\(\int \frac {x^m}{1+3 x^4+x^8} \, dx\) [367]

Optimal result

Integrand size = 16, antiderivative size = 117 \[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{3-\sqrt {5}}\right )}{\sqrt {5} \left (3-\sqrt {5}\right ) (1+m)}-\frac {2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 x^4}{3+\sqrt {5}}\right )}{\sqrt {5} \left (3+\sqrt {5}\right ) (1+m)} \]

[Out]

2/5*x^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-2*x^4/(3-5^(1/2)))/(1+m)/(3-5^(1/2))*5^(1/2)-2/5*x^(1+m)*hyp
ergeom([1, 1/4+1/4*m],[5/4+1/4*m],-2*x^4/(3+5^(1/2)))/(1+m)*5^(1/2)/(3+5^(1/2))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1389, 371} \[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\frac {2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{3-\sqrt {5}}\right )}{\sqrt {5} \left (3-\sqrt {5}\right ) (m+1)}-\frac {2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 x^4}{3+\sqrt {5}}\right )}{\sqrt {5} \left (3+\sqrt {5}\right ) (m+1)} \]

[In]

Int[x^m/(1 + 3*x^4 + x^8),x]

[Out]

(2*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*x^4)/(3 - Sqrt[5])])/(Sqrt[5]*(3 - Sqrt[5])*(1 + m
)) - (2*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*x^4)/(3 + Sqrt[5])])/(Sqrt[5]*(3 + Sqrt[5])*(
1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1389

Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]
}, Dist[c/q, Int[(d*x)^m/(b/2 - q/2 + c*x^n), x], x] - Dist[c/q, Int[(d*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; F
reeQ[{a, b, c, d, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {x^m}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^4} \, dx}{\sqrt {5}}-\frac {\int \frac {x^m}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^4} \, dx}{\sqrt {5}} \\ & = \frac {2 x^{1+m} \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};-\frac {2 x^4}{3-\sqrt {5}}\right )}{\sqrt {5} \left (3-\sqrt {5}\right ) (1+m)}-\frac {2 x^{1+m} \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};-\frac {2 x^4}{3+\sqrt {5}}\right )}{\sqrt {5} \left (3+\sqrt {5}\right ) (1+m)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\frac {x^m \text {RootSum}\left [1+3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ]}{4 m} \]

[In]

Integrate[x^m/(1 + 3*x^4 + x^8),x]

[Out]

(x^m*RootSum[1 + 3*#1^4 + #1^8 & , Hypergeometric2F1[-m, -m, 1 - m, -(#1/(x - #1))]/((x/(x - #1))^m*(3*#1^3 +
2*#1^7)) & ])/(4*m)

Maple [F]

\[\int \frac {x^{m}}{x^{8}+3 x^{4}+1}d x\]

[In]

int(x^m/(x^8+3*x^4+1),x)

[Out]

int(x^m/(x^8+3*x^4+1),x)

Fricas [F]

\[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

integrate(x^m/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

integral(x^m/(x^8 + 3*x^4 + 1), x)

Sympy [F]

\[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\int \frac {x^{m}}{x^{8} + 3 x^{4} + 1}\, dx \]

[In]

integrate(x**m/(x**8+3*x**4+1),x)

[Out]

Integral(x**m/(x**8 + 3*x**4 + 1), x)

Maxima [F]

\[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

integrate(x^m/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

integrate(x^m/(x^8 + 3*x^4 + 1), x)

Giac [F]

\[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\int { \frac {x^{m}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

integrate(x^m/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

integrate(x^m/(x^8 + 3*x^4 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m}{1+3 x^4+x^8} \, dx=\int \frac {x^m}{x^8+3\,x^4+1} \,d x \]

[In]

int(x^m/(3*x^4 + x^8 + 1),x)

[Out]

int(x^m/(3*x^4 + x^8 + 1), x)