\(\int \frac {1+x^8}{\sqrt {1+x^4} (-1+x^8)} \, dx\) [890]

Optimal result

Integrand size = 22, antiderivative size = 67 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {6857, 226, 1418, 425, 537, 418, 1225, 1713, 212, 209} \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {x}{2 \sqrt {x^4+1}} \]

[In]

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

[Out]

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

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 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 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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 1225

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[1/(2*d), Int[1/Sqrt[a + c*x^4], x],
x] + Dist[1/(2*d), Int[(d - e*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] && EqQ[c*d^2 - a*e^2, 0]

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 1713

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> Dist[A, Subst[Int[1/
(d + 2*a*e*x^2), x], x, x/Sqrt[a + c*x^4]], x] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ
[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 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 {1}{\sqrt {1+x^4}}+\frac {2}{\sqrt {1+x^4} \left (-1+x^8\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx+\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}}+2 \int \frac {1}{\left (-1+x^4\right ) \left (1+x^4\right )^{3/2}} \, dx \\ & = -\frac {x}{2 \sqrt {1+x^4}}+\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 {3-x^4}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x}{2 \sqrt {1+x^4}}+\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}{\sqrt {1+x^4}} \, dx+\int \frac {1}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x}{2 \sqrt {1+x^4}}+\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 )}{4 \sqrt {1+x^4}}-\frac {1}{2} \int \frac {1}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{2} \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x}{2 \sqrt {1+x^4}}+\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 )}{4 \sqrt {1+x^4}}-2 \left (\frac {1}{4} \int \frac {1}{\sqrt {1+x^4}} \, dx\right )-\frac {1}{4} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {1}{4} \int \frac {1+x^2}{\left (1-x^2\right ) \sqrt {1+x^4}} \, dx \\ & = -\frac {x}{2 \sqrt {1+x^4}}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right ) \\ & = -\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 5.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09

method	result	size
risch	\(-\frac {x}{2 \sqrt {x^{4}+1}}-\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16}\)	\(73\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}-1\right )}{8}-\frac {\sqrt {2}\, x}{2 \sqrt {x^{4}+1}}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{8}\right ) \sqrt {2}}{2}\)	\(78\)
trager	\(-\frac {x}{2 \sqrt {x^{4}+1}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (x -1\right ) \left (1+x \right )}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{8}\)	\(83\)
default	\(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) x^{4}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+4 \sqrt {2}\, x \sqrt {x^{4}+1}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16 \left (x^{2}+x \sqrt {2}+1\right ) \left (x \sqrt {2}-x^{2}-1\right )}\)	\(166\)
pseudoelliptic	\(\frac {\sqrt {2}\, \left (2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) x^{4}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right ) x^{4}+2 \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )+4 \sqrt {2}\, x \sqrt {x^{4}+1}-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right )}{16 \left (x^{2}+x \sqrt {2}+1\right ) \left (x \sqrt {2}-x^{2}-1\right )}\)	\(166\)

[In]

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

[Out]

-1/2*x/(x^4+1)^(1/2)-1/16*2^(1/2)*(2*arctan(2^(1/2)*x/(x^4+1)^(1/2))+arctanh((x^2-x+1)*2^(1/2)/(x^4+1)^(1/2))-
arctanh((x^2+x+1)*2^(1/2)/(x^4+1)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{16 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

-1/16*(2*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) - sqrt(2)*(x^4 + 1)*log((x^4 - 2*sqrt(2)*sqrt(x^4 +
1)*x + 2*x^2 + 1)/(x^4 - 2*x^2 + 1)) + 8*sqrt(x^4 + 1)*x)/(x^4 + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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