Integrand size = 16, antiderivative size = 88 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\frac {x^2}{2}-\frac {\arctan \left (\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9-4 \sqrt {5}\right )}}+\frac {\arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9+4 \sqrt {5}\right )}} \] Output:
1/2*x^2-1/2*arctan((1/2*5^(1/2)-1/2)*x^2)/(5-2*5^(1/2))+1/2*arctan((1/2+1/ 2*5^(1/2))*x^2)/(5+2*5^(1/2))
Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\frac {1}{40} \left (20 x^2-\sqrt {6-2 \sqrt {5}} \left (15+7 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\sqrt {2 \left (3+\sqrt {5}\right )} \left (-15+7 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )\right ) \] Input:
Integrate[x^9/(1 + 3*x^4 + x^8),x]
Output:
(20*x^2 - Sqrt[6 - 2*Sqrt[5]]*(15 + 7*Sqrt[5])*ArcTan[Sqrt[2/(3 + Sqrt[5]) ]*x^2] + Sqrt[2*(3 + Sqrt[5])]*(-15 + 7*Sqrt[5])*ArcTan[Sqrt[(3 + Sqrt[5]) /2]*x^2])/40
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1695, 1442, 1480, 216}
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.
\(\displaystyle \int \frac {x^9}{x^8+3 x^4+1} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{x^8+3 x^4+1}dx^2\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {1}{2} \left (x^2-\int \frac {3 x^4+1}{x^8+3 x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{10} \left (15-7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3-\sqrt {5}\right )}dx^2-\frac {1}{10} \left (15+7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (3+\sqrt {5}\right )}dx^2+x^2\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (15+7 \sqrt {5}\right ) \arctan \left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{5 \sqrt {2 \left (3+\sqrt {5}\right )}}-\frac {1}{10} \left (15-7 \sqrt {5}\right ) \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )+x^2\right )\) |
Input:
Int[x^9/(1 + 3*x^4 + x^8),x]
Output:
(x^2 - ((15 + 7*Sqrt[5])*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2])/(5*Sqrt[2*(3 + Sqrt[5])]) - ((15 - 7*Sqrt[5])*Sqrt[(3 + Sqrt[5])/2]*ArcTan[Sqrt[(3 + Sqr t[5])/2]*x^2])/10)/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
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 = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48
method | result | size |
risch | \(\frac {x^{2}}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+90 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (15 \textit {\_R}^{3}+8 x^{2}+47 \textit {\_R} \right )\right )}{4}\) | \(42\) |
default | \(\frac {x^{2}}{2}-\frac {\left (-7+3 \sqrt {5}\right ) \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{5 \left (-2+2 \sqrt {5}\right )}-\frac {\sqrt {5}\, \left (7+3 \sqrt {5}\right ) \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{5 \left (2+2 \sqrt {5}\right )}\) | \(79\) |
Input:
int(x^9/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)
Output:
1/2*x^2+1/4*sum(_R*ln(15*_R^3+8*x^2+47*_R),_R=RootOf(25*_Z^4+90*_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} - \frac {1}{10} \, {\left (2 \, \sqrt {5} - 5\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} + \frac {1}{2} \, x^{2}\right ) - \frac {1}{10} \, {\left (2 \, \sqrt {5} + 5\right )} \arctan \left (\frac {1}{2} \, \sqrt {5} x^{2} - \frac {1}{2} \, x^{2}\right ) \] Input:
integrate(x^9/(x^8+3*x^4+1),x, algorithm="fricas")
Output:
1/2*x^2 - 1/10*(2*sqrt(5) - 5)*arctan(1/2*sqrt(5)*x^2 + 1/2*x^2) - 1/10*(2 *sqrt(5) + 5)*arctan(1/2*sqrt(5)*x^2 - 1/2*x^2)
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\frac {x^{2}}{2} + 2 \cdot \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} - 2 \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} \] Input:
integrate(x**9/(x**8+3*x**4+1),x)
Output:
x**2/2 + 2*(1/4 - sqrt(5)/10)*atan(2*x**2/(-1 + sqrt(5))) - 2*(sqrt(5)/10 + 1/4)*atan(2*x**2/(1 + sqrt(5)))
\[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\int { \frac {x^{9}}{x^{8} + 3 \, x^{4} + 1} \,d x } \] Input:
integrate(x^9/(x^8+3*x^4+1),x, algorithm="maxima")
Output:
1/2*x^2 - integrate((3*x^4 + 1)*x/(x^8 + 3*x^4 + 1), x)
Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} - \frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} - 5\right )} + \sqrt {5} - 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) - \frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} + 5\right )} + \sqrt {5} + 5\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) \] Input:
integrate(x^9/(x^8+3*x^4+1),x, algorithm="giac")
Output:
1/2*x^2 - 1/20*(3*x^4*(sqrt(5) - 5) + sqrt(5) - 5)*arctan(2*x^2/(sqrt(5) + 1)) - 1/20*(3*x^4*(sqrt(5) + 5) + sqrt(5) + 5)*arctan(2*x^2/(sqrt(5) - 1) )
Time = 0.13 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.48 \[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=2\,\mathrm {atanh}\left (\frac {1280\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{64\,\sqrt {5}-192}+\frac {768\,\sqrt {5}\,x^2\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}}{64\,\sqrt {5}-192}\right )\,\sqrt {\frac {\sqrt {5}}{20}-\frac {9}{80}}-2\,\mathrm {atanh}\left (\frac {1280\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{64\,\sqrt {5}+192}-\frac {768\,\sqrt {5}\,x^2\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}}{64\,\sqrt {5}+192}\right )\,\sqrt {-\frac {\sqrt {5}}{20}-\frac {9}{80}}+\frac {x^2}{2} \] Input:
int(x^9/(3*x^4 + x^8 + 1),x)
Output:
2*atanh((1280*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(64*5^(1/2) - 192) + (768*5^( 1/2)*x^2*(5^(1/2)/20 - 9/80)^(1/2))/(64*5^(1/2) - 192))*(5^(1/2)/20 - 9/80 )^(1/2) - 2*atanh((1280*x^2*(- 5^(1/2)/20 - 9/80)^(1/2))/(64*5^(1/2) + 192 ) - (768*5^(1/2)*x^2*(- 5^(1/2)/20 - 9/80)^(1/2))/(64*5^(1/2) + 192))*(- 5 ^(1/2)/20 - 9/80)^(1/2) + x^2/2
\[ \int \frac {x^9}{1+3 x^4+x^8} \, dx=-3 \left (\int \frac {x^{5}}{x^{8}+3 x^{4}+1}d x \right )-\left (\int \frac {x}{x^{8}+3 x^{4}+1}d x \right )+\frac {x^{2}}{2} \] Input:
int(x^9/(x^8+3*x^4+1),x)
Output:
( - 6*int(x**5/(x**8 + 3*x**4 + 1),x) - 2*int(x/(x**8 + 3*x**4 + 1),x) + x **2)/2