\(\int \frac {x^m}{a+b x^4+c x^8} \, dx\) [309]

Optimal result

Integrand size = 18, antiderivative size = 163 \[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\frac {2 c x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 c x^4}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) (1+m)}-\frac {2 c x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-\frac {2 c x^4}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m)} \]

[Out]

2*c*x^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-2*c*x^4/(b-(-4*a*c+b^2)^(1/2)))/(1+m)/(b-(-4*a*c+b^2)^(1/2))
/(-4*a*c+b^2)^(1/2)-2*c*x^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-2*c*x^4/(b+(-4*a*c+b^2)^(1/2)))/(1+m)/(-
4*a*c+b^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1389, 371} \[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\frac {2 c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 c x^4}{b-\sqrt {b^2-4 a c}}\right )}{(m+1) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-\frac {2 c x^4}{b+\sqrt {b^2-4 a c}}\right )}{(m+1) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )} \]

[In]

Int[x^m/(a + b*x^4 + c*x^8),x]

[Out]

(2*c*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*c*x^4)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a
*c]*(b - Sqrt[b^2 - 4*a*c])*(1 + m)) - (2*c*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, (-2*c*x^4)/(b
+ Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(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 {c \int \frac {x^m}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^4} \, dx}{\sqrt {b^2-4 a c}}-\frac {c \int \frac {x^m}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^4} \, dx}{\sqrt {b^2-4 a c}} \\ & = \frac {2 c x^{1+m} \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};-\frac {2 c x^4}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) (1+m)}-\frac {2 c x^{1+m} \, _2F_1\left (1,\frac {1+m}{4};\frac {5+m}{4};-\frac {2 c x^4}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\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 = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.50 \[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\frac {x^m \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {\operatorname {Hypergeometric2F1}\left (-m,-m,1-m,-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{4 m} \]

[In]

Integrate[x^m/(a + b*x^4 + c*x^8),x]

[Out]

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

Maple [F]

\[\int \frac {x^{m}}{c \,x^{8}+b \,x^{4}+a}d x\]

[In]

int(x^m/(c*x^8+b*x^4+a),x)

[Out]

int(x^m/(c*x^8+b*x^4+a),x)

Fricas [F]

\[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\int { \frac {x^{m}}{c x^{8} + b x^{4} + a} \,d x } \]

[In]

integrate(x^m/(c*x^8+b*x^4+a),x, algorithm="fricas")

[Out]

integral(x^m/(c*x^8 + b*x^4 + a), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\text {Timed out} \]

[In]

integrate(x**m/(c*x**8+b*x**4+a),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\int { \frac {x^{m}}{c x^{8} + b x^{4} + a} \,d x } \]

[In]

integrate(x^m/(c*x^8+b*x^4+a),x, algorithm="maxima")

[Out]

integrate(x^m/(c*x^8 + b*x^4 + a), x)

Giac [F]

\[ \int \frac {x^m}{a+b x^4+c x^8} \, dx=\int { \frac {x^{m}}{c x^{8} + b x^{4} + a} \,d x } \]

[In]

integrate(x^m/(c*x^8+b*x^4+a),x, algorithm="giac")

[Out]

integrate(x^m/(c*x^8 + b*x^4 + a), x)

Mupad [F(-1)]

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

[In]

int(x^m/(a + b*x^4 + c*x^8),x)

[Out]

int(x^m/(a + b*x^4 + c*x^8), x)