
*ArcTan[1 + Sqrt[2]*x] - 5*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] + 5*Sqrt[2]*Lo 
g[1 + Sqrt[2]*x + x^2])/32
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.37, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {1380, 819, 847, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x^2 \left (x^8+2 x^4+1\right )} \, dx\)

\(\Big \downarrow \) 1380

\(\displaystyle \int \frac {1}{x^2 \left (x^4+1\right )^2}dx\)

\(\Big \downarrow \) 819

\(\displaystyle \frac {5}{4} \int \frac {1}{x^2 \left (x^4+1\right )}dx+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {5}{4} \left (-\int \frac {x^2}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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 819

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] + Simp[(m + n*(p + 
1) + 1)/(a*n*(p + 1))   Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a 
, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 847

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] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
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 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.49

method	result	size
risch	\(\frac {-\frac {5 x^{4}}{4}-1}{x \left (x^{4}+1\right )}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R}^{3}+x \right )\right )}{16}\)	\(41\)
default	\(-\frac {x^{3}}{4 \left (x^{4}+1\right )}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-x \sqrt {2}+1}{x^{2}+x \sqrt {2}+1}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{32}-\frac {1}{x}\)	\(70\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {40 \, x^{4} + 10 \, \sqrt {2} {\left (x^{5} + x\right )} \arctan \left (\sqrt {2} x + 1\right ) + 10 \, \sqrt {2} {\left (x^{5} + x\right )} \arctan \left (\sqrt {2} x - 1\right ) - 5 \, \sqrt {2} {\left (x^{5} + x\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) + 5 \, \sqrt {2} {\left (x^{5} + x\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + 32}{32 \, {\left (x^{5} + x\right )}} \] Input:

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

Output:

-1/32*(40*x^4 + 10*sqrt(2)*(x^5 + x)*arctan(sqrt(2)*x + 1) + 10*sqrt(2)*(x 
^5 + x)*arctan(sqrt(2)*x - 1) - 5*sqrt(2)*(x^5 + x)*log(x^2 + sqrt(2)*x + 
1) + 5*sqrt(2)*(x^5 + x)*log(x^2 - sqrt(2)*x + 1) + 32)/(x^5 + x)
 

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 5 x^{4} - 4}{4 x^{5} + 4 x} - \frac {5 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} + \frac {5 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \] Input:

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

Output:

(-5*x**4 - 4)/(4*x**5 + 4*x) - 5*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 + 5* 
sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 - 5*sqrt(2)*atan(sqrt(2)*x - 1)/16 - 
5*sqrt(2)*atan(sqrt(2)*x + 1)/16
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {5 \, x^{4} + 4}{4 \, {\left (x^{5} + x\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {5 \, x^{4} + 4}{4 \, {\left (x^{5} + x\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 19.85 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {\frac {5\,x^4}{4}+1}{x^5+x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}+\frac {5}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}-\frac {5}{16}{}\mathrm {i}\right ) \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.89 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=\frac {10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{5}+10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x -10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{5}-10 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x -5 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{5}-5 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x +5 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{5}+5 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x -40 x^{4}-32}{32 x \left (x^{4}+1\right )} \] Input:

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

Output:

(10*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**5 + 10*sqrt(2)*atan((sqrt(2) 
- 2*x)/sqrt(2))*x - 10*sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**5 - 10*sqr 
t(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x - 5*sqrt(2)*log( - sqrt(2)*x + x**2 + 
 1)*x**5 - 5*sqrt(2)*log( - sqrt(2)*x + x**2 + 1)*x + 5*sqrt(2)*log(sqrt(2 
)*x + x**2 + 1)*x**5 + 5*sqrt(2)*log(sqrt(2)*x + x**2 + 1)*x - 40*x**4 - 3 
2)/(32*x*(x**4 + 1))