\(\int \frac {x^5}{1-x^4+x^8} \, dx\) [351]

Optimal result

Integrand size = 16, antiderivative size = 82 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )+\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \]

[Out]

1/4*arctan(2*x^2-3^(1/2))+1/4*arctan(2*x^2+3^(1/2))+1/24*ln(1+x^4-x^2*3^(1/2))*3^(1/2)-1/24*ln(1+x^4+x^2*3^(1/
2))*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1373, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (2 x^2+\sqrt {3}\right )+\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{8 \sqrt {3}}-\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{8 \sqrt {3}} \]

[In]

Int[x^5/(1 - x^4 + x^8),x]

[Out]

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.20 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {\sqrt {-1-i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )+\sqrt {-1+i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{4 \sqrt {6}} \]

[In]

Integrate[x^5/(1 - x^4 + x^8),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.39

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (6 \textit {\_R}^{3}+x^{2}+\textit {\_R} \right )\right )}{4}\)	\(32\)
default	\(-\frac {\sqrt {3}\, \left (-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}-\sqrt {3}\right )\right )}{12}-\frac {\sqrt {3}\, \left (\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}+\sqrt {3}\right )\right )}{12}\)	\(77\)

[In]

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

[Out]

1/4*sum(_R*ln(6*_R^3+x^2+_R),_R=RootOf(9*_Z^4+3*_Z^2+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.87 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) \]

[In]

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

[Out]

1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(6*x^2 + I*sqrt(6)*sqrt(3)*sqrt(I*sqrt(3) - 1)) - 1/24*sqrt(6)*sqrt(I*sqrt
(3) - 1)*log(6*x^2 - I*sqrt(6)*sqrt(3)*sqrt(I*sqrt(3) - 1)) - 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log(6*x^2 + I*
sqrt(6)*sqrt(3)*sqrt(-I*sqrt(3) - 1)) + 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log(6*x^2 - I*sqrt(6)*sqrt(3)*sqrt(-
I*sqrt(3) - 1))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} + \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} \]

[In]

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

[Out]

sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 - sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 + atan(2*x**2 - sqrt(3))/4 +
atan(2*x**2 + sqrt(3))/4

Maxima [F]

\[ \int \frac {x^5}{1-x^4+x^8} \, dx=\int { \frac {x^{5}}{x^{8} - x^{4} + 1} \,d x } \]

[In]

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

[Out]

integrate(x^5/(x^8 - x^4 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) - \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) + \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) + \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) \]

[In]

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

[Out]

1/24*sqrt(3)*x^4*log(x^4 + sqrt(3)*x^2 + 1) - 1/24*sqrt(3)*x^4*log(x^4 - sqrt(3)*x^2 + 1) + 1/4*x^4*arctan(2*x
^2 + sqrt(3)) + 1/4*x^4*arctan(2*x^2 - sqrt(3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {x^5}{1-x^4+x^8} \, dx=-\mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

[In]

int(x^5/(x^8 - x^4 + 1),x)

[Out]

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