\(\int \frac {x^2}{1-3 x^4+x^8} \, dx\) [399]

Optimal result

Integrand size = 16, antiderivative size = 145 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=\frac {1}{20} \sqrt {-10+10 \sqrt {5}} \arctan \left (\frac {1}{2} \sqrt {-2+2 \sqrt {5}} x\right )-\frac {1}{20} \sqrt {10+10 \sqrt {5}} \arctan \left (\frac {1}{2} \sqrt {2+2 \sqrt {5}} x\right )-\frac {1}{20} \sqrt {-10+10 \sqrt {5}} \text {arctanh}\left (\frac {1}{2} \sqrt {-2+2 \sqrt {5}} x\right )+\frac {1}{20} \sqrt {10+10 \sqrt {5}} \text {arctanh}\left (\frac {1}{2} \sqrt {2+2 \sqrt {5}} x\right ) \]

[Out]

1/20*arctan(1/2*x*(-2+2*5^(1/2))^(1/2))*(-10+10*5^(1/2))^(1/2)-1/20*arctanh(1/2*x*(-2+2*5^(1/2))^(1/2))*(-10+1
0*5^(1/2))^(1/2)-1/20*arctan(1/2*x*(2+2*5^(1/2))^(1/2))*(10+10*5^(1/2))^(1/2)+1/20*arctanh(1/2*x*(2+2*5^(1/2))
^(1/2))*(10+10*5^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1389, 304, 209, 212} \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=\frac {\arctan \left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}}-\frac {\text {arctanh}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}+\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}} \]

[In]

Int[x^2/(1 - 3*x^4 + x^8),x]

[Out]

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

Rule 209

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

Rule 212

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 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^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 {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx}{\sqrt {10}}-\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx}{\sqrt {10}}-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx}{\sqrt {10}}+\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx}{\sqrt {10}} \\ & = \frac {\tan ^{-1}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}}-\frac {\tanh ^{-1}\left (\sqrt [4]{\frac {2}{3+\sqrt {5}}} x\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}+\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \tanh ^{-1}\left (\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x\right )}{2 \sqrt {5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.90 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=-\frac {\arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}+\frac {\arctan \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{-1+\sqrt {5}}} x\right )}{\sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right )}{\sqrt {10 \left (1+\sqrt {5}\right )}} \]

[In]

Integrate[x^2/(1 - 3*x^4 + x^8),x]

[Out]

-(ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[10*(-1 + Sqrt[5])]) + ArcTan[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[10*(1 + Sqr
t[5])] + ArcTanh[Sqrt[2/(-1 + Sqrt[5])]*x]/Sqrt[10*(-1 + Sqrt[5])] - ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x]/Sqrt[10*
(1 + Sqrt[5])]

Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}-5 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-5 \textit {\_R}^{3}+3 \textit {\_R} +x \right )\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+5 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (-5 \textit {\_R}^{3}-3 \textit {\_R} +x \right )\right )}{4}\)	\(64\)
default	\(-\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}-\frac {\sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}+\frac {\sqrt {5}\, \arctan \left (\frac {2 x}{\sqrt {2 \sqrt {5}+2}}\right )}{5 \sqrt {2 \sqrt {5}+2}}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x}{\sqrt {2 \sqrt {5}-2}}\right )}{5 \sqrt {2 \sqrt {5}-2}}\)	\(110\)

[In]

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

[Out]

1/4*sum(_R*ln(-5*_R^3+3*_R+x),_R=RootOf(25*_Z^4-5*_Z^2-1))+1/4*sum(_R*ln(-5*_R^3-3*_R+x),_R=RootOf(25*_Z^4+5*_
Z^2-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (97) = 194\).

Time = 0.28 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.97 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=-\frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {\sqrt {5} - 1} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (-\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {\sqrt {5} - 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (\sqrt {10} \sqrt {\sqrt {5} + 1} {\left (\sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (-\sqrt {10} \sqrt {\sqrt {5} + 1} {\left (\sqrt {5} - 5\right )} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {-\sqrt {5} + 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (-\sqrt {10} {\left (\sqrt {5} + 5\right )} \sqrt {-\sqrt {5} + 1} + 20 \, x\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (\sqrt {10} {\left (\sqrt {5} - 5\right )} \sqrt {-\sqrt {5} - 1} + 20 \, x\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (-\sqrt {10} {\left (\sqrt {5} - 5\right )} \sqrt {-\sqrt {5} - 1} + 20 \, x\right ) \]

[In]

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

[Out]

-1/40*sqrt(10)*sqrt(sqrt(5) - 1)*log(sqrt(10)*(sqrt(5) + 5)*sqrt(sqrt(5) - 1) + 20*x) + 1/40*sqrt(10)*sqrt(sqr
t(5) - 1)*log(-sqrt(10)*(sqrt(5) + 5)*sqrt(sqrt(5) - 1) + 20*x) - 1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log(sqrt(10)
*sqrt(sqrt(5) + 1)*(sqrt(5) - 5) + 20*x) + 1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log(-sqrt(10)*sqrt(sqrt(5) + 1)*(sq
rt(5) - 5) + 20*x) + 1/40*sqrt(10)*sqrt(-sqrt(5) + 1)*log(sqrt(10)*(sqrt(5) + 5)*sqrt(-sqrt(5) + 1) + 20*x) -
1/40*sqrt(10)*sqrt(-sqrt(5) + 1)*log(-sqrt(10)*(sqrt(5) + 5)*sqrt(-sqrt(5) + 1) + 20*x) + 1/40*sqrt(10)*sqrt(-
sqrt(5) - 1)*log(sqrt(10)*(sqrt(5) - 5)*sqrt(-sqrt(5) - 1) + 20*x) - 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log(-sqr
t(10)*(sqrt(5) - 5)*sqrt(-sqrt(5) - 1) + 20*x)

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.37 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=\operatorname {RootSum} {\left (6400 t^{4} - 80 t^{2} - 1, \left ( t \mapsto t \log {\left (6144000 t^{7} - 2240 t^{3} + x \right )} \right )\right )} + \operatorname {RootSum} {\left (6400 t^{4} + 80 t^{2} - 1, \left ( t \mapsto t \log {\left (6144000 t^{7} - 2240 t^{3} + x \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(6400*_t**4 - 80*_t**2 - 1, Lambda(_t, _t*log(6144000*_t**7 - 2240*_t**3 + x))) + RootSum(6400*_t**4 +
80*_t**2 - 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 = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=\frac {1}{20} \, \sqrt {10 \, \sqrt {5} - 10} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{20} \, \sqrt {10 \, \sqrt {5} + 10} \arctan \left (\frac {x}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{40} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) + \frac {1}{40} \, \sqrt {10 \, \sqrt {5} + 10} \log \left ({\left | x + \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) - \frac {1}{40} \, \sqrt {10 \, \sqrt {5} + 10} \log \left ({\left | x - \sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} \right |}\right ) \]

[In]

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

[Out]

1/20*sqrt(10*sqrt(5) - 10)*arctan(x/sqrt(1/2*sqrt(5) + 1/2)) - 1/20*sqrt(10*sqrt(5) + 10)*arctan(x/sqrt(1/2*sq
rt(5) - 1/2)) - 1/40*sqrt(10*sqrt(5) - 10)*log(abs(x + sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(10*sqrt(5) - 10)*
log(abs(x - sqrt(1/2*sqrt(5) + 1/2))) + 1/40*sqrt(10*sqrt(5) + 10)*log(abs(x + sqrt(1/2*sqrt(5) - 1/2))) - 1/4
0*sqrt(10*sqrt(5) + 10)*log(abs(x - sqrt(1/2*sqrt(5) - 1/2)))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.86 \[ \int \frac {x^2}{1-3 x^4+x^8} \, dx=\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {\sqrt {5}-1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}-1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}-7\right )}\right )\,\sqrt {\sqrt {5}-1}\,1{}\mathrm {i}}{20}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {\sqrt {5}+1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}+7\right )}\right )\,\sqrt {\sqrt {5}+1}\,1{}\mathrm {i}}{20}+\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}-7\right )}-\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {1-\sqrt {5}}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}-7\right )}\right )\,\sqrt {1-\sqrt {5}}\,1{}\mathrm {i}}{20}-\frac {\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,x\,\sqrt {-\sqrt {5}-1}\,3{}\mathrm {i}}{2\,\left (3\,\sqrt {5}+7\right )}+\frac {\sqrt {5}\,\sqrt {10}\,x\,\sqrt {-\sqrt {5}-1}\,7{}\mathrm {i}}{10\,\left (3\,\sqrt {5}+7\right )}\right )\,\sqrt {-\sqrt {5}-1}\,1{}\mathrm {i}}{20} \]

[In]

int(x^2/(x^8 - 3*x^4 + 1),x)

[Out]

(10^(1/2)*atan((10^(1/2)*x*(5^(1/2) - 1)^(1/2)*3i)/(2*(3*5^(1/2) - 7)) - (5^(1/2)*10^(1/2)*x*(5^(1/2) - 1)^(1/
2)*7i)/(10*(3*5^(1/2) - 7)))*(5^(1/2) - 1)^(1/2)*1i)/20 - (10^(1/2)*atan((10^(1/2)*x*(5^(1/2) + 1)^(1/2)*3i)/(
2*(3*5^(1/2) + 7)) + (5^(1/2)*10^(1/2)*x*(5^(1/2) + 1)^(1/2)*7i)/(10*(3*5^(1/2) + 7)))*(5^(1/2) + 1)^(1/2)*1i)
/20 + (10^(1/2)*atan((10^(1/2)*x*(1 - 5^(1/2))^(1/2)*3i)/(2*(3*5^(1/2) - 7)) - (5^(1/2)*10^(1/2)*x*(1 - 5^(1/2
))^(1/2)*7i)/(10*(3*5^(1/2) - 7)))*(1 - 5^(1/2))^(1/2)*1i)/20 - (10^(1/2)*atan((10^(1/2)*x*(- 5^(1/2) - 1)^(1/
2)*3i)/(2*(3*5^(1/2) + 7)) + (5^(1/2)*10^(1/2)*x*(- 5^(1/2) - 1)^(1/2)*7i)/(10*(3*5^(1/2) + 7)))*(- 5^(1/2) -
1)^(1/2)*1i)/20