\(\int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} (-2-x^4+x^8)} \, dx\) [1519]

Optimal result

Integrand size = 34, antiderivative size = 105 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \] Output:

-5/3*x/(x^4+1)^(1/4)+arctan(x/(x^4+1)^(1/4))-5/36*arctan(1/2*3^(1/4)*2^(3/ 
4)*x/(x^4+1)^(1/4))*3^(3/4)*2^(1/4)+arctanh(x/(x^4+1)^(1/4))-5/36*arctanh( 
1/2*3^(1/4)*2^(3/4)*x/(x^4+1)^(1/4))*3^(3/4)*2^(1/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {5 x}{3 \sqrt [4]{1+x^4}}+\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{1+x^4}}\right )}{6\ 2^{3/4} \sqrt [4]{3}} \] Input:

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

Output:

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^ 
(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4) 
] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1387, 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.

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

\(\Big \downarrow \) 1387

\(\displaystyle \int \frac {2 x^8-2 x^4+1}{\left (x^4-2\right ) \left (x^4+1\right )^{5/4}}dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (\frac {2 x^4}{\left (x^4+1\right )^{5/4}}+\frac {5}{\left (x^4-2\right ) \left (x^4+1\right )^{5/4}}+\frac {2}{\left (x^4+1\right )^{5/4}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \arctan \left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {5 \text {arctanh}\left (\frac {\sqrt [4]{\frac {3}{2}} x}{\sqrt [4]{x^4+1}}\right )}{6\ 2^{3/4} \sqrt [4]{3}}-\frac {5 x}{3 \sqrt [4]{x^4+1}}\)

Input:

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

Output:

(-5*x)/(3*(1 + x^4)^(1/4)) + ArcTan[x/(1 + x^4)^(1/4)] - (5*ArcTan[((3/2)^ 
(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4) 
] - (5*ArcTanh[((3/2)^(1/4)*x)/(1 + x^4)^(1/4)])/(6*2^(3/4)*3^(1/4))
 

Defintions of rubi rules used

rule 1387

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(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, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 - 
b*d*e + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(172\) vs. \(2(81)=162\).

Time = 1.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65

method	result	size
pseudoelliptic	\(\frac {10 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}} 2^{\frac {1}{4}} 3^{\frac {3}{4}}}{3 x}\right ) 2^{\frac {1}{4}} 3^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}-5 \ln \left (\frac {2^{\frac {3}{4}} 3^{\frac {1}{4}} x +2 \left (x^{4}+1\right )^{\frac {1}{4}}}{-2^{\frac {3}{4}} 3^{\frac {1}{4}} x +2 \left (x^{4}+1\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} 3^{\frac {3}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}-72 \arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}+36 \ln \left (\frac {x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-36 \ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}-x}{x}\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-120 x}{72 \left (x^{4}+1\right )^{\frac {1}{4}}}\)	\(173\)

Input:

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

Output:

1/72*(10*arctan(1/3*(x^4+1)^(1/4)/x*2^(1/4)*3^(3/4))*2^(1/4)*3^(3/4)*(x^4+ 
1)^(1/4)-5*ln((2^(3/4)*3^(1/4)*x+2*(x^4+1)^(1/4))/(-2^(3/4)*3^(1/4)*x+2*(x 
^4+1)^(1/4)))*2^(1/4)*3^(3/4)*(x^4+1)^(1/4)-72*arctan((x^4+1)^(1/4)/x)*(x^ 
4+1)^(1/4)+36*ln((x+(x^4+1)^(1/4))/x)*(x^4+1)^(1/4)-36*ln(((x^4+1)^(1/4)-x 
)/x)*(x^4+1)^(1/4)-120*x)/(x^4+1)^(1/4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (81) = 162\).

Time = 0.10 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.59 \[ \int \frac {1-2 x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-2-x^4+x^8\right )} \, dx=-\frac {10 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \arctan \left (\frac {24^{\frac {1}{4}} x}{2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\right ) + 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (\frac {24^{\frac {1}{4}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 5 \cdot 24^{\frac {3}{4}} {\left (x^{4} + 1\right )} \log \left (-\frac {24^{\frac {1}{4}} x - 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - 288 \, {\left (x^{4} + 1\right )} \arctan \left (\frac {x}{{\left (x^{4} + 1\right )}^{\frac {1}{4}}}\right ) - 144 \, {\left (x^{4} + 1\right )} \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 144 \, {\left (x^{4} + 1\right )} \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + 480 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{288 \, {\left (x^{4} + 1\right )}} \] Input:

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

Output:

-1/288*(10*24^(3/4)*(x^4 + 1)*arctan(1/2*24^(1/4)*x/(x^4 + 1)^(1/4)) + 5*2 
4^(3/4)*(x^4 + 1)*log((24^(1/4)*x + 2*(x^4 + 1)^(1/4))/x) - 5*24^(3/4)*(x^ 
4 + 1)*log(-(24^(1/4)*x - 2*(x^4 + 1)^(1/4))/x) - 288*(x^4 + 1)*arctan(x/( 
x^4 + 1)^(1/4)) - 144*(x^4 + 1)*log((x + (x^4 + 1)^(1/4))/x) + 144*(x^4 + 
1)*log(-(x - (x^4 + 1)^(1/4))/x) + 480*(x^4 + 1)^(3/4)*x)/(x^4 + 1)
 

Sympy [F]

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

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

Output:

Integral((2*x**8 - 2*x**4 + 1)/((x**4 - 2)*(x**4 + 1)**(5/4)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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