Integrand size = 16, antiderivative size = 59 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x^2}{1+x^4}\right )}{4 \sqrt {3}} \] Output:
1/4*arctan(2*x^2-3^(1/2))+1/4*arctan(3^(1/2)+2*x^2)-1/12*arctanh(3^(1/2)*x ^2/(x^4+1))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.66 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {\sqrt {-1-i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )+\sqrt {-1+i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{4 \sqrt {6}} \] Input:
Integrate[x^5/(1 - x^4 + x^8),x]
Output:
(Sqrt[-1 - I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*x^2)/2] + Sqrt [-1 + I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan[((1 + I*Sqrt[3])*x^2)/2])/(4*Sqrt[6 ])
Time = 0.30 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.53, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^4-\sqrt {3} x^2+1}dx^2+\frac {1}{2} \int \frac {1}{x^4+\sqrt {3} x^2+1}dx^2\right )-\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\int \frac {1}{-x^4-1}d\left (2 x^2-\sqrt {3}\right )-\int \frac {1}{-x^4-1}d\left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (2 x^2+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x^2\right )\right )-\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\int -\frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x^2+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x^2\right )\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}-\frac {\int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x^2+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x^2\right )\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (2 x^2+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x^2\right )\right )+\frac {1}{2} \left (\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{2 \sqrt {3}}\right )\right )\) |
Input:
Int[x^5/(1 - x^4 + x^8),x]
Output:
((-ArcTan[Sqrt[3] - 2*x^2] + ArcTan[Sqrt[3] + 2*x^2])/2 + (Log[1 - Sqrt[3] *x^2 + x^4]/(2*Sqrt[3]) - Log[1 + Sqrt[3]*x^2 + x^4]/(2*Sqrt[3]))/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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.54
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (6 \textit {\_R}^{3}+x^{2}+\textit {\_R} \right )\right )}{4}\) | \(32\) |
default | \(-\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{4}+\sqrt {3}\, x^{2}+1\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}+\sqrt {3}\right )\right )}{12}-\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{4}-\sqrt {3}\, x^{2}+1\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}-\sqrt {3}\right )\right )}{12}\) | \(77\) |
Input:
int(x^5/(x^8-x^4+1),x,method=_RETURNVERBOSE)
Output:
1/4*sum(_R*ln(6*_R^3+x^2+_R),_R=RootOf(9*_Z^4+3*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\frac {1}{24} \, \sqrt {3} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, \arctan \left (-2 \, x^{2} + \sqrt {3}\right ) \] Input:
integrate(x^5/(x^8-x^4+1),x, algorithm="fricas")
Output:
-1/24*sqrt(3)*log(x^4 + sqrt(3)*x^2 + 1) + 1/24*sqrt(3)*log(x^4 - sqrt(3)* x^2 + 1) + 1/4*arctan(2*x^2 + sqrt(3)) - 1/4*arctan(-2*x^2 + sqrt(3))
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} + \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} \] Input:
integrate(x**5/(x**8-x**4+1),x)
Output:
sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 - sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 + atan(2*x**2 - sqrt(3))/4 + atan(2*x**2 + sqrt(3))/4
\[ \int \frac {x^5}{1-x^4+x^8} \, dx=\int { \frac {x^{5}}{x^{8} - x^{4} + 1} \,d x } \] Input:
integrate(x^5/(x^8-x^4+1),x, algorithm="maxima")
Output:
integrate(x^5/(x^8 - x^4 + 1), x)
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) - \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) + \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) + \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) \] Input:
integrate(x^5/(x^8-x^4+1),x, algorithm="giac")
Output:
1/24*sqrt(3)*x^4*log(x^4 + sqrt(3)*x^2 + 1) - 1/24*sqrt(3)*x^4*log(x^4 - s qrt(3)*x^2 + 1) + 1/4*x^4*arctan(2*x^2 + sqrt(3)) + 1/4*x^4*arctan(2*x^2 - sqrt(3))
Time = 18.68 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \] Input:
int(x^5/(x^8 - x^4 + 1),x)
Output:
- atan((2*x^2)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/12 + 1/4) - atan((2*x^2)/(3 ^(1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1/4)
Time = 0.17 (sec) , antiderivative size = 336, normalized size of antiderivative = 5.69 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}-4 x}{\sqrt {6}+\sqrt {2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}-4 x}{\sqrt {6}+\sqrt {2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}+4 x}{\sqrt {6}+\sqrt {2}}\right )}{8}-\frac {\sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}+4 x}{\sqrt {6}+\sqrt {2}}\right )}{8}-\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right )}{24}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right )}{24}+\frac {\sqrt {3}\, \mathrm {log}\left (-\frac {\sqrt {6}\, x}{2}-\frac {\sqrt {2}\, x}{2}+x^{2}+1\right )}{24}+\frac {\sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {6}\, x}{2}+\frac {\sqrt {2}\, x}{2}+x^{2}+1\right )}{24} \] Input:
int(x^5/(x^8-x^4+1),x)
Output:
( - 3*sqrt( - sqrt(3) + 2)*sqrt(6)*atan((sqrt(6) + sqrt(2) - 4*x)/(2*sqrt( - sqrt(3) + 2))) - 3*sqrt( - sqrt(3) + 2)*sqrt(2)*atan((sqrt(6) + sqrt(2) - 4*x)/(2*sqrt( - sqrt(3) + 2))) - 3*sqrt( - sqrt(3) + 2)*sqrt(6)*atan((s qrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2))) - 3*sqrt( - sqrt(3) + 2) *sqrt(2)*atan((sqrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2))) - 3*sqrt ( - sqrt(3) + 2)*sqrt(6)*atan((2*sqrt( - sqrt(3) + 2) - 4*x)/(sqrt(6) + sq rt(2))) - 3*sqrt( - sqrt(3) + 2)*sqrt(2)*atan((2*sqrt( - sqrt(3) + 2) - 4* x)/(sqrt(6) + sqrt(2))) - 3*sqrt( - sqrt(3) + 2)*sqrt(6)*atan((2*sqrt( - s qrt(3) + 2) + 4*x)/(sqrt(6) + sqrt(2))) - 3*sqrt( - sqrt(3) + 2)*sqrt(2)*a tan((2*sqrt( - sqrt(3) + 2) + 4*x)/(sqrt(6) + sqrt(2))) - sqrt(3)*log( - s qrt( - sqrt(3) + 2)*x + x**2 + 1) - sqrt(3)*log(sqrt( - sqrt(3) + 2)*x + x **2 + 1) + sqrt(3)*log(( - sqrt(6)*x - sqrt(2)*x + 2*x**2 + 2)/2) + sqrt(3 )*log((sqrt(6)*x + sqrt(2)*x + 2*x**2 + 2)/2))/24