\(\int \frac {x^m}{1+x^4+x^8} \, dx\) [140]

Optimal result

Integrand size = 14, antiderivative size = 127 \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{1-i \sqrt {3}}\right )}{\sqrt {3} \left (i+\sqrt {3}\right ) (1+m)}-\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{1+i \sqrt {3}}\right )}{\sqrt {3} \left (i-\sqrt {3}\right ) (1+m)} \] Output:

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)))*3^( 
1/2)/(3^(1/2)+I)/(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)))*3^(1/2)/(I-3^(1/2))/(1+m)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 1.09 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.84 \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\frac {x^m \left (-\frac {i \left (\left (\frac {x}{-\sqrt [3]{-1}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {\sqrt [3]{-1}}{\sqrt [3]{-1}-x}\right )+\left (\frac {x}{-(-1)^{2/3}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {(-1)^{2/3}}{(-1)^{2/3}-x}\right )-\left (\frac {x}{\sqrt [3]{-1}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {\sqrt [3]{-1}}{\sqrt [3]{-1}+x}\right )-\left (\frac {x}{(-1)^{2/3}+x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {(-1)^{2/3}}{(-1)^{2/3}+x}\right )\right )}{\sqrt {3}}+\text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{-\text {$\#$1}+2 \text {$\#$1}^3}\&\right ]-\frac {\text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{-\text {$\#$1}+2 \text {$\#$1}^3}\&\right ]}{2+3 m+m^2}\right )}{4 m} \] Input:

Integrate[x^m/(1 + x^4 + x^8),x]
 

Output:

(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)
 

Rubi [A] (verified)

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

\(\displaystyle \frac {2 i x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{1+i \sqrt {3}}\right )}{\sqrt {3} \left (1+i \sqrt {3}\right ) (m+1)}-\frac {2 i x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{1-i \sqrt {3}}\right )}{\sqrt {3} \left (1-i \sqrt {3}\right ) (m+1)}\)

Input:

Int[x^m/(1 + x^4 + x^8),x]
 

Output:

((-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))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 888

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])
 

rule 1711

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]
 

Maple [F]

\[\int \frac {x^{m}}{x^{8}+x^{4}+1}d x\]

Input:

int(x^m/(x^8+x^4+1),x)
 

Output:

int(x^m/(x^8+x^4+1),x)
 

Fricas [F]

\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \] Input:

integrate(x^m/(x^8+x^4+1),x, algorithm="fricas")
 

Output:

integral(x^m/(x^8 + x^4 + 1), x)
 

Sympy [F]

\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int \frac {x^{m}}{\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right ) \left (x^{4} - x^{2} + 1\right )}\, dx \] Input:

integrate(x**m/(x**8+x**4+1),x)
 

Output:

Integral(x**m/((x**2 - x + 1)*(x**2 + x + 1)*(x**4 - x**2 + 1)), x)
 

Maxima [F]

\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \] Input:

integrate(x^m/(x^8+x^4+1),x, algorithm="maxima")
 

Output:

integrate(x^m/(x^8 + x^4 + 1), x)
 

Giac [F]

\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + x^{4} + 1} \,d x } \] Input:

integrate(x^m/(x^8+x^4+1),x, algorithm="giac")
 

Output:

integrate(x^m/(x^8 + x^4 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int \frac {x^m}{x^8+x^4+1} \,d x \] Input:

int(x^m/(x^4 + x^8 + 1),x)
 

Output:

int(x^m/(x^4 + x^8 + 1), x)
 

Reduce [F]

\[ \int \frac {x^m}{1+x^4+x^8} \, dx=\int \frac {x^{m}}{x^{8}+x^{4}+1}d x \] Input:

int(x^m/(x^8+x^4+1),x)
 

Output:

int(x**m/(x**8 + x**4 + 1),x)