Integrand size = 14, antiderivative size = 63 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \text {arctanh}\left (\frac {x^2}{1+x^4}\right ) \] Output:
-1/12*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))
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.49 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {\sqrt {1-i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x^2\right )+\sqrt {1+i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x^2\right )}{4 \sqrt {6}} \] Input:
Integrate[x^5/(1 + x^4 + x^8),x]
Output:
(Sqrt[1 - I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan[((-I + Sqrt[3])*x^2)/2] + Sqrt[ 1 + I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[((I + Sqrt[3])*x^2)/2])/(4*Sqrt[6])
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.32, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1695, 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.
\(\displaystyle \int \frac {x^5}{x^8+x^4+1} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {x^4}{x^8+x^4+1}dx^2\) |
\(\Big \downarrow \) 1447 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {x^4+1}{x^8+x^4+1}dx^2-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-x^2+1}dx^2+\frac {1}{2} \int \frac {1}{x^4+x^2+1}dx^2\right )-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\int \frac {1}{-x^4-3}d\left (2 x^2-1\right )-\int \frac {1}{-x^4-3}d\left (2 x^2+1\right )\right )-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {1-x^4}{x^8+x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx^2+\frac {1}{2} \int -\frac {2 x^2+1}{x^4+x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-\frac {1}{2} \int \frac {2 x^2+1}{x^4+x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \log \left (x^4-x^2+1\right )-\frac {1}{2} \log \left (x^4+x^2+1\right )\right )\right )\) |
Input:
Int[x^5/(1 + x^4 + x^8),x]
Output:
((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
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[(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]
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[(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]
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\) | \(62\) |
risch | \(\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (4 x^{4}-4 x^{2}+4\right )}{8}\) | \(68\) |
Input:
int(x^5/(x^8+x^4+1),x,method=_RETURNVERBOSE)
Output:
1/8*ln(x^4-x^2+1)+1/12*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))-1/8*ln(x^4+x^ 2+1)+1/12*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x^5/(x^8+x^4+1),x, algorithm="fricas")
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)) - 1/8*log(x^4 + x^2 + 1) + 1/8*log(x^4 - x^2 + 1)
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.21 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \] Input:
integrate(x**5/(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
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x^5/(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)) - 1/8*log(x^4 + x^2 + 1) + 1/8*log(x^4 - x^2 + 1)
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \] Input:
integrate(x^5/(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)) - 1/8*log(x^4 + x^2 + 1) + 1/8*log(x^4 - x^2 + 1)
Time = 19.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\mathrm {atanh}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atanh}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \] Input:
int(x^5/(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)
Time = 0.15 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}-2 x \right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\sqrt {3}+2 x \right )}{12}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{12}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{12}-\frac {\mathrm {log}\left (x^{2}-x +1\right )}{8}-\frac {\mathrm {log}\left (x^{2}+x +1\right )}{8}+\frac {\mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{8}+\frac {\mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{8} \] Input:
int(x^5/(x^8+x^4+1),x)
Output:
( - 2*sqrt(3)*atan(sqrt(3) - 2*x) - 2*sqrt(3)*atan(sqrt(3) + 2*x) + 2*sqrt (3)*atan((2*x - 1)/sqrt(3)) - 2*sqrt(3)*atan((2*x + 1)/sqrt(3)) - 3*log(x* *2 - x + 1) - 3*log(x**2 + x + 1) + 3*log( - sqrt(3)*x + x**2 + 1) + 3*log (sqrt(3)*x + x**2 + 1))/24