\(\int \frac {x^5}{1+x^4+x^8} \, dx\) [331]

Optimal result

Integrand size = 14, antiderivative size = 75 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right ) \]

[Out]

1/8*ln(x^4-x^2+1)-1/8*ln(x^4+x^2+1)-1/12*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)+1/12*arctan(1/3*(2*x^2+1)*3^(1
/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1373, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {x^5}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (x^4-x^2+1\right )-\frac {1}{8} \log \left (x^4+x^2+1\right ) \]

[In]

Int[x^5/(1 + x^4 + x^8),x]

[Out]

-1/4*ArcTan[(1 - 2*x^2)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 2*x^2)/Sqrt[3]]/(4*Sqrt[3]) + Log[1 - x^2 + x^4]/8 - Lo
g[1 + x^2 + x^4]/8

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1141

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 1373

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[
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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{1+x^2+x^4} \, dx,x,x^2\right ) \\ & = -\left (\frac {1}{4} \text {Subst}\left (\int \frac {1-x^2}{1+x^2+x^4} \, dx,x,x^2\right )\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1+x^2}{1+x^2+x^4} \, dx,x,x^2\right ) \\ & = \frac {1}{8} \text {Subst}\left (\int \frac {1+2 x}{-1-x-x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1-2 x}{-1+x-x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {\sqrt {1-i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x^2\right )+\sqrt {1+i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x^2\right )}{4 \sqrt {6}} \]

[In]

Integrate[x^5/(1 + x^4 + x^8),x]

[Out]

(Sqrt[1 - I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan[((-I + Sqrt[3])*x^2)/2] + Sqrt[1 + I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[
((I + Sqrt[3])*x^2)/2])/(4*Sqrt[6])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\)	\(62\)
risch	\(-\frac {\ln \left (4 x^{4}+4 x^{2}+4\right )}{8}+\frac {\arctan \left (\frac {\left (2 x^{2}+1\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (4 x^{4}-4 x^{2}+4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}\)	\(68\)

[In]

int(x^5/(x^8+x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/8*ln(x^4-x^2+1)+1/12*3^(1/2)*arctan(1/3*(2*x^2-1)*3^(1/2))-1/8*ln(x^4+x^2+1)+1/12*arctan(1/3*(2*x^2+1)*3^(1/
2))*3^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

integrate(x^5/(x^8+x^4+1),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/8*log(x^4 + x^
2 + 1) + 1/8*log(x^4 - x^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \]

[In]

integrate(x**5/(x**8+x**4+1),x)

[Out]

log(x**4 - x**2 + 1)/8 - log(x**4 + x**2 + 1)/8 + sqrt(3)*atan(2*sqrt(3)*x**2/3 - sqrt(3)/3)/12 + sqrt(3)*atan
(2*sqrt(3)*x**2/3 + sqrt(3)/3)/12

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

integrate(x^5/(x^8+x^4+1),x, algorithm="maxima")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/8*log(x^4 + x^
2 + 1) + 1/8*log(x^4 - x^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]

[In]

integrate(x^5/(x^8+x^4+1),x, algorithm="giac")

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/8*log(x^4 + x^
2 + 1) + 1/8*log(x^4 - x^2 + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68 \[ \int \frac {x^5}{1+x^4+x^8} \, dx=\mathrm {atanh}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atanh}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

[In]

int(x^5/(x^4 + x^8 + 1),x)

[Out]

atanh((2*x^2)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/12 + 1/4) + atanh((2*x^2)/(3^(1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1
/4)