3.24.1 \(\int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} (-1-2 x^4+x^8)} \, dx\) [2301]

3.24.1.1 Optimal result

Integrand size = 34, antiderivative size = 176 \[ \int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1-2 x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}+\frac {3 \arctan \left (\frac {2^{5/8} x \sqrt [4]{1+x^4}}{\sqrt [4]{2} x^2-\sqrt {1+x^4}}\right )}{4\ 2^{5/8}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}-\frac {3 \text {arctanh}\left (\frac {2\ 2^{3/8} x \sqrt [4]{1+x^4}}{2 x^2+2^{3/4} \sqrt {1+x^4}}\right )}{4\ 2^{5/8}} \]

output

arctan(x/(x^4+1)^(1/4))-3/8*arctan(2^(1/8)*x/(x^4+1)^(1/4))*2^(7/8)+3/8*ar 
ctan(2^(5/8)*x*(x^4+1)^(1/4)/(x^2*2^(1/4)-(x^4+1)^(1/2)))*2^(3/8)+arctanh( 
x/(x^4+1)^(1/4))-3/8*arctanh(2^(1/8)*x/(x^4+1)^(1/4))*2^(7/8)-3/8*arctanh( 
2*2^(3/8)*x*(x^4+1)^(1/4)/(2*x^2+2^(3/4)*(x^4+1)^(1/2)))*2^(3/8)
 

3.24.1.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00 \[ \int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1-2 x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}+\frac {3 \arctan \left (\frac {2^{5/8} x \sqrt [4]{1+x^4}}{\sqrt [4]{2} x^2-\sqrt {1+x^4}}\right )}{4\ 2^{5/8}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}-\frac {3 \text {arctanh}\left (\frac {2\ 2^{3/8} x \sqrt [4]{1+x^4}}{2 x^2+2^{3/4} \sqrt {1+x^4}}\right )}{4\ 2^{5/8}} \]

input

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

output

ArcTan[x/(1 + x^4)^(1/4)] - (3*ArcTan[(2^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^( 
1/8)) + (3*ArcTan[(2^(5/8)*x*(1 + x^4)^(1/4))/(2^(1/4)*x^2 - Sqrt[1 + x^4] 
)])/(4*2^(5/8)) + ArcTanh[x/(1 + x^4)^(1/4)] - (3*ArcTanh[(2^(1/8)*x)/(1 + 
 x^4)^(1/4)])/(4*2^(1/8)) - (3*ArcTanh[(2*2^(3/8)*x*(1 + x^4)^(1/4))/(2*x^ 
2 + 2^(3/4)*Sqrt[1 + x^4])])/(4*2^(5/8))
 

3.24.1.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, 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.

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

\(\displaystyle \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \arctan \left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}-\frac {3 \left (1-\sqrt {2}\right ) \arctan \left (1-\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\frac {3 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}+1\right )}{4 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}-\frac {3 \left (1-\sqrt {2}\right ) \log \left (-\frac {2 x}{\sqrt [4]{x^4+1}}+\frac {2^{5/8} x^2}{\sqrt {x^4+1}}+2^{3/8}\right )}{8 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\frac {3 \left (1-\sqrt {2}\right ) \log \left (\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}+\frac {\sqrt [4]{2} x^2}{\sqrt {x^4+1}}+1\right )}{8 \sqrt [8]{2} \left (2-\sqrt {2}\right )}\)

input

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

output

ArcTan[x/(1 + x^4)^(1/4)] - (3*ArcTan[(2^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^( 
1/8)) - (3*(1 - Sqrt[2])*ArcTan[1 - (2^(5/8)*x)/(1 + x^4)^(1/4)])/(4*2^(1/ 
8)*(2 - Sqrt[2])) + (3*(1 - Sqrt[2])*ArcTan[1 + (2^(5/8)*x)/(1 + x^4)^(1/4 
)])/(4*2^(1/8)*(2 - Sqrt[2])) + ArcTanh[x/(1 + x^4)^(1/4)] - (3*ArcTanh[(2 
^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^(1/8)) - (3*(1 - Sqrt[2])*Log[2^(3/8) + ( 
2^(5/8)*x^2)/Sqrt[1 + x^4] - (2*x)/(1 + x^4)^(1/4)])/(8*2^(1/8)*(2 - Sqrt[ 
2])) + (3*(1 - Sqrt[2])*Log[1 + (2^(1/4)*x^2)/Sqrt[1 + x^4] + (2^(5/8)*x)/ 
(1 + x^4)^(1/4)])/(8*2^(1/8)*(2 - Sqrt[2]))
 

3.24.1.3.1 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]
 

3.24.1.4 Maple [C] (verified)

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

Time = 7.94 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.47

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{2}-\frac {\ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}-x}{x}\right )}{2}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{8}-\arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )\)	\(82\)

input

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

output

1/2*ln((x+(x^4+1)^(1/4))/x)-1/2*ln(((x^4+1)^(1/4)-x)/x)+3/8*sum(ln((-_R*x+ 
(x^4+1)^(1/4))/x)/_R,_R=RootOf(_Z^8-2))-arctan((x^4+1)^(1/4)/x)
 

3.24.1.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.43 \[ \int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1-2 x^4+x^8\right )} \, dx=-\frac {3}{16} \cdot 2^{\frac {7}{8}} \log \left (\frac {2^{\frac {1}{8}} x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{16} \cdot 2^{\frac {7}{8}} \log \left (-\frac {2^{\frac {1}{8}} x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{16} i \cdot 2^{\frac {7}{8}} \log \left (\frac {i \cdot 2^{\frac {1}{8}} x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} i \cdot 2^{\frac {7}{8}} \log \left (\frac {-i \cdot 2^{\frac {1}{8}} x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \left (\frac {3}{16} i - \frac {3}{16}\right ) \cdot 2^{\frac {3}{8}} \log \left (\frac {\left (i + 1\right ) \cdot 2^{\frac {5}{8}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \left (\frac {3}{16} i + \frac {3}{16}\right ) \cdot 2^{\frac {3}{8}} \log \left (\frac {-\left (i - 1\right ) \cdot 2^{\frac {5}{8}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \left (\frac {3}{16} i + \frac {3}{16}\right ) \cdot 2^{\frac {3}{8}} \log \left (\frac {\left (i - 1\right ) \cdot 2^{\frac {5}{8}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \left (\frac {3}{16} i - \frac {3}{16}\right ) \cdot 2^{\frac {3}{8}} \log \left (\frac {-\left (i + 1\right ) \cdot 2^{\frac {5}{8}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \]

input

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

output

-3/16*2^(7/8)*log((2^(1/8)*x + (x^4 + 1)^(1/4))/x) + 3/16*2^(7/8)*log(-(2^ 
(1/8)*x - (x^4 + 1)^(1/4))/x) + 3/16*I*2^(7/8)*log((I*2^(1/8)*x + (x^4 + 1 
)^(1/4))/x) - 3/16*I*2^(7/8)*log((-I*2^(1/8)*x + (x^4 + 1)^(1/4))/x) + (3/ 
16*I - 3/16)*2^(3/8)*log(((I + 1)*2^(5/8)*x + 2*(x^4 + 1)^(1/4))/x) - (3/1 
6*I + 3/16)*2^(3/8)*log((-(I - 1)*2^(5/8)*x + 2*(x^4 + 1)^(1/4))/x) + (3/1 
6*I + 3/16)*2^(3/8)*log(((I - 1)*2^(5/8)*x + 2*(x^4 + 1)^(1/4))/x) - (3/16 
*I - 3/16)*2^(3/8)*log((-(I + 1)*2^(5/8)*x + 2*(x^4 + 1)^(1/4))/x) - arcta 
n((x^4 + 1)^(1/4)/x) + 1/2*log((x + (x^4 + 1)^(1/4))/x) - 1/2*log(-(x - (x 
^4 + 1)^(1/4))/x)
 

3.24.1.6 Sympy [F]

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

input

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

output

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

3.24.1.7 Maxima [F]

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

input

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

output

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

3.24.1.8 Giac [F]

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

input

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

output

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

3.24.1.9 Mupad [F(-1)]

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

input

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

output

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