\(\int \frac {1}{x^7 (1+x^4+x^8)} \, dx\) [72]

Optimal result

Integrand size = 14, antiderivative size = 77 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}+\frac {1}{2 x^2}-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \text {arctanh}\left (\frac {x^2}{1+x^4}\right ) \] Output:

-1/6/x^6+1/2/x^2-1/12*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)+1/12*arctan(1 
/3*(2*x^2+1)*3^(1/2))*3^(1/2)-1/4*arctanh(x^2/(x^4+1))
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{24} \left (-\frac {4}{x^6}+\frac {12}{x^2}-2 \sqrt {3} \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )-3 \log \left (1-x+x^2\right )-3 \log \left (1+x+x^2\right )+3 \log \left (1-x^2+x^4\right )\right ) \] Input:

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

Output:

(-4/x^6 + 12/x^2 - 2*Sqrt[3]*ArcTan[(1 - 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[ 
(1 + 2*x)/Sqrt[3]] - 2*Sqrt[3]*ArcTan[(1 - 2*x^2)/Sqrt[3]] - 3*Log[1 - x + 
 x^2] - 3*Log[1 + x + x^2] + 3*Log[1 - x^2 + x^4])/24
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {1695, 1443, 27, 1604, 1447, 1475, 1083, 217, 1478, 25, 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.

\(\displaystyle \int \frac {1}{x^7 \left (x^8+x^4+1\right )} \, dx\)

\(\Big \downarrow \) 1695

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

\(\Big \downarrow \) 1443

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int -\frac {3 \left (x^4+1\right )}{x^4 \left (x^8+x^4+1\right )}dx^2-\frac {1}{3 x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\int \frac {x^4+1}{x^4 \left (x^8+x^4+1\right )}dx^2-\frac {1}{3 x^6}\right )\)

\(\Big \downarrow \) 1604

\(\displaystyle \frac {1}{2} \left (\int \frac {x^4}{x^8+x^4+1}dx^2-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1447

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2+\frac {1}{2} \int \frac {x^4+1}{x^8+x^4+1}dx^2-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1475

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-x^2+1}dx^2+\frac {1}{2} \int \frac {1}{x^4+x^2+1}dx^2\right )-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\int \frac {1}{-x^4-3}d\left (2 x^2-1\right )-\int \frac {1}{-x^4-3}d\left (2 x^2+1\right )\right )-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1478

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx^2+\frac {1}{2} \int -\frac {2 x^2+1}{x^4+x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{3 x^6}+\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{3 x^6}+\frac {1}{x^2}+\frac {1}{2} \left (\frac {1}{2} \log \left (x^4-x^2+1\right )-\frac {1}{2} \log \left (x^4+x^2+1\right )\right )\right )\)

Input:

Int[1/(x^7*(1 + x^4 + x^8)),x]
 

Output:

(-1/3*1/x^6 + x^(-2) + (ArcTan[(-1 + 2*x^2)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 
 2*x^2)/Sqrt[3]]/Sqrt[3])/2 + (Log[1 - x^2 + x^4]/2 - Log[1 + x^2 + x^4]/2 
)/2)/2
 

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 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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 1443

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] - Sim 
p[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] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1447

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a/c, 2]}, Simp[1/2   Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 
 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b 
^2 - 4*a*c, 0] && PosQ[a*c]
 

rule 1475

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

rule 1478

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[-2*(d/e) - b/c, 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]] /; 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 1604

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] + Simp[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 1695

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] 
 :> With[{k = GCD[m + 1, n]}, Simp[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]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\frac {x^{4}}{2}-\frac {1}{6}}{x^{6}}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \sqrt {3}\, \left (x^{2}+\frac {1}{2}\right )}{3}\right )}{12}-\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}\)	\(69\)
default	\(-\frac {\ln \left (x^{2}-x +1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {1}{6 x^{6}}+\frac {1}{2 x^{2}}\)	\(95\)

Input:

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

Output:

(1/2*x^4-1/6)/x^6+1/12*3^(1/2)*arctan(2/3*(x^2-1/2)*3^(1/2))+1/8*ln(x^4-x^ 
2+1)+1/12*3^(1/2)*arctan(2/3*3^(1/2)*(x^2+1/2))-1/8*ln(x^4+x^2+1)
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + 2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - 3 \, x^{6} \log \left (x^{4} + x^{2} + 1\right ) + 3 \, x^{6} \log \left (x^{4} - x^{2} + 1\right ) + 12 \, x^{4} - 4}{24 \, x^{6}} \] Input:

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

Output:

1/24*(2*sqrt(3)*x^6*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 2*sqrt(3)*x^6*arctan 
(1/3*sqrt(3)*(2*x^2 - 1)) - 3*x^6*log(x^4 + x^2 + 1) + 3*x^6*log(x^4 - x^2 
 + 1) + 12*x^4 - 4)/x^6
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} + \frac {3 x^{4} - 1}{6 x^{6}} \] Input:

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

Output:

log(x**4 - x**2 + 1)/8 - log(x**4 + x**2 + 1)/8 + sqrt(3)*atan(2*sqrt(3)*x 
**2/3 - sqrt(3)/3)/12 + sqrt(3)*atan(2*sqrt(3)*x**2/3 + sqrt(3)/3)/12 + (3 
*x**4 - 1)/(6*x**6)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:

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

Output:

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqr 
t(3)*(2*x^2 - 1)) + 1/6*(3*x^4 - 1)/x^6 - 1/8*log(x^4 + x^2 + 1) + 1/8*log 
(x^4 - x^2 + 1)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:

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

Output:

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqr 
t(3)*(2*x^2 - 1)) + 1/6*(3*x^4 - 1)/x^6 - 1/8*log(x^4 + x^2 + 1) + 1/8*log 
(x^4 - x^2 + 1)
 

Mupad [B] (verification not implemented)

Time = 19.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\mathrm {atanh}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atanh}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\frac {\frac {x^4}{2}-\frac {1}{6}}{x^6} \] Input:

int(1/(x^7*(x^4 + x^8 + 1)),x)
 

Output:

atanh((2*x^2)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/12 + 1/4) + atanh((2*x^2)/(3 
^(1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1/4) + (x^4/2 - 1/6)/x^6
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.73 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\sqrt {3}-2 x \right ) x^{6}-2 \sqrt {3}\, \mathit {atan} \left (\sqrt {3}+2 x \right ) x^{6}+2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) x^{6}-2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) x^{6}-3 \,\mathrm {log}\left (x^{2}-x +1\right ) x^{6}-3 \,\mathrm {log}\left (x^{2}+x +1\right ) x^{6}+3 \,\mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right ) x^{6}+3 \,\mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right ) x^{6}+12 x^{4}-4}{24 x^{6}} \] Input:

int(1/x^7/(x^8+x^4+1),x)
 

Output:

( - 2*sqrt(3)*atan(sqrt(3) - 2*x)*x**6 - 2*sqrt(3)*atan(sqrt(3) + 2*x)*x** 
6 + 2*sqrt(3)*atan((2*x - 1)/sqrt(3))*x**6 - 2*sqrt(3)*atan((2*x + 1)/sqrt 
(3))*x**6 - 3*log(x**2 - x + 1)*x**6 - 3*log(x**2 + x + 1)*x**6 + 3*log( - 
 sqrt(3)*x + x**2 + 1)*x**6 + 3*log(sqrt(3)*x + x**2 + 1)*x**6 + 12*x**4 - 
 4)/(24*x**6)