\(\int \frac {1}{\sqrt [4]{1+x^4} (-1+x^8)} \, dx\) [888]

Optimal result

Integrand size = 17, antiderivative size = 67 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1418, 390, 385, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [4]{2}}-\frac {x}{2 \sqrt [4]{x^4+1}} \]

[In]

Int[1/((1 + x^4)^(1/4)*(-1 + x^8)),x]

[Out]

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

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 390

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[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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]

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(-\frac {-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}+\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{4}+1\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}+8 x}{16 \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(84\)
risch	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}-4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(237\)
trager	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (x -1\right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(238\)

[In]

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

[Out]

-1/16*(-2*arctan(1/2*2^(3/4)/x*(x^4+1)^(1/4))*2^(3/4)*(x^4+1)^(1/4)+ln((2^(1/4)*x+(x^4+1)^(1/4))/(-2^(1/4)*x+(
x^4+1)^(1/4)))*2^(3/4)*(x^4+1)^(1/4)+8*x)/(x^4+1)^(1/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.04 (sec) , antiderivative size = 312, normalized size of antiderivative = 4.66 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=-\frac {2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - 2^{\frac {3}{4}} {\left (-i \, x^{4} - i\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{4} + i\right )} - 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - 2^{\frac {3}{4}} {\left (i \, x^{4} + i\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{4} - i\right )} - 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} + 1\right )}} \]

[In]

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

[Out]

-1/32*(2^(3/4)*(x^4 + 1)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4)*sqrt(x^4 + 1)*x^2 + 2^(3/4)*(3*x^4 + 1
) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 2^(3/4)*(-I*x^4 - I)*log(-(4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*I*2^(1/4)*s
qrt(x^4 + 1)*x^2 - 2^(3/4)*(3*I*x^4 + I) - 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 2^(3/4)*(I*x^4 + I)*log(-(4*sqrt(
2)*(x^4 + 1)^(1/4)*x^3 - 4*I*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(-3*I*x^4 - I) - 4*(x^4 + 1)^(3/4)*x)/(x^4 -
1)) - 2^(3/4)*(x^4 + 1)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(3*x^4 + 1)
+ 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) + 16*(x^4 + 1)^(3/4)*x)/(x^4 + 1)

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {1}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {5}{4}}}\, dx \]

[In]

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

[Out]

Integral(1/((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)**(5/4)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/((x^4 + 1)^(1/4)*(x^8 - 1)),x)

[Out]

int(1/((x^4 + 1)^(1/4)*(x^8 - 1)), x)