Integrand size = 10, antiderivative size = 67 \[ \int \frac {1}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:
-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3 ^(1/2)+1/6*arctanh(3^(1/2)*x/(x^2+1))*3^(1/2)
Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {2 \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-\log \left (-1+\sqrt {3} x-x^2\right )+\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \] Input:
Integrate[(1 + x^4 + x^8)^(-1),x]
Output:
(2*ArcTan[(-1 + 2*x)/Sqrt[3]] + 2*ArcTan[(1 + 2*x)/Sqrt[3]] - Log[-1 + Sqr t[3]*x - x^2] + Log[1 + Sqrt[3]*x + x^2])/(4*Sqrt[3])
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.37, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {1684, 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.
\(\displaystyle \int \frac {1}{x^8+x^4+1} \, dx\) |
\(\Big \downarrow \) 1684 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )\) |
Input:
Int[(1 + x^4 + x^8)^(-1),x]
Output:
(ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3])/2 + (-1/2*Log[1 - Sqrt[3]*x + x^2]/Sqrt[3] + Log[1 + Sqrt[3]*x + x^2]/(2*Sq rt[3]))/2
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])
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]
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]
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]))
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]
Int[((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) Int[(r - x^( n/2))/(q - r*x^(n/2) + x^n), x], x] + Simp[1/(2*c*q*r) Int[(r + x^(n/2))/ (q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && N eQ[b^2 - 4*a*c, 0] && IGtQ[n/2, 0] && NegQ[b^2 - 4*a*c]
Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}-\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}\) | \(67\) |
risch | \(\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}-\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, x}{3}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, x^{3}}{3}+\frac {2 \sqrt {3}\, x}{3}\right )}{6}\) | \(68\) |
Input:
int(1/(x^8+x^4+1),x,method=_RETURNVERBOSE)
Output:
1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/6*arctan(1/3*(2*x+1)*3^(1/2))*3^ (1/2)+1/12*3^(1/2)*ln(x^2+3^(1/2)*x+1)-1/12*3^(1/2)*ln(x^2-3^(1/2)*x+1)
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.04 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + \frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) \] Input:
integrate(1/(x^8+x^4+1),x, algorithm="fricas")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(x^3 + 2*x)) + 1/6*sqrt(3)*arctan(1/3*sqrt( 3)*x) + 1/12*sqrt(3)*log((x^4 + 5*x^2 + 2*sqrt(3)*(x^3 + x) + 1)/(x^4 - x^ 2 + 1))
Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {\sqrt {3} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{12} - \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} \] Input:
integrate(1/(x**8+x**4+1),x)
Output:
sqrt(3)*(2*atan(sqrt(3)*x/3) + 2*atan(sqrt(3)*x**3/3 + 2*sqrt(3)*x/3))/12 - sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 + sqrt(3)*log(x**2 + sqrt(3)*x + 1) /12
\[ \int \frac {1}{1+x^4+x^8} \, dx=\int { \frac {1}{x^{8} + x^{4} + 1} \,d x } \] Input:
integrate(1/(x^8+x^4+1),x, algorithm="maxima")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) - 1/2*integrate((x^2 - 1)/(x^4 - x^2 + 1), x)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) \] Input:
integrate(1/(x^8+x^4+1),x, algorithm="giac")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) + 1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) - 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1)
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.60 \[ \int \frac {1}{1+x^4+x^8} \, dx=-\frac {\sqrt {3}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}-\frac {2}{3}\right )}\right )-\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )\right )}{6} \] Input:
int(1/(x^4 + x^8 + 1),x)
Output:
-(3^(1/2)*(atan((2*3^(1/2)*x)/(3*((2*x^2)/3 - 2/3))) - atanh((2*3^(1/2)*x) /(3*((2*x^2)/3 + 2/3)))))/6
Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {\sqrt {3}\, \left (2 \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )+2 \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )-\mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )+\mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )\right )}{12} \] Input:
int(1/(x^8+x^4+1),x)
Output:
(sqrt(3)*(2*atan((2*x - 1)/sqrt(3)) + 2*atan((2*x + 1)/sqrt(3)) - log( - s qrt(3)*x + x**2 + 1) + log(sqrt(3)*x + x**2 + 1)))/12