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

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {x}{2 \sqrt {1+x^4}}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, 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

\(\displaystyle \int \frac {x^8+1}{\left (x^4-1\right ) \left (x^4+1\right )^{3/2}}dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (\frac {x^4}{\left (x^4+1\right )^{3/2}}+\frac {2}{\left (x^4-1\right ) \left (x^4+1\right )^{3/2}}+\frac {1}{\left (x^4+1\right )^{3/2}}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 = 2.23 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34 \[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=-\frac {2 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \, \sqrt {x^{4} + 1} x}{16 \, {\left (x^{4} + 1\right )}} \] Input:

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

Output:

-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]

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

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

Output:

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

Maxima [F]

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

\[ \int \frac {1+x^8}{\sqrt {1+x^4} \left (-1+x^8\right )} \, dx=\int \frac {\sqrt {x^{4}+1}}{x^{12}+x^{8}-x^{4}-1}d x +\int \frac {\sqrt {x^{4}+1}\, x^{8}}{x^{12}+x^{8}-x^{4}-1}d x \] Input:

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

Output:

int(sqrt(x**4 + 1)/(x**12 + x**8 - x**4 - 1),x) + int((sqrt(x**4 + 1)*x**8 
)/(x**12 + x**8 - x**4 - 1),x)