Integrand size = 14, antiderivative size = 127 \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{1-i \sqrt {3}}\right )}{\sqrt {3} \left (i+\sqrt {3}\right ) (1+m)}-\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{1+i \sqrt {3}}\right )}{\sqrt {3} \left (i-\sqrt {3}\right ) (1+m)} \]
-2/3*x^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-2*x^4/(1+I*3^(1/2)))/(1 +m)/(I-3^(1/2))*3^(1/2)+2/3*x^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],- 2*x^4/(1-I*3^(1/2)))/(1+m)*3^(1/2)/(3^(1/2)+I)
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 1.06 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.84 \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\frac {x^m \left (-\frac {i \left (\left (\frac {x}{-\sqrt [3]{-1}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {\sqrt [3]{-1}}{\sqrt [3]{-1}-x}\right )+\left (\frac {x}{-(-1)^{2/3}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {(-1)^{2/3}}{(-1)^{2/3}-x}\right )-\left (\frac {x}{\sqrt [3]{-1}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {\sqrt [3]{-1}}{\sqrt [3]{-1}+x}\right )-\left (\frac {x}{(-1)^{2/3}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {(-1)^{2/3}}{(-1)^{2/3}+x}\right )\right )}{\sqrt {3}}+\text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{-\text {$\#$1}+2 \text {$\#$1}^3}\&\right ]-\frac {\text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{-\text {$\#$1}+2 \text {$\#$1}^3}\&\right ]}{2+3 m+m^2}\right )}{4 m} \]
(x^m*(((-I)*(Hypergeometric2F1[-m, -m, 1 - m, (-1)^(1/3)/((-1)^(1/3) - x)] /(x/(-(-1)^(1/3) + x))^m + Hypergeometric2F1[-m, -m, 1 - m, (-1)^(2/3)/((- 1)^(2/3) - x)]/(x/(-(-1)^(2/3) + x))^m - Hypergeometric2F1[-m, -m, 1 - m, (-1)^(1/3)/((-1)^(1/3) + x)]/(x/((-1)^(1/3) + x))^m - Hypergeometric2F1[-m , -m, 1 - m, (-1)^(2/3)/((-1)^(2/3) + x)]/(x/((-1)^(2/3) + x))^m))/Sqrt[3] + RootSum[1 - #1^2 + #1^4 & , Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]/((x/(x - #1))^m*(-#1 + 2*#1^3)) & ] - RootSum[1 - #1^2 + #1^4 & , (m *x^2 + m^2*x^2 + 2*m*x*#1 + m^2*x*#1 + (2*Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (3*m*Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (m^2*Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]*#1^2)/(x/(x - #1))^m + (m*#1^2)/(x/#1)^m)/(-#1 + 2* #1^3) & ]/(2 + 3*m + m^2)))/(4*m)
Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1711, 27, 888}
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.
\(\displaystyle \int \frac {x^m}{x^8+x^4+1} \, dx\) |
\(\Big \downarrow \) 1711 |
\(\displaystyle \frac {i \int \frac {2 x^m}{2 x^4+i \sqrt {3}+1}dx}{\sqrt {3}}-\frac {i \int \frac {2 x^m}{2 x^4-i \sqrt {3}+1}dx}{\sqrt {3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 i \int \frac {x^m}{2 x^4+i \sqrt {3}+1}dx}{\sqrt {3}}-\frac {2 i \int \frac {x^m}{2 x^4-i \sqrt {3}+1}dx}{\sqrt {3}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 i x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{1+i \sqrt {3}}\right )}{\sqrt {3} \left (1+i \sqrt {3}\right ) (m+1)}-\frac {2 i x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{1-i \sqrt {3}}\right )}{\sqrt {3} \left (1-i \sqrt {3}\right ) (m+1)}\) |
((-2*I)*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*x^4)/(1 - I*Sqrt[3])])/(Sqrt[3]*(1 - I*Sqrt[3])*(1 + m)) + ((2*I)*x^(1 + m)*Hyperge ometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*x^4)/(1 + I*Sqrt[3])])/(Sqrt[3]*(1 + I*Sqrt[3])*(1 + m))
3.4.27.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symb ol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[c/q Int[(d*x)^m/(b/2 - q/2 + c *x^n), x], x] - Simp[c/q Int[(d*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; Free Q[{a, b, c, d, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]
\[\int \frac {x^{m}}{x^{8}+x^{4}+1}d x\]
\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \]
\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int \frac {x^{m}}{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}\, dx \]
\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \]
\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \]
Timed out. \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int \frac {x^m}{x^8+x^4+1} \,d x \]