\(\int \frac {-1+x^8}{\sqrt {1+x^4} (1+x^8)} \, dx\) [683]

Optimal result

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

-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)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x^8}{\sqrt {1+x^4} \left (1+x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt [4]{2}} \] Input:

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

Output:

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

Rubi [C] (verified)

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

Time = 0.77 (sec) , antiderivative size = 314, normalized size of antiderivative = 5.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1388, 7276, 2009}

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.

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

\(\Big \downarrow \) 1388

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}+\frac {i \left (\sqrt {2}+(1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {i \left (\sqrt {2}+(-1+i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}-\frac {i \left (\sqrt {2}+(-1-i)\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {\left ((-1-i)-i \sqrt {2}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{8 \sqrt {x^4+1}}+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt [4]{2}}\)

Input:

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

Output:

-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)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2 
*ArcTan[x], 1/2])/(2*Sqrt[1 + x^4]) + (((-1 - I) - I*Sqrt[2])*(1 + x^2)*Sq 
rt[(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]*Ellipti 
cF[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] + 
 ((I/8)*((1 + I) + Sqrt[2])*(1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*Elliptic 
F[2*ArcTan[x], 1/2])/Sqrt[1 + x^4]
 

Defintions of rubi rules used

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]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7276

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

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )-\ln \left (\frac {2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}{-2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}\right )\right )}{8}\)	\(58\)
pseudoelliptic	\(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \sqrt {x^{4}+1}}{2 x}\right )-\ln \left (\frac {2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}{-2^{\frac {1}{4}} x +\sqrt {x^{4}+1}}\right )\right )}{8}\)	\(58\)
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	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x^{2}+2 x^{4} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )+8 \sqrt {x^{4}+1}\, x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}-2 x^{4}-2}\right )}{8}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{2} \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{4}+8 \sqrt {x^{4}+1}\, x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{2 x^{4}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x^{2}+2}\right )}{8}\)	\(184\)

Input:

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

Output:

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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (37) = 74\).

Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 3.11 \[ \int \frac {-1+x^8}{\sqrt {1+x^4} \left (1+x^8\right )} \, dx=-\frac {1}{8} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {2 \, \sqrt {x^{4} + 1} {\left (2^{\frac {3}{4}} x^{3} + 2^{\frac {1}{4}} {\left (x^{5} + x\right )}\right )}}{x^{8} + 1}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} + 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {3}{4}} {\left (x^{8} + 4 \, x^{4} + 1\right )} - 4 \, {\left (x^{5} + \sqrt {2} x^{3} + x\right )} \sqrt {x^{4} + 1} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}}{x^{8} + 1}\right ) \] Input:

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

Output:

-1/8*2^(3/4)*arctan(2*sqrt(x^4 + 1)*(2^(3/4)*x^3 + 2^(1/4)*(x^5 + x))/(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*2^(3/4)*l 
og((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))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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