\(\int \frac {(-1+x^4) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx\) [996]

Optimal result

Integrand size = 27, antiderivative size = 75 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1-x^2+x^4}\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1+x^2+x^4}\right )}{2 \sqrt {2}} \]

[Out]

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

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.12 (sec) , antiderivative size = 1128, normalized size of antiderivative = 15.04, number of steps used = 20, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6860, 415, 226, 418, 1231, 1721} \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\frac {\sqrt [4]{-1} \left (2 i-\sqrt {6-2 \sqrt {5}}\right ) \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \arctan \left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{16 \sqrt {5}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \arctan \left (\frac {\sqrt [4]{-1} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) \left (2 i+\sqrt {6-2 \sqrt {5}}\right ) \left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \arctan \left (\frac {(-1)^{3/4} x}{\sqrt {x^4+1}}\right )}{\sqrt {10}}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \arctan \left (\frac {(-1)^{3/4} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}}-\frac {\left (3-\sqrt {5}\right ) \left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \left (5-\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3-\sqrt {5}\right ) \left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \left (5-\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3+\sqrt {5}\right ) \left (1+i \sqrt {\frac {2}{3+\sqrt {5}}}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \left (5+\sqrt {5}\right ) \sqrt {x^4+1}}-\frac {\left (3+\sqrt {5}\right ) \left (1-i \sqrt {\frac {2}{3+\sqrt {5}}}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \left (5+\sqrt {5}\right ) \sqrt {x^4+1}}+\frac {\left (1+\sqrt {5}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}+\frac {\left (1-\sqrt {5}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {x^4+1}}+\frac {\left (5-2 \sqrt {5}\right ) \left (3+\sqrt {5}\right ) \left (1+\sqrt {5}-2 i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i-\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{160 \sqrt {x^4+1}}+\frac {\left (5-2 \sqrt {5}\right ) \left (3+\sqrt {5}\right ) \left (1+\sqrt {5}+2 i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {1}{4} i \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (i+\sqrt {\frac {2}{3+\sqrt {5}}}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{160 \sqrt {x^4+1}}-\frac {\left (2+i \sqrt {6-2 \sqrt {5}}\right ) \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{32 \sqrt {5} \sqrt {x^4+1}}+\frac {\left (2 i+\sqrt {6-2 \sqrt {5}}\right ) \left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (-\frac {1}{16} i \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \left (2 i+\sqrt {2 \left (3+\sqrt {5}\right )}\right )^2,2 \arctan (x),\frac {1}{2}\right )}{32 \sqrt {5} \sqrt {x^4+1}} \]

[In]

Int[((-1 + x^4)*Sqrt[1 + x^4])/(1 + 3*x^4 + x^8),x]

[Out]

((-1/4 + I/4)*ArcTan[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/Sqrt[2] - ((-1)^(1/4)*(2*I - Sqrt[6 - 2*Sqrt[5]])*(2*I - S
qrt[2*(3 + Sqrt[5])])*ArcTan[((-1)^(1/4)*x)/Sqrt[1 + x^4]])/(16*Sqrt[5]) + ((1/4 + I/4)*ArcTan[((-1)^(3/4)*x)/
Sqrt[1 + x^4]])/Sqrt[2] + ((1/16 + I/16)*(2*I + Sqrt[6 - 2*Sqrt[5]])*(2 - I*Sqrt[2*(3 + Sqrt[5])])*ArcTan[((-1
)^(3/4)*x)/Sqrt[1 + x^4]])/Sqrt[10] + ((1 - Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[
x], 1/2])/(4*Sqrt[1 + x^4]) + ((1 + Sqrt[5])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2]
)/(4*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 - I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ellip
ticF[2*ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 + Sqrt[5])*(1 + I*Sqrt[2/(3 + Sqrt[5])])*(1 + x^
2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(4*(5 + Sqrt[5])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(
2 - I*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(8*(5 - Sqrt[5
])*Sqrt[1 + x^4]) - ((3 - Sqrt[5])*(2 + I*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Ellipti
cF[2*ArcTan[x], 1/2])/(8*(5 - Sqrt[5])*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sqrt[5])*(1 + Sqrt[5] - (2*I)*Sq
rt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(I/4)*Sqrt[(3 + Sqrt[5])/2]*(I - Sqrt[2/
(3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(160*Sqrt[1 + x^4]) + ((5 - 2*Sqrt[5])*(3 + Sqrt[5])*(1 + Sqrt[5] + (2*I
)*Sqrt[2*(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(-1/4*I)*Sqrt[(3 + Sqrt[5])/2]*(I +
Sqrt[2/(3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(160*Sqrt[1 + x^4]) - ((2 + I*Sqrt[6 - 2*Sqrt[5]])*(2*I + Sqrt[2*
(3 + Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(I/16)*Sqrt[(3 - Sqrt[5])/2]*(2*I - Sqrt[2*(3
+ Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(32*Sqrt[5]*Sqrt[1 + x^4]) + ((2*I + Sqrt[6 - 2*Sqrt[5]])*(2 + I*Sqrt[2*(3
+ Sqrt[5])])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticPi[(-1/16*I)*Sqrt[(3 - Sqrt[5])/2]*(2*I + Sqrt[2*(
3 + Sqrt[5])])^2, 2*ArcTan[x], 1/2])/(32*Sqrt[5]*Sqrt[1 + x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 415

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

Rule 418

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

Rule 1231

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

Rule 1721

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, Simp[(-(B*d - A*e))*(ArcTan[Rt[c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + c*x^4])]/(2*d*e*Rt[c*(d/e) + a*(e/d), 2]))
, x] + Simp[(B*d + A*e)*(A + B*x^2)*(Sqrt[A^2*((a + c*x^4)/(a*(A + B*x^2)^2))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

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 {\left (1-\sqrt {5}\right ) \sqrt {1+x^4}}{3-\sqrt {5}+2 x^4}+\frac {\left (1+\sqrt {5}\right ) \sqrt {1+x^4}}{3+\sqrt {5}+2 x^4}\right ) \, dx \\ & = \left (1-\sqrt {5}\right ) \int \frac {\sqrt {1+x^4}}{3-\sqrt {5}+2 x^4} \, dx+\left (1+\sqrt {5}\right ) \int \frac {\sqrt {1+x^4}}{3+\sqrt {5}+2 x^4} \, dx \\ & = \left (-3-\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4} \left (3+\sqrt {5}+2 x^4\right )} \, dx+\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx+\left (-3+\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4} \left (3-\sqrt {5}+2 x^4\right )} \, dx+\frac {1}{2} \left (1+\sqrt {5}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = \frac {\left (1-\sqrt {5}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}+\frac {\left (1+\sqrt {5}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx \\ & = \frac {\left (1-\sqrt {5}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}+\frac {\left (1+\sqrt {5}\right ) \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}}-\frac {\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (1+\frac {2}{3-\sqrt {5}}\right )}-\frac {\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (1+\frac {2}{3-\sqrt {5}}\right )}-\frac {\left (\left (3+\sqrt {5}\right ) \left (1-i \sqrt {\frac {2}{3+\sqrt {5}}}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (\left (3+\sqrt {5}\right ) \left (1+i \sqrt {\frac {2}{3+\sqrt {5}}}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (i+\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1-i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \sqrt {5}}+\frac {\left (2 i-\sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1+i \sqrt {\frac {2}{3-\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{4 \sqrt {5}}-\frac {\left (2-i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1+i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )}-\frac {\left (2+i \sqrt {2 \left (3+\sqrt {5}\right )}\right ) \int \frac {1+x^2}{\left (1-i \sqrt {\frac {2}{3+\sqrt {5}}} x^2\right ) \sqrt {1+x^4}} \, dx}{2 \left (5+\sqrt {5}\right )} \\ & = \text {Too large to display} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1-x^2+x^4}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt {1+x^4}}{1+x^2+x^4}\right )}{2 \sqrt {2}} \]

[In]

Integrate[((-1 + x^4)*Sqrt[1 + x^4])/(1 + 3*x^4 + x^8),x]

[Out]

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

Maple [A] (verified)

Time = 9.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {x^{4}-\sqrt {2}\, x \sqrt {x^{4}+1}+x^{2}+1}{x^{4}+\sqrt {2}\, x \sqrt {x^{4}+1}+x^{2}+1}\right )+2 \arctan \left (\frac {\sqrt {x^{4}+1}\, \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {\sqrt {x^{4}+1}\, \sqrt {2}-x}{x}\right )\right )}{8}\)	\(94\)
default	\(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{2}\right ) \sqrt {2}}{2}\)	\(102\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {x^{4}+1}{x^{2}}-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}\right )}{2}-\frac {\ln \left (\frac {x^{4}+1}{x^{2}}+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}+1\right )}{2}\right ) \sqrt {2}}{2}\)	\(102\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) x^{4}+2 \sqrt {x^{4}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}+x^{4}+1}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{5} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3}-2 \sqrt {x^{4}+1}\, x}{\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x^{2}-x^{4}-1}\right )}{4}\)	\(150\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.35 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.48 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=-\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i + 1\right ) \, x^{8} - \left (2 i - 2\right ) \, x^{6} + \left (i + 1\right ) \, x^{4} - \left (2 i - 2\right ) \, x^{2} + i + 1\right )} + 4 \, {\left (x^{5} - i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i - 1\right ) \, x^{8} + \left (2 i + 2\right ) \, x^{6} - \left (i - 1\right ) \, x^{4} + \left (2 i + 2\right ) \, x^{2} - i + 1\right )} + 4 \, {\left (x^{5} + i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (\left (i - 1\right ) \, x^{8} - \left (2 i + 2\right ) \, x^{6} + \left (i - 1\right ) \, x^{4} - \left (2 i + 2\right ) \, x^{2} + i - 1\right )} + 4 \, {\left (x^{5} + i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (-\left (i + 1\right ) \, x^{8} + \left (2 i - 2\right ) \, x^{6} - \left (i + 1\right ) \, x^{4} + \left (2 i - 2\right ) \, x^{2} - i - 1\right )} + 4 \, {\left (x^{5} - i \, x^{3} + x\right )} \sqrt {x^{4} + 1}}{x^{8} + 3 \, x^{4} + 1}\right ) \]

[In]

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

[Out]

-(1/16*I + 1/16)*sqrt(2)*log((sqrt(2)*((I + 1)*x^8 - (2*I - 2)*x^6 + (I + 1)*x^4 - (2*I - 2)*x^2 + I + 1) + 4*
(x^5 - I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1)) + (1/16*I - 1/16)*sqrt(2)*log((sqrt(2)*(-(I - 1)*x^8 + (2*
I + 2)*x^6 - (I - 1)*x^4 + (2*I + 2)*x^2 - I + 1) + 4*(x^5 + I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1)) - (1
/16*I - 1/16)*sqrt(2)*log((sqrt(2)*((I - 1)*x^8 - (2*I + 2)*x^6 + (I - 1)*x^4 - (2*I + 2)*x^2 + I - 1) + 4*(x^
5 + I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1)) + (1/16*I + 1/16)*sqrt(2)*log((sqrt(2)*(-(I + 1)*x^8 + (2*I -
2)*x^6 - (I + 1)*x^4 + (2*I - 2)*x^2 - I - 1) + 4*(x^5 - I*x^3 + x)*sqrt(x^4 + 1))/(x^8 + 3*x^4 + 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{4} - 1\right )}}{x^{8} + 3 \, x^{4} + 1} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+3 x^4+x^8} \, dx=\int \frac {\left (x^4-1\right )\,\sqrt {x^4+1}}{x^8+3\,x^4+1} \,d x \]

[In]

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

[Out]

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