Integrand size = 14, antiderivative size = 81 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=x-\frac {x^5}{5}+\frac {x^7}{7}-\frac {x^{11}}{11}+\frac {x^{13}}{13}-\frac {x^{17}}{17}+\frac {x^{19}}{19}+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}} \] Output:
x-1/5*x^5+1/7*x^7-1/11*x^11+1/13*x^13-1/17*x^17+1/19*x^19+1/3*arctan(1/3*( 1-2*x)*3^(1/2))*3^(1/2)-1/3*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=x-\frac {x^5}{5}+\frac {x^7}{7}-\frac {x^{11}}{11}+\frac {x^{13}}{13}-\frac {x^{17}}{17}+\frac {x^{19}}{19}-\frac {\left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )}{\sqrt {6+6 i \sqrt {3}}}-\frac {\left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{\sqrt {6-6 i \sqrt {3}}} \] Input:
Integrate[x^22/(1 + x^2 + x^4),x]
Output:
x - x^5/5 + x^7/7 - x^11/11 + x^13/13 - x^17/17 + x^19/19 - ((I + Sqrt[3]) *ArcTan[((-I + Sqrt[3])*x)/2])/Sqrt[6 + (6*I)*Sqrt[3]] - ((-I + Sqrt[3])*A rcTan[((I + Sqrt[3])*x)/2])/Sqrt[6 - (6*I)*Sqrt[3]]
Time = 0.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {1442, 27, 1602, 27, 1442, 27, 1602, 27, 1442, 27, 1602, 27, 1442, 1475, 1083, 217}
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.
\(\displaystyle \int \frac {x^{22}}{x^4+x^2+1} \, dx\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {x^{19}}{19}-\frac {1}{19} \int \frac {19 x^{18} \left (x^2+1\right )}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^{19}}{19}-\int \frac {x^{18} \left (x^2+1\right )}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \frac {1}{17} \int \frac {17 x^{16}}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{16}}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle -\frac {1}{13} \int \frac {13 x^{12} \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^{12} \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \frac {1}{11} \int \frac {11 x^{10}}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{10}}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle -\frac {1}{7} \int \frac {7 x^6 \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {x^6 \left (x^2+1\right )}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle \frac {1}{5} \int \frac {5 x^4}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^4}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle -\int \frac {x^2+1}{x^4+x^2+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle -\frac {1}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1}{x^2+x+1}dx+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)+\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}+x\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^7}{7}-\frac {x^5}{5}+x\) |
Input:
Int[x^22/(1 + x^2 + x^4),x]
Output:
x - x^5/5 + x^7/7 - x^11/11 + x^13/13 - x^17/17 + x^19/19 - ArcTan[(-1 + 2 *x)/Sqrt[3]]/Sqrt[3] - ArcTan[(1 + 2*x)/Sqrt[3]]/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_.)*(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[((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[((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])
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^{7}}{7}-\frac {x^{5}}{5}+x -\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\) | \(65\) |
risch | \(\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^{7}}{7}-\frac {x^{5}}{5}+x -\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, x}{3}\right )}{3}-\frac {\sqrt {3}\, \arctan \left (\frac {x^{3} \sqrt {3}}{3}+\frac {2 \sqrt {3}\, x}{3}\right )}{3}\) | \(66\) |
Input:
int(x^22/(x^4+x^2+1),x,method=_RETURNVERBOSE)
Output:
1/19*x^19-1/17*x^17+1/13*x^13-1/11*x^11+1/7*x^7-1/5*x^5+x-1/3*arctan(1/3*( 1+2*x)*3^(1/2))*3^(1/2)-1/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=\frac {1}{19} \, x^{19} - \frac {1}{17} \, x^{17} + \frac {1}{13} \, x^{13} - \frac {1}{11} \, x^{11} + \frac {1}{7} \, x^{7} - \frac {1}{5} \, x^{5} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + x \] Input:
integrate(x^22/(x^4+x^2+1),x, algorithm="fricas")
Output:
1/19*x^19 - 1/17*x^17 + 1/13*x^13 - 1/11*x^11 + 1/7*x^7 - 1/5*x^5 - 1/3*sq rt(3)*arctan(1/3*sqrt(3)*(x^3 + 2*x)) - 1/3*sqrt(3)*arctan(1/3*sqrt(3)*x) + x
Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=\frac {x^{19}}{19} - \frac {x^{17}}{17} + \frac {x^{13}}{13} - \frac {x^{11}}{11} + \frac {x^{7}}{7} - \frac {x^{5}}{5} + x + \frac {\sqrt {3} \left (- 2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} - 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{6} \] Input:
integrate(x**22/(x**4+x**2+1),x)
Output:
x**19/19 - x**17/17 + x**13/13 - x**11/11 + x**7/7 - x**5/5 + x + sqrt(3)* (-2*atan(sqrt(3)*x/3) - 2*atan(sqrt(3)*x**3/3 + 2*sqrt(3)*x/3))/6
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=\frac {1}{19} \, x^{19} - \frac {1}{17} \, x^{17} + \frac {1}{13} \, x^{13} - \frac {1}{11} \, x^{11} + \frac {1}{7} \, x^{7} - \frac {1}{5} \, x^{5} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x \] Input:
integrate(x^22/(x^4+x^2+1),x, algorithm="maxima")
Output:
1/19*x^19 - 1/17*x^17 + 1/13*x^13 - 1/11*x^11 + 1/7*x^7 - 1/5*x^5 - 1/3*sq rt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + x
Time = 0.10 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.79 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=\frac {1}{19} \, x^{19} - \frac {1}{17} \, x^{17} + \frac {1}{13} \, x^{13} - \frac {1}{11} \, x^{11} + \frac {1}{7} \, x^{7} - \frac {1}{5} \, x^{5} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x \] Input:
integrate(x^22/(x^4+x^2+1),x, algorithm="giac")
Output:
1/19*x^19 - 1/17*x^17 + 1/13*x^13 - 1/11*x^11 + 1/7*x^7 - 1/5*x^5 - 1/3*sq rt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + x
Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=x-\frac {\sqrt {3}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x^3}{3}+\frac {2\,\sqrt {3}\,x}{3}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {3}\,x}{3}\right )\right )}{6}-\frac {x^5}{5}+\frac {x^7}{7}-\frac {x^{11}}{11}+\frac {x^{13}}{13}-\frac {x^{17}}{17}+\frac {x^{19}}{19} \] Input:
int(x^22/(x^2 + x^4 + 1),x)
Output:
x - (3^(1/2)*(2*atan((2*3^(1/2)*x)/3 + (3^(1/2)*x^3)/3) + 2*atan((3^(1/2)* x)/3)))/6 - x^5/5 + x^7/7 - x^11/11 + x^13/13 - x^17/17 + x^19/19
Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \frac {x^{22}}{1+x^2+x^4} \, dx=-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{3}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{3}+\frac {x^{19}}{19}-\frac {x^{17}}{17}+\frac {x^{13}}{13}-\frac {x^{11}}{11}+\frac {x^{7}}{7}-\frac {x^{5}}{5}+x \] Input:
int(x^22/(x^4+x^2+1),x)
Output:
( - 1616615*sqrt(3)*atan((2*x - 1)/sqrt(3)) - 1616615*sqrt(3)*atan((2*x + 1)/sqrt(3)) + 255255*x**19 - 285285*x**17 + 373065*x**13 - 440895*x**11 + 692835*x**7 - 969969*x**5 + 4849845*x)/4849845