\(\int \frac {x^2}{1+3 x^4+x^8} \, dx\) [381]

Optimal result

Integrand size = 16, antiderivative size = 427 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=-\frac {\arctan \left (1-\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}+\frac {\arctan \left (1+\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}+\frac {\arctan \left (1-\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}-\frac {\arctan \left (1+\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\log \left (\sqrt {2 \left (3-\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}-\frac {\log \left (\sqrt {2 \left (3-\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}-\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}} \]

[Out]

1/20*arctan(-1+2^(3/4)*x/(3-5^(1/2))^(1/4))*2^(3/4)/(3-5^(1/2))^(1/4)*5^(1/2)+1/20*arctan(1+2^(3/4)*x/(3-5^(1/
2))^(1/4))*2^(3/4)/(3-5^(1/2))^(1/4)*5^(1/2)+1/40*ln(2*x^2-2*2^(1/4)*x*(3-5^(1/2))^(1/4)+5^(1/2)-1)*2^(3/4)/(3
-5^(1/2))^(1/4)*5^(1/2)-1/40*ln(2*x^2+2*2^(1/4)*x*(3-5^(1/2))^(1/4)+5^(1/2)-1)*2^(3/4)/(3-5^(1/2))^(1/4)*5^(1/
2)-1/20*arctan(-1+2^(3/4)*x/(3+5^(1/2))^(1/4))*2^(3/4)*5^(1/2)/(3+5^(1/2))^(1/4)-1/20*arctan(1+2^(3/4)*x/(3+5^
(1/2))^(1/4))*2^(3/4)*5^(1/2)/(3+5^(1/2))^(1/4)-1/40*ln(2*x^2-2*2^(1/4)*x*(3+5^(1/2))^(1/4)+5^(1/2)+1)*2^(3/4)
*5^(1/2)/(3+5^(1/2))^(1/4)+1/40*ln(2*x^2+2*2^(1/4)*x*(3+5^(1/2))^(1/4)+5^(1/2)+1)*2^(3/4)*5^(1/2)/(3+5^(1/2))^
(1/4)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1389, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=-\frac {\sqrt [4]{3+\sqrt {5}} \arctan \left (1-\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2\ 2^{3/4} \sqrt {5}}+\frac {\sqrt [4]{3+\sqrt {5}} \arctan \left (\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}+1\right )}{2\ 2^{3/4} \sqrt {5}}+\frac {\arctan \left (1-\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}-\frac {\arctan \left (\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}+1\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\sqrt [4]{3+\sqrt {5}} \log \left (2 x^2-2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+\sqrt {2 \left (3-\sqrt {5}\right )}\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\sqrt [4]{3+\sqrt {5}} \log \left (2 x^2+2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+\sqrt {2 \left (3-\sqrt {5}\right )}\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\log \left (2 x^2-2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+\sqrt {2 \left (3+\sqrt {5}\right )}\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\log \left (2 x^2+2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+\sqrt {2 \left (3+\sqrt {5}\right )}\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}} \]

[In]

Int[x^2/(1 + 3*x^4 + x^8),x]

[Out]

-1/2*((3 + Sqrt[5])^(1/4)*ArcTan[1 - (2^(3/4)*x)/(3 - Sqrt[5])^(1/4)])/(2^(3/4)*Sqrt[5]) + ((3 + Sqrt[5])^(1/4
)*ArcTan[1 + (2^(3/4)*x)/(3 - Sqrt[5])^(1/4)])/(2*2^(3/4)*Sqrt[5]) + ArcTan[1 - (2^(3/4)*x)/(3 + Sqrt[5])^(1/4
)]/(2*Sqrt[5]*(2*(3 + Sqrt[5]))^(1/4)) - ArcTan[1 + (2^(3/4)*x)/(3 + Sqrt[5])^(1/4)]/(2*Sqrt[5]*(2*(3 + Sqrt[5
]))^(1/4)) + ((3 + Sqrt[5])^(1/4)*Log[Sqrt[2*(3 - Sqrt[5])] - 2*(2*(3 - Sqrt[5]))^(1/4)*x + 2*x^2])/(4*2^(3/4)
*Sqrt[5]) - ((3 + Sqrt[5])^(1/4)*Log[Sqrt[2*(3 - Sqrt[5])] + 2*(2*(3 - Sqrt[5]))^(1/4)*x + 2*x^2])/(4*2^(3/4)*
Sqrt[5]) - Log[Sqrt[2*(3 + Sqrt[5])] - 2*(2*(3 + Sqrt[5]))^(1/4)*x + 2*x^2]/(4*Sqrt[5]*(2*(3 + Sqrt[5]))^(1/4)
) + Log[Sqrt[2*(3 + Sqrt[5])] + 2*(2*(3 + Sqrt[5]))^(1/4)*x + 2*x^2]/(4*Sqrt[5]*(2*(3 + Sqrt[5]))^(1/4))

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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[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, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 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, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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^2}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^4} \, dx}{\sqrt {5}}-\frac {\int \frac {x^2}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^4} \, dx}{\sqrt {5}} \\ & = -\frac {\int \frac {\sqrt {3-\sqrt {5}}-\sqrt {2} x^2}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^4} \, dx}{2 \sqrt {10}}+\frac {\int \frac {\sqrt {3-\sqrt {5}}+\sqrt {2} x^2}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^4} \, dx}{2 \sqrt {10}}+\frac {\int \frac {\sqrt {3+\sqrt {5}}-\sqrt {2} x^2}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^4} \, dx}{2 \sqrt {10}}-\frac {\int \frac {\sqrt {3+\sqrt {5}}+\sqrt {2} x^2}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^4} \, dx}{2 \sqrt {10}} \\ & = \frac {\int \frac {1}{\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )}-\sqrt [4]{2 \left (3-\sqrt {5}\right )} x+x^2} \, dx}{4 \sqrt {5}}+\frac {\int \frac {1}{\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )}+\sqrt [4]{2 \left (3-\sqrt {5}\right )} x+x^2} \, dx}{4 \sqrt {5}}-\frac {\int \frac {1}{\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}-\sqrt [4]{2 \left (3+\sqrt {5}\right )} x+x^2} \, dx}{4 \sqrt {5}}-\frac {\int \frac {1}{\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}+\sqrt [4]{2 \left (3+\sqrt {5}\right )} x+x^2} \, dx}{4 \sqrt {5}}+\frac {\sqrt [4]{3+\sqrt {5}} \int \frac {\sqrt [4]{2 \left (3-\sqrt {5}\right )}+2 x}{-\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )}-\sqrt [4]{2 \left (3-\sqrt {5}\right )} x-x^2} \, dx}{4\ 2^{3/4} \sqrt {5}}+\frac {\sqrt [4]{3+\sqrt {5}} \int \frac {\sqrt [4]{2 \left (3-\sqrt {5}\right )}-2 x}{-\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )}+\sqrt [4]{2 \left (3-\sqrt {5}\right )} x-x^2} \, dx}{4\ 2^{3/4} \sqrt {5}}-\frac {\int \frac {\sqrt [4]{2 \left (3+\sqrt {5}\right )}+2 x}{-\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}-\sqrt [4]{2 \left (3+\sqrt {5}\right )} x-x^2} \, dx}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}-\frac {\int \frac {\sqrt [4]{2 \left (3+\sqrt {5}\right )}-2 x}{-\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}+\sqrt [4]{2 \left (3+\sqrt {5}\right )} x-x^2} \, dx}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}} \\ & = \frac {\sqrt [4]{3+\sqrt {5}} \log \left (\sqrt {2 \left (3-\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\sqrt [4]{3+\sqrt {5}} \log \left (\sqrt {2 \left (3-\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}} \\ & = -\frac {\tan ^{-1}\left (1-\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}+\frac {\tan ^{-1}\left (1+\frac {2^{3/4} x}{\sqrt [4]{3-\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3-\sqrt {5}\right )}}+\frac {\tan ^{-1}\left (1-\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}-\frac {\tan ^{-1}\left (1+\frac {2^{3/4} x}{\sqrt [4]{3+\sqrt {5}}}\right )}{2 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\sqrt [4]{3+\sqrt {5}} \log \left (\sqrt {2 \left (3-\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\sqrt [4]{3+\sqrt {5}} \log \left (\sqrt {2 \left (3-\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3-\sqrt {5}\right )} x+2 x^2\right )}{4\ 2^{3/4} \sqrt {5}}-\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}-2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}}+\frac {\log \left (\sqrt {2 \left (3+\sqrt {5}\right )}+2 \sqrt [4]{2 \left (3+\sqrt {5}\right )} x+2 x^2\right )}{4 \sqrt {5} \sqrt [4]{2 \left (3+\sqrt {5}\right )}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.09 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=\frac {1}{4} \text {RootSum}\left [1+3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1})}{3 \text {$\#$1}+2 \text {$\#$1}^5}\&\right ] \]

[In]

Integrate[x^2/(1 + 3*x^4 + x^8),x]

[Out]

RootSum[1 + 3*#1^4 + #1^8 & , Log[x - #1]/(3*#1 + 2*#1^5) & ]/4

Maple [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.09

method	result	size
default	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+3 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7}+3 \textit {\_R}^{3}}\right )}{4}\)	\(40\)
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+3 \textit {\_Z}^{4}+1\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7}+3 \textit {\_R}^{3}}\right )}{4}\)	\(40\)

[In]

int(x^2/(x^8+3*x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/4*sum(_R^2/(2*_R^7+3*_R^3)*ln(x-_R),_R=RootOf(_Z^8+3*_Z^4+1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.23 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=-\frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {2} \sqrt {\sqrt {5} - 3}} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {\sqrt {2} \sqrt {\sqrt {5} - 3}} \sqrt {\sqrt {5} - 3} + 40 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {2} \sqrt {\sqrt {5} - 3}} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {\sqrt {2} \sqrt {\sqrt {5} - 3}} \sqrt {\sqrt {5} - 3} + 40 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {2} \sqrt {\sqrt {5} - 3}} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {2} \sqrt {\sqrt {5} - 3}} \sqrt {\sqrt {5} - 3} + 40 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {2} \sqrt {\sqrt {5} - 3}} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {2} \sqrt {\sqrt {5} - 3}} \sqrt {\sqrt {5} - 3} + 40 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {2} \sqrt {-\sqrt {5} - 3}} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {\sqrt {2} \sqrt {-\sqrt {5} - 3}} \sqrt {-\sqrt {5} - 3} + 40 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {2} \sqrt {-\sqrt {5} - 3}} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {\sqrt {2} \sqrt {-\sqrt {5} - 3}} \sqrt {-\sqrt {5} - 3} + 40 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {2} \sqrt {-\sqrt {5} - 3}} \log \left (\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {2} \sqrt {-\sqrt {5} - 3}} \sqrt {-\sqrt {5} - 3} + 40 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {2} \sqrt {-\sqrt {5} - 3}} \log \left (-\sqrt {10} {\left (3 \, \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {2} \sqrt {-\sqrt {5} - 3}} \sqrt {-\sqrt {5} - 3} + 40 \, x\right ) \]

[In]

integrate(x^2/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

-1/40*sqrt(10)*sqrt(sqrt(2)*sqrt(sqrt(5) - 3))*log(sqrt(10)*(3*sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(sqrt(2)*sqrt(
sqrt(5) - 3))*sqrt(sqrt(5) - 3) + 40*x) + 1/40*sqrt(10)*sqrt(sqrt(2)*sqrt(sqrt(5) - 3))*log(-sqrt(10)*(3*sqrt(
5)*sqrt(2) + 5*sqrt(2))*sqrt(sqrt(2)*sqrt(sqrt(5) - 3))*sqrt(sqrt(5) - 3) + 40*x) + 1/40*sqrt(10)*sqrt(-sqrt(2
)*sqrt(sqrt(5) - 3))*log(sqrt(10)*(3*sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(-sqrt(2)*sqrt(sqrt(5) - 3))*sqrt(sqrt(5
) - 3) + 40*x) - 1/40*sqrt(10)*sqrt(-sqrt(2)*sqrt(sqrt(5) - 3))*log(-sqrt(10)*(3*sqrt(5)*sqrt(2) + 5*sqrt(2))*
sqrt(-sqrt(2)*sqrt(sqrt(5) - 3))*sqrt(sqrt(5) - 3) + 40*x) + 1/40*sqrt(10)*sqrt(sqrt(2)*sqrt(-sqrt(5) - 3))*lo
g(sqrt(10)*(3*sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(sqrt(2)*sqrt(-sqrt(5) - 3))*sqrt(-sqrt(5) - 3) + 40*x) - 1/40*
sqrt(10)*sqrt(sqrt(2)*sqrt(-sqrt(5) - 3))*log(-sqrt(10)*(3*sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(sqrt(2)*sqrt(-sqr
t(5) - 3))*sqrt(-sqrt(5) - 3) + 40*x) - 1/40*sqrt(10)*sqrt(-sqrt(2)*sqrt(-sqrt(5) - 3))*log(sqrt(10)*(3*sqrt(5
)*sqrt(2) - 5*sqrt(2))*sqrt(-sqrt(2)*sqrt(-sqrt(5) - 3))*sqrt(-sqrt(5) - 3) + 40*x) + 1/40*sqrt(10)*sqrt(-sqrt
(2)*sqrt(-sqrt(5) - 3))*log(-sqrt(10)*(3*sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(-sqrt(2)*sqrt(-sqrt(5) - 3))*sqrt(-
sqrt(5) - 3) + 40*x)

Sympy [A] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.06 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=\operatorname {RootSum} {\left (40960000 t^{8} + 19200 t^{4} + 1, \left ( t \mapsto t \log {\left (- 6144000 t^{7} - 2240 t^{3} + x \right )} \right )\right )} \]

[In]

integrate(x**2/(x**8+3*x**4+1),x)

[Out]

RootSum(40960000*_t**8 + 19200*_t**4 + 1, Lambda(_t, _t*log(-6144000*_t**7 - 2240*_t**3 + x)))

Maxima [F]

\[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=\int { \frac {x^{2}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

integrate(x^2/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

integrate(x^2/(x^8 + 3*x^4 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.56 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=\frac {1}{80} \, {\left (\pi + 4 \, \arctan \left (x \sqrt {\sqrt {5} + 1} - 1\right )\right )} \sqrt {5 \, \sqrt {5} + 5} - \frac {1}{80} \, {\left (\pi + 4 \, \arctan \left (-x \sqrt {\sqrt {5} + 1} - 1\right )\right )} \sqrt {5 \, \sqrt {5} + 5} - \frac {1}{80} \, {\left (\pi + 4 \, \arctan \left (x \sqrt {\sqrt {5} - 1} + 1\right )\right )} \sqrt {5 \, \sqrt {5} - 5} + \frac {1}{80} \, {\left (\pi + 4 \, \arctan \left (-x \sqrt {\sqrt {5} - 1} + 1\right )\right )} \sqrt {5 \, \sqrt {5} - 5} + \frac {1}{40} \, \sqrt {5 \, \sqrt {5} - 5} \log \left (16900 \, {\left (x + \sqrt {\sqrt {5} + 1}\right )}^{2} + 16900 \, x^{2}\right ) - \frac {1}{40} \, \sqrt {5 \, \sqrt {5} - 5} \log \left (16900 \, {\left (x - \sqrt {\sqrt {5} + 1}\right )}^{2} + 16900 \, x^{2}\right ) - \frac {1}{40} \, \sqrt {5 \, \sqrt {5} + 5} \log \left (2500 \, {\left (x + \sqrt {\sqrt {5} - 1}\right )}^{2} + 2500 \, x^{2}\right ) + \frac {1}{40} \, \sqrt {5 \, \sqrt {5} + 5} \log \left (2500 \, {\left (x - \sqrt {\sqrt {5} - 1}\right )}^{2} + 2500 \, x^{2}\right ) \]

[In]

integrate(x^2/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

1/80*(pi + 4*arctan(x*sqrt(sqrt(5) + 1) - 1))*sqrt(5*sqrt(5) + 5) - 1/80*(pi + 4*arctan(-x*sqrt(sqrt(5) + 1) -
1))*sqrt(5*sqrt(5) + 5) - 1/80*(pi + 4*arctan(x*sqrt(sqrt(5) - 1) + 1))*sqrt(5*sqrt(5) - 5) + 1/80*(pi + 4*ar
ctan(-x*sqrt(sqrt(5) - 1) + 1))*sqrt(5*sqrt(5) - 5) + 1/40*sqrt(5*sqrt(5) - 5)*log(16900*(x + sqrt(sqrt(5) + 1
))^2 + 16900*x^2) - 1/40*sqrt(5*sqrt(5) - 5)*log(16900*(x - sqrt(sqrt(5) + 1))^2 + 16900*x^2) - 1/40*sqrt(5*sq
rt(5) + 5)*log(2500*(x + sqrt(sqrt(5) - 1))^2 + 2500*x^2) + 1/40*sqrt(5*sqrt(5) + 5)*log(2500*(x - sqrt(sqrt(5
) - 1))^2 + 2500*x^2)

Mupad [B] (verification not implemented)

Time = 8.29 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.64 \[ \int \frac {x^2}{1+3 x^4+x^8} \, dx=\frac {2^{3/4}\,\sqrt {5}\,\mathrm {atan}\left (\frac {7\,2^{3/4}\,x\,{\left (\sqrt {5}-3\right )}^{1/4}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {3\,2^{3/4}\,\sqrt {5}\,x\,{\left (\sqrt {5}-3\right )}^{1/4}}{2\,\left (3\,\sqrt {5}-7\right )}\right )\,{\left (\sqrt {5}-3\right )}^{1/4}}{20}+\frac {2^{3/4}\,\sqrt {5}\,\mathrm {atan}\left (\frac {2^{3/4}\,x\,{\left (\sqrt {5}-3\right )}^{1/4}\,7{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {2^{3/4}\,\sqrt {5}\,x\,{\left (\sqrt {5}-3\right )}^{1/4}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}\right )\,{\left (\sqrt {5}-3\right )}^{1/4}\,1{}\mathrm {i}}{20}+\frac {2^{3/4}\,\sqrt {5}\,\mathrm {atan}\left (\frac {7\,2^{3/4}\,x\,{\left (-\sqrt {5}-3\right )}^{1/4}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {3\,2^{3/4}\,\sqrt {5}\,x\,{\left (-\sqrt {5}-3\right )}^{1/4}}{2\,\left (3\,\sqrt {5}+7\right )}\right )\,{\left (-\sqrt {5}-3\right )}^{1/4}}{20}+\frac {2^{3/4}\,\sqrt {5}\,\mathrm {atan}\left (\frac {2^{3/4}\,x\,{\left (-\sqrt {5}-3\right )}^{1/4}\,7{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {2^{3/4}\,\sqrt {5}\,x\,{\left (-\sqrt {5}-3\right )}^{1/4}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}\right )\,{\left (-\sqrt {5}-3\right )}^{1/4}\,1{}\mathrm {i}}{20} \]

[In]

int(x^2/(3*x^4 + x^8 + 1),x)

[Out]

(2^(3/4)*5^(1/2)*atan((7*2^(3/4)*x*(5^(1/2) - 3)^(1/4))/(2*(3*5^(1/2) - 7)) - (3*2^(3/4)*5^(1/2)*x*(5^(1/2) -
3)^(1/4))/(2*(3*5^(1/2) - 7)))*(5^(1/2) - 3)^(1/4))/20 + (2^(3/4)*5^(1/2)*atan((2^(3/4)*x*(5^(1/2) - 3)^(1/4)*
7i)/(2*(3*5^(1/2) - 7)) - (2^(3/4)*5^(1/2)*x*(5^(1/2) - 3)^(1/4)*3i)/(2*(3*5^(1/2) - 7)))*(5^(1/2) - 3)^(1/4)*
1i)/20 + (2^(3/4)*5^(1/2)*atan((7*2^(3/4)*x*(- 5^(1/2) - 3)^(1/4))/(2*(3*5^(1/2) + 7)) + (3*2^(3/4)*5^(1/2)*x*
(- 5^(1/2) - 3)^(1/4))/(2*(3*5^(1/2) + 7)))*(- 5^(1/2) - 3)^(1/4))/20 + (2^(3/4)*5^(1/2)*atan((2^(3/4)*x*(- 5^
(1/2) - 3)^(1/4)*7i)/(2*(3*5^(1/2) + 7)) + (2^(3/4)*5^(1/2)*x*(- 5^(1/2) - 3)^(1/4)*3i)/(2*(3*5^(1/2) + 7)))*(
- 5^(1/2) - 3)^(1/4)*1i)/20