\(\int \frac {(-1+a x^8) (1+a x^8)^{3/4}}{1+x^8+a^2 x^{16}} \, dx\) [3028]

Optimal result

Integrand size = 33, antiderivative size = 432 \[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\frac {\left (1+\sqrt [4]{-1}\right ) \arctan \left (\frac {(-1)^{7/8} \sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt [4]{-1+2 a} x^2+\sqrt {1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}}-\frac {i \left (-i \sqrt {2}+\sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \arctan \left (\frac {(-1)^{7/8} \left (-2+\sqrt {2}\right ) \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}{(-1)^{3/4} \sqrt {2-\sqrt {2}} \sqrt [4]{-1+2 a} x^2+\sqrt {2-\sqrt {2}} \sqrt {1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (\sqrt {2}+i \sqrt {2 \left (3-2 \sqrt {2}\right )}\right ) \text {arctanh}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2-\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{16 \sqrt [8]{-1+2 a}}+\frac {\left (1+\sqrt [4]{-1}\right ) \text {arctanh}\left (\frac {(-1)^{7/8} \sqrt [4]{-1+2 a} x^2-\sqrt [8]{-1} \sqrt {1+a x^8}}{\sqrt {2+\sqrt {2}} \sqrt [8]{-1+2 a} x \sqrt [4]{1+a x^8}}\right )}{8 \sqrt [8]{-1+2 a}} \] Output:

1/8*(1+(-1)^(1/4))*arctan((-1)^(7/8)*(2+2^(1/2))^(1/2)*(-1+2*a)^(1/8)*x*(a 
*x^8+1)^(1/4)/((-1)^(3/4)*(-1+2*a)^(1/4)*x^2+(a*x^8+1)^(1/2)))/(-1+2*a)^(1 
/8)-1/16*I*(-I*2^(1/2)+2-2^(1/2))*arctan((-1)^(7/8)*(-2+2^(1/2))*(-1+2*a)^ 
(1/8)*x*(a*x^8+1)^(1/4)/((-1)^(3/4)*(2-2^(1/2))^(1/2)*(-1+2*a)^(1/4)*x^2+( 
2-2^(1/2))^(1/2)*(a*x^8+1)^(1/2)))/(-1+2*a)^(1/8)+1/16*(2^(1/2)+I*(2-2^(1/ 
2)))*arctanh(((-1)^(7/8)*(-1+2*a)^(1/4)*x^2-(-1)^(1/8)*(a*x^8+1)^(1/2))/(2 
-2^(1/2))^(1/2)/(-1+2*a)^(1/8)/x/(a*x^8+1)^(1/4))/(-1+2*a)^(1/8)+1/8*(1+(- 
1)^(1/4))*arctanh(((-1)^(7/8)*(-1+2*a)^(1/4)*x^2-(-1)^(1/8)*(a*x^8+1)^(1/2 
))/(2+2^(1/2))^(1/2)/(-1+2*a)^(1/8)/x/(a*x^8+1)^(1/4))/(-1+2*a)^(1/8)
 

Mathematica [F]

\[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx \] Input:

Integrate[((-1 + a*x^8)*(1 + a*x^8)^(3/4))/(1 + x^8 + a^2*x^16),x]
 

Output:

Integrate[((-1 + a*x^8)*(1 + a*x^8)^(3/4))/(1 + x^8 + a^2*x^16), x]
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.63 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.37, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {7279, 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.

\(\displaystyle \int \frac {\left (a x^8-1\right ) \left (a x^8+1\right )^{3/4}}{a^2 x^{16}+x^8+1} \, dx\)

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (1-\frac {2 a+1}{\sqrt {1-4 a^2}}\right ) x \operatorname {AppellF1}\left (\frac {1}{8},-\frac {3}{4},1,\frac {9}{8},-a x^8,-\frac {2 a^2 x^8}{1-\sqrt {1-4 a^2}}\right )}{1-\sqrt {1-4 a^2}}+\frac {a \left (\frac {2 a+1}{\sqrt {1-4 a^2}}+1\right ) x \operatorname {AppellF1}\left (\frac {1}{8},-\frac {3}{4},1,\frac {9}{8},-a x^8,-\frac {2 a^2 x^8}{\sqrt {1-4 a^2}+1}\right )}{\sqrt {1-4 a^2}+1}\)

Input:

Int[((-1 + a*x^8)*(1 + a*x^8)^(3/4))/(1 + x^8 + a^2*x^16),x]
 

Output:

(a*(1 - (1 + 2*a)/Sqrt[1 - 4*a^2])*x*AppellF1[1/8, -3/4, 1, 9/8, -(a*x^8), 
 (-2*a^2*x^8)/(1 - Sqrt[1 - 4*a^2])])/(1 - Sqrt[1 - 4*a^2]) + (a*(1 + (1 + 
 2*a)/Sqrt[1 - 4*a^2])*x*AppellF1[1/8, -3/4, 1, 9/8, -(a*x^8), (-2*a^2*x^8 
)/(1 + Sqrt[1 - 4*a^2])])/(1 + Sqrt[1 - 4*a^2])
 

Defintions of rubi rules used

rule 2009

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

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.09

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a +1\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{8}+1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{8}\)	\(38\)

Input:

int((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x,method=_RETURNVERBOSE)
 

Output:

1/8*sum(ln((-_R*x+(a*x^8+1)^(1/4))/x)/_R,_R=RootOf(_Z^8-2*a+1))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\text {Timed out} \] Input:

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="fricas" 
)
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\text {Timed out} \] Input:

integrate((a*x**8-1)*(a*x**8+1)**(3/4)/(a**2*x**16+x**8+1),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\int { \frac {{\left (a x^{8} + 1\right )}^{\frac {3}{4}} {\left (a x^{8} - 1\right )}}{a^{2} x^{16} + x^{8} + 1} \,d x } \] Input:

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="maxima" 
)
 

Output:

integrate((a*x^8 + 1)^(3/4)*(a*x^8 - 1)/(a^2*x^16 + x^8 + 1), x)
 

Giac [F]

\[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\int { \frac {{\left (a x^{8} + 1\right )}^{\frac {3}{4}} {\left (a x^{8} - 1\right )}}{a^{2} x^{16} + x^{8} + 1} \,d x } \] Input:

integrate((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x, algorithm="giac")
 

Output:

integrate((a*x^8 + 1)^(3/4)*(a*x^8 - 1)/(a^2*x^16 + x^8 + 1), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=\int \frac {\left (a\,x^8-1\right )\,{\left (a\,x^8+1\right )}^{3/4}}{a^2\,x^{16}+x^8+1} \,d x \] Input:

int(((a*x^8 - 1)*(a*x^8 + 1)^(3/4))/(x^8 + a^2*x^16 + 1),x)
 

Output:

int(((a*x^8 - 1)*(a*x^8 + 1)^(3/4))/(x^8 + a^2*x^16 + 1), x)
 

Reduce [F]

\[ \int \frac {\left (-1+a x^8\right ) \left (1+a x^8\right )^{3/4}}{1+x^8+a^2 x^{16}} \, dx=-\left (\int \frac {\left (a \,x^{8}+1\right )^{\frac {3}{4}}}{a^{2} x^{16}+x^{8}+1}d x \right )+\left (\int \frac {\left (a \,x^{8}+1\right )^{\frac {3}{4}} x^{8}}{a^{2} x^{16}+x^{8}+1}d x \right ) a \] Input:

int((a*x^8-1)*(a*x^8+1)^(3/4)/(a^2*x^16+x^8+1),x)
 

Output:

 - int((a*x**8 + 1)**(3/4)/(a**2*x**16 + x**8 + 1),x) + int(((a*x**8 + 1)* 
*(3/4)*x**8)/(a**2*x**16 + x**8 + 1),x)*a