\(\int \frac {(-1+x^4) \sqrt {1+x^4}}{1+x^8} \, dx\) [682]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}} \]

[Out]

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

Rubi [C] (verified)

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

Time = 0.50 (sec) , antiderivative size = 314, normalized size of antiderivative = 5.92, number of steps used = 20, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6857, 415, 226, 418, 1231, 1721} \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {\left ((-1-i)-i \sqrt {2}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{\sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}} \]

[In]

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

[Out]

-1/2*ArcTan[(2^(1/4)*x)/Sqrt[1 + x^4]]/2^(1/4) - ArcTanh[(2^(1/4)*x)/Sqrt[1 + x^4]]/(2*2^(1/4)) + ((1 + x^2)*S
qrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4]) - ((1/8 - I/8)*(1 + (-1)^(3/4))*(1 +
x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + (((-1 - I) - I*Sqrt[2])*(1 + x^
2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/(8*Sqrt[1 + x^4]) + ((I/8)*((-1 + I) + Sqrt[2])*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/Sqrt[1 + x^4] + ((I/8)*((1 + I) + Sqrt[2])*(1
+ x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 1/2])/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 6857

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1+x^4}}{i-x^4}+\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+x^4}}{i+x^4}\right ) \, dx \\ & = \left (-\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i-x^4} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {\sqrt {1+x^4}}{i+x^4} \, dx \\ & = -\left (i \int \frac {1}{\left (i-x^4\right ) \sqrt {1+x^4}} \, dx\right )-i \int \frac {1}{\left (i+x^4\right ) \sqrt {1+x^4}} \, dx+\left (\frac {1}{2}-\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx+\left (\frac {1}{2}+\frac {i}{2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = \frac {\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 )}{2 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx \\ & = \frac {\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 )}{2 \sqrt {1+x^4}}-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}-\frac {i}{4}\right ) \sqrt [4]{-1} \left (1+\sqrt [4]{-1}\right )\right ) \int \frac {1+x^2}{\left (1+\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (-1)^{3/4} \left (1-(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1-(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\left (\left (\frac {1}{4}-\frac {i}{4}\right ) \left (1+(-1)^{3/4}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\left (\left (-\frac {1}{4}+\frac {i}{4}\right ) (-1)^{3/4} \left (1+(-1)^{3/4}\right )\right ) \int \frac {1+x^2}{\left (1+(-1)^{3/4} x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \left ((1+i)-i \sqrt {2}\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx-\frac {1}{4} \left (i \left ((-1-i)+\sqrt {2}\right )\right ) \int \frac {1+x^2}{\left (1-\sqrt [4]{-1} x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{4} \left (i \left ((1+i)+\sqrt {2}\right )\right ) \int \frac {1}{\sqrt {1+x^4}} \, dx \\ & = -\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}}+\frac {\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 )}{2 \sqrt {1+x^4}}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (1+\sqrt [4]{-1}\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 )}{\sqrt {1+x^4}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (1+(-1)^{3/4}\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 )}{\sqrt {1+x^4}}-\frac {\left ((1+i)-i \sqrt {2}\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 )}{8 \sqrt {1+x^4}}+\frac {i \left ((1+i)+\sqrt {2}\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 )}{8 \sqrt {1+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.53 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {3}{4}} \left (-2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )+\ln \left (\frac {2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}{-2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}\right )\right )}{8}\)	\(56\)
default	\(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )-\ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )\right )}{8}\)	\(73\)
elliptic	\(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )-\ln \left (\frac {\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}+\frac {2^{\frac {3}{4}}}{2}}{\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-\frac {2^{\frac {3}{4}}}{2}}\right )\right )}{8}\)	\(73\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+8 \sqrt {x^{4}+1}\, x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{2 x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+8 \sqrt {x^{4}+1}\, x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 x^{4}-2}\right )}{8}\)	\(185\)

[In]

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

[Out]

-1/8*2^(3/4)*(-2*arctan(1/2*2^(3/4)/x*(x^4+1)^(1/2))+ln((2^(1/4)*x+(x^4+1)^(1/2))/(-2^(1/4)*x+(x^4+1)^(1/2))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 264, normalized size of antiderivative = 4.98 \[ \int \frac {\left (-1+x^4\right ) \sqrt {1+x^4}}{1+x^8} \, dx=-\frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} + 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} - 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) + \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (i \, x^{8} + 4 i \, x^{4} + i\right )} - 4 \, {\left (x^{5} - \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} - 4 \cdot 2^{\frac {1}{4}} {\left (i \, x^{6} + i \, x^{2}\right )}}{x^{8} + 1}\right ) - \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (-i \, x^{8} - 4 i \, x^{4} - i\right )} - 4 \, {\left (x^{5} - \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} - 4 \cdot 2^{\frac {1}{4}} {\left (-i \, x^{6} - i \, x^{2}\right )}}{x^{8} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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