\(\int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} (-1+x^4+x^8)} \, dx\) [2471]

Optimal result

Integrand size = 27, antiderivative size = 201 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {1}{2} \sqrt {\frac {1}{10} \left (11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-11+5 \sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-11+5 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \]

[Out]

1/20*(110+50*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))-1/20*(-110+50*5^(1/2))^(1/2)*arct
an(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))+1/20*(110+50*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^
4+1)^(1/4))-1/20*(-110+50*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/4))

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6860, 385, 218, 212, 209} \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt {5}} \]

[In]

Int[(-1 + 2*x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + x^8)),x]

[Out]

(((123 + 55*Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) - (((123 - 55*S
qrt[5])/2)^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) + (((123 + 55*Sqrt[5])/2)^(1
/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)])/(2*Sqrt[5]) - (((123 - 55*Sqrt[5])/2)^(1/4)*ArcTanh[
(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2-\frac {4}{\sqrt {5}}}{\sqrt [4]{1+x^4} \left (1-\sqrt {5}+2 x^4\right )}+\frac {2+\frac {4}{\sqrt {5}}}{\sqrt [4]{1+x^4} \left (1+\sqrt {5}+2 x^4\right )}\right ) \, dx \\ & = \frac {1}{5} \left (2 \left (5-2 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt [4]{1+x^4} \left (1-\sqrt {5}+2 x^4\right )} \, dx+\frac {1}{5} \left (2 \left (5+2 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt [4]{1+x^4} \left (1+\sqrt {5}+2 x^4\right )} \, dx \\ & = \frac {1}{5} \left (2 \left (5-2 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {5}-\left (-1-\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{5} \left (2 \left (5+2 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {5}-\left (-1+\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right ) \\ & = \frac {\left (2-\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{\sqrt {10}}+\frac {\left (2-\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{\sqrt {10}}+\frac {\left (2+\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{\sqrt {10}}+\frac {\left (2+\sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{\sqrt {10}} \\ & = \frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {5}}+\frac {\sqrt [4]{\frac {1}{2} \left (123+55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (123-55 \sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\frac {\sqrt {11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {-11+5 \sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {11+5 \sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )-\sqrt {-11+5 \sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {10}} \]

[In]

Integrate[(-1 + 2*x^4)/((1 + x^4)^(1/4)*(-1 + x^4 + x^8)),x]

[Out]

(Sqrt[11 + 5*Sqrt[5]]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] - Sqrt[-11 + 5*Sqrt[5]]*ArcTan[(Sqrt[
(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sqrt[11 + 5*Sqrt[5]]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)
] - Sqrt[-11 + 5*Sqrt[5]]*ArcTanh[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)])/(2*Sqrt[10])

Maple [A] (verified)

Time = 11.98 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {\sqrt {5}\, \left (\left (\sqrt {5}-3\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )\right ) \sqrt {-2+2 \sqrt {5}}+\sqrt {2+2 \sqrt {5}}\, \left (3+\sqrt {5}\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )\right )\right )}{40}\)	\(131\)
trager	\(\text {Expression too large to display}\)	\(1571\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1309 vs. \(2 (125) = 250\).

Time = 17.46 (sec) , antiderivative size = 1309, normalized size of antiderivative = 6.51 \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/80*sqrt(10)*sqrt(-5*sqrt(5) + 11)*log((sqrt(10)*(10*x^6 + 15*x^2 + sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1)*sq
rt(-5*sqrt(5) + 11) - sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt(5) + 11) + 10*
(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 -
1)) - 1/80*sqrt(10)*sqrt(-5*sqrt(5) + 11)*log(-(sqrt(10)*(10*x^6 + 15*x^2 + sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4
+ 1)*sqrt(-5*sqrt(5) + 11) - sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt(5) + 1
1) - 10*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8
+ x^4 - 1)) - 1/80*sqrt(10)*sqrt(-5*sqrt(5) - 11)*log((sqrt(10)*(10*x^6 + 15*x^2 - sqrt(5)*(4*x^6 + 7*x^2))*s
qrt(x^4 + 1)*sqrt(-5*sqrt(5) - 11) + sqrt(10)*(10*x^8 + 20*x^4 - sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt(-5*sqrt
(5) - 11) + 10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) + 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4
))/(x^8 + x^4 - 1)) + 1/80*sqrt(10)*sqrt(-5*sqrt(5) - 11)*log(-(sqrt(10)*(10*x^6 + 15*x^2 - sqrt(5)*(4*x^6 + 7
*x^2))*sqrt(x^4 + 1)*sqrt(-5*sqrt(5) - 11) + sqrt(10)*(10*x^8 + 20*x^4 - sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5)*sqrt
(-5*sqrt(5) - 11) - 10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 +
1)^(1/4))/(x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(5*sqrt(5) - 11)*log((10*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/
4) + (sqrt(10)*(10*x^6 + 15*x^2 + sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1) + sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)
*(5*x^8 + 9*x^4 + 2) + 5))*sqrt(5*sqrt(5) - 11) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8
+ x^4 - 1)) + 1/80*sqrt(10)*sqrt(5*sqrt(5) - 11)*log((10*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) - (sqrt(10)*
(10*x^6 + 15*x^2 + sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1) + sqrt(10)*(10*x^8 + 20*x^4 + sqrt(5)*(5*x^8 + 9*x^4
+ 2) + 5))*sqrt(5*sqrt(5) - 11) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) +
1/80*sqrt(10)*sqrt(5*sqrt(5) + 11)*log((10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) + (sqrt(10)*(10*x^6 + 15*x^
2 - sqrt(5)*(4*x^6 + 7*x^2))*sqrt(x^4 + 1) - sqrt(10)*(10*x^8 + 20*x^4 - sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5))*sqr
t(5*sqrt(5) + 11) - 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) - 1/80*sqrt(10)*s
qrt(5*sqrt(5) + 11)*log((10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) - (sqrt(10)*(10*x^6 + 15*x^2 - sqrt(5)*(4*
x^6 + 7*x^2))*sqrt(x^4 + 1) - sqrt(10)*(10*x^8 + 20*x^4 - sqrt(5)*(5*x^8 + 9*x^4 + 2) + 5))*sqrt(5*sqrt(5) + 1
1) - 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1))

Sympy [F(-1)]

Timed out. \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{4} - 1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+2 x^4}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {2\,x^4-1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \]

[In]

int((2*x^4 - 1)/((x^4 + 1)^(1/4)*(x^4 + x^8 - 1)),x)

[Out]

int((2*x^4 - 1)/((x^4 + 1)^(1/4)*(x^4 + x^8 - 1)), x)