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

Optimal result

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

-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*arcta 
nh(2^(1/4)*x/(x^4+1)^(1/4))*2^(3/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt [4]{1+x^4}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [4]{2}} \] Input:

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

Output:

-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))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {1388, 907, 25, 902, 756, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1388

\(\displaystyle \int \frac {1}{\left (x^4-1\right ) \left (x^4+1\right )^{5/4}}dx\)

\(\Big \downarrow \) 907

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 902

\(\displaystyle -\frac {1}{2} \int \frac {1}{1-\frac {2 x^4}{x^4+1}}d\frac {x}{\sqrt [4]{x^4+1}}-\frac {x}{2 \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{2}}\right )-\frac {x}{2 \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{2}}\right )-\frac {x}{2 \sqrt [4]{x^4+1}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 756

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

rule 902

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su 
bst[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 907

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] + Simp[(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] && EqQ[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  ! 
LtQ[q, -1]) && NeQ[p, -1]
 

rule 1388

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

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.25

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{4}+1\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}+8 x}{16 \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(84\)
risch	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {-\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}+3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(237\)
trager	\(-\frac {x}{2 \left (x^{4}+1\right )^{\frac {1}{4}}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {-\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{16}\)	\(241\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (47) = 94\).

Time = 2.76 (sec) , antiderivative size = 209, normalized size of antiderivative = 3.12 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx=\frac {2 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {2 \, {\left (2^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x\right )}}{x^{4} - 1}\right ) - 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) + 2^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{32 \, {\left (x^{4} + 1\right )}} \] Input:

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

Output:

1/32*(2*2^(3/4)*(x^4 + 1)*arctan(2*(2^(3/4)*(x^4 + 1)^(1/4)*x^3 + 2^(1/4)* 
(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)) + 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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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