Integrand size = 16, antiderivative size = 61 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=x+\frac {x^3}{3}+\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \arctan \left (\sqrt {3}+2 x\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:
x+1/3*x^3-1/2*arctan(-3^(1/2)+2*x)-1/2*arctan(3^(1/2)+2*x)-1/6*arctanh(3^( 1/2)*x/(x^2+1))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.39 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=\frac {1}{6} \left (2 x \left (3+x^2\right )-i \sqrt {-6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )+i \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )\right ) \] Input:
Integrate[x^6/(1 - x^2 + x^4),x]
Output:
(2*x*(3 + x^2) - I*Sqrt[-6 - (6*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] + I*Sqrt[-6 + (6*I)*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x)/2])/6
Time = 0.51 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.59, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {1442, 27, 1602, 25, 1407, 1142, 25, 1083, 217, 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.
\(\displaystyle \int \frac {x^6}{x^4-x^2+1} \, dx\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {x^3}{3}-\frac {1}{3} \int \frac {3 x^2 \left (1-x^2\right )}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^3}{3}-\int \frac {x^2 \left (1-x^2\right )}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \int -\frac {1}{x^4-x^2+1}dx+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{x^4-x^2+1}dx+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 1407 |
\(\displaystyle -\frac {\int \frac {\sqrt {3}-x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle -\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx-\frac {1}{2} \int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2-\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\frac {1}{2} \sqrt {3} \int \frac {1}{x^2+\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx-\sqrt {3} \int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )}{2 \sqrt {3}}-\frac {\frac {1}{2} \int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx+\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle -\frac {-\sqrt {3} \arctan \left (\sqrt {3}-2 x\right )-\frac {1}{2} \log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\sqrt {3} \arctan \left (2 x+\sqrt {3}\right )+\frac {1}{2} \log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}+\frac {x^3}{3}+x\) |
Input:
Int[x^6/(1 - x^2 + x^4),x]
Output:
x + x^3/3 - (-(Sqrt[3]*ArcTan[Sqrt[3] - 2*x]) - Log[1 - Sqrt[3]*x + x^2]/2 )/(2*Sqrt[3]) - (Sqrt[3]*ArcTan[Sqrt[3] + 2*x] + Log[1 + Sqrt[3]*x + x^2]/ 2)/(2*Sqrt[3])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-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)/(q - r* x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(r + x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[d^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[d^4/(c*(m + 4*p + 1)) Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x ] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2* p] && (IntegerQ[p] || IntegerQ[m])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69
method | result | size |
risch | \(\frac {x^{3}}{3}+x -\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3}-\textit {\_R}}\right )}{2}\) | \(42\) |
default | \(\frac {x^{3}}{3}+x +\frac {\sqrt {3}\, \ln \left (x^{2}-\sqrt {3}\, x +1\right )}{12}-\frac {\arctan \left (2 x -\sqrt {3}\right )}{2}-\frac {\sqrt {3}\, \ln \left (x^{2}+\sqrt {3}\, x +1\right )}{12}-\frac {\arctan \left (2 x +\sqrt {3}\right )}{2}\) | \(63\) |
Input:
int(x^6/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
1/3*x^3+x-1/2*sum(1/(2*_R^3-_R)*ln(x-_R),_R=RootOf(_Z^4-_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=\frac {1}{3} \, x^{3} - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + x - \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{2} \, \arctan \left (-2 \, x + \sqrt {3}\right ) \] Input:
integrate(x^6/(x^4-x^2+1),x, algorithm="fricas")
Output:
1/3*x^3 - 1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - s qrt(3)*x + 1) + x - 1/2*arctan(2*x + sqrt(3)) + 1/2*arctan(-2*x + sqrt(3))
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=\frac {x^{3}}{3} + x + \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} - \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{2} - \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{2} \] Input:
integrate(x**6/(x**4-x**2+1),x)
Output:
x**3/3 + x + sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqr t(3)*x + 1)/12 - atan(2*x - sqrt(3))/2 - atan(2*x + sqrt(3))/2
\[ \int \frac {x^6}{1-x^2+x^4} \, dx=\int { \frac {x^{6}}{x^{4} - x^{2} + 1} \,d x } \] Input:
integrate(x^6/(x^4-x^2+1),x, algorithm="maxima")
Output:
1/3*x^3 + x - integrate(1/(x^4 - x^2 + 1), x)
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=\frac {1}{3} \, x^{3} - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + x - \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{2} \, \arctan \left (2 \, x - \sqrt {3}\right ) \] Input:
integrate(x^6/(x^4-x^2+1),x, algorithm="giac")
Output:
1/3*x^3 - 1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - s qrt(3)*x + 1) + x - 1/2*arctan(2*x + sqrt(3)) - 1/2*arctan(2*x - sqrt(3))
Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=x-\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {x^3}{3} \] Input:
int(x^6/(x^4 - x^2 + 1),x)
Output:
x - atan((2*x)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/6 - 1/2) - atan((2*x)/(3^(1 /2)*1i + 1))*((3^(1/2)*1i)/6 + 1/2) + x^3/3
Time = 0.15 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {x^6}{1-x^2+x^4} \, dx=\frac {\mathit {atan} \left (\sqrt {3}-2 x \right )}{2}-\frac {\mathit {atan} \left (\sqrt {3}+2 x \right )}{2}+\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{12}+\frac {x^{3}}{3}+x \] Input:
int(x^6/(x^4-x^2+1),x)
Output:
(6*atan(sqrt(3) - 2*x) - 6*atan(sqrt(3) + 2*x) + sqrt(3)*log( - sqrt(3)*x + x**2 + 1) - sqrt(3)*log(sqrt(3)*x + x**2 + 1) + 4*x**3 + 12*x)/12