\(\int \frac {-1+x^2}{(1+x^2) \sqrt [4]{x^2+x^6}} \, dx\) [1426]

Optimal result

Integrand size = 24, antiderivative size = 101 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{2^{3/4}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{2^{3/4}} \] Output:

1/2*arctan(2^(3/4)*x*(x^6+x^2)^(1/4)/(2^(1/2)*x^2-(x^6+x^2)^(1/2)))*2^(1/4 
)-1/2*arctanh((1/2*x^2*2^(3/4)+1/2*(x^6+x^2)^(1/2)*2^(1/4))/x/(x^6+x^2)^(1 
/4))*2^(1/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.18 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\frac {\sqrt {x} \sqrt [4]{1+x^4} \left (\arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )-\text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{2^{3/4} \sqrt [4]{x^2+x^6}} \] Input:

Integrate[(-1 + x^2)/((1 + x^2)*(x^2 + x^6)^(1/4)),x]
 

Output:

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

Rubi [C] (warning: unable to verify)

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

Time = 0.51 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2467, 25, 2035, 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.

\(\displaystyle \int \frac {x^2-1}{\left (x^2+1\right ) \sqrt [4]{x^6+x^2}} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x} \sqrt [4]{x^4+1} \int -\frac {1-x^2}{\sqrt {x} \left (x^2+1\right ) \sqrt [4]{x^4+1}}dx}{\sqrt [4]{x^6+x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4+1} \int \frac {1-x^2}{\sqrt {x} \left (x^2+1\right ) \sqrt [4]{x^4+1}}dx}{\sqrt [4]{x^6+x^2}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \frac {1-x^2}{\left (x^2+1\right ) \sqrt [4]{x^4+1}}d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\)

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \left (\frac {2}{\left (x^2+1\right ) \sqrt [4]{x^4+1}}-\frac {1}{\sqrt [4]{x^4+1}}\right )d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \left (2 \sqrt {x} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},x^4,-x^4\right )-\frac {2}{5} x^{5/2} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},x^4,-x^4\right )-\sqrt {x} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )\right )}{\sqrt [4]{x^6+x^2}}\)

Input:

Int[(-1 + x^2)/((1 + x^2)*(x^2 + x^6)^(1/4)),x]
 

Output:

(-2*Sqrt[x]*(1 + x^4)^(1/4)*(2*Sqrt[x]*AppellF1[1/8, 1, 1/4, 9/8, x^4, -x^ 
4] - (2*x^(5/2)*AppellF1[5/8, 1, 1/4, 13/8, x^4, -x^4])/5 - Sqrt[x]*Hyperg 
eometric2F1[1/8, 1/4, 9/8, -x^4]))/(x^2 + x^6)^(1/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

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 = 1.92 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.31

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )\right )}{4}\)	\(132\)

Input:

int((x^2-1)/(x^2+1)/(x^6+x^2)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/4*2^(1/4)*(ln((-2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^ 
(1/2))/(2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^(1/2)))+2* 
arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x)+2*arctan((2^(1/4)*(x^2*(x^4+1))^ 
(1/4)-x)/x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (83) = 166\).

Time = 13.98 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.96 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {1}{8} \cdot 8^{\frac {3}{4}} \arctan \left (-\frac {8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )} - 4 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{8 \, {\left (x^{5} + x\right )}}\right ) - \frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (\frac {x^{5} + 8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, x^{3} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + 2 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}\right ) + \frac {1}{32} \cdot 8^{\frac {3}{4}} \log \left (\frac {x^{5} - 8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, x^{3} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x - 2 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}\right ) \] Input:

integrate((x^2-1)/(x^2+1)/(x^6+x^2)^(1/4),x, algorithm="fricas")
 

Output:

-1/8*8^(3/4)*arctan(-1/8*(8^(3/4)*(x^6 + x^2)^(1/4)*(x^4 + 1) - 4*8^(1/4)* 
(x^6 + x^2)^(3/4))/(x^5 + x)) - 1/32*8^(3/4)*log((x^5 + 8^(3/4)*(x^6 + x^2 
)^(1/4)*x^2 + 2*x^3 + 4*sqrt(2)*sqrt(x^6 + x^2)*x + 2*8^(1/4)*(x^6 + x^2)^ 
(3/4) + x)/(x^5 + 2*x^3 + x)) + 1/32*8^(3/4)*log((x^5 - 8^(3/4)*(x^6 + x^2 
)^(1/4)*x^2 + 2*x^3 + 4*sqrt(2)*sqrt(x^6 + x^2)*x - 2*8^(1/4)*(x^6 + x^2)^ 
(3/4) + x)/(x^5 + 2*x^3 + x))
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )}{\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x^{2} + 1\right )}\, dx \] Input:

integrate((x**2-1)/(x**2+1)/(x**6+x**2)**(1/4),x)
 

Output:

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

Maxima [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate((x^2-1)/(x^2+1)/(x^6+x^2)^(1/4),x, algorithm="maxima")
 

Output:

integrate((x^2 - 1)/((x^6 + x^2)^(1/4)*(x^2 + 1)), x)
 

Giac [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate((x^2-1)/(x^2+1)/(x^6+x^2)^(1/4),x, algorithm="giac")
 

Output:

integrate((x^2 - 1)/((x^6 + x^2)^(1/4)*(x^2 + 1)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {x^2-1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^2+1\right )} \,d x \] Input:

int((x^2 - 1)/((x^2 + x^6)^(1/4)*(x^2 + 1)),x)
 

Output:

int((x^2 - 1)/((x^2 + x^6)^(1/4)*(x^2 + 1)), x)
 

Reduce [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [4]{x^2+x^6}} \, dx=\frac {4 \sqrt {x}\, \left (x^{4}+1\right )^{\frac {5}{4}}+\frac {2 \sqrt {x}\, \left (x^{4}+1\right )^{\frac {1}{4}} x^{4}}{3}+\frac {2 \sqrt {x}\, \left (x^{4}+1\right )^{\frac {1}{4}}}{3}-\frac {10 \sqrt {x^{4}+1}\, \left (\int \frac {\left (x^{4}+1\right )^{\frac {3}{4}}}{\sqrt {x}\, x^{10}+\sqrt {x}\, x^{8}+2 \sqrt {x}\, x^{6}+2 \sqrt {x}\, x^{4}+\sqrt {x}\, x^{2}+\sqrt {x}}d x \right ) x^{4}}{3}-\frac {10 \sqrt {x^{4}+1}\, \left (\int \frac {\left (x^{4}+1\right )^{\frac {3}{4}}}{\sqrt {x}\, x^{10}+\sqrt {x}\, x^{8}+2 \sqrt {x}\, x^{6}+2 \sqrt {x}\, x^{4}+\sqrt {x}\, x^{2}+\sqrt {x}}d x \right )}{3}+2 \sqrt {x^{4}+1}\, \left (\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x^{5}}{x^{10}+x^{8}+2 x^{6}+2 x^{4}+x^{2}+1}d x \right ) x^{4}+2 \sqrt {x^{4}+1}\, \left (\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x^{5}}{x^{10}+x^{8}+2 x^{6}+2 x^{4}+x^{2}+1}d x \right )-\frac {4 \sqrt {x^{4}+1}\, \left (\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x}{x^{10}+x^{8}+2 x^{6}+2 x^{4}+x^{2}+1}d x \right ) x^{4}}{3}-\frac {4 \sqrt {x^{4}+1}\, \left (\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x}{x^{10}+x^{8}+2 x^{6}+2 x^{4}+x^{2}+1}d x \right )}{3}}{\sqrt {x^{4}+1}\, \left (x^{4}+1\right )} \] Input:

int((x^2-1)/(x^2+1)/(x^6+x^2)^(1/4),x)
 

Output:

(2*(6*sqrt(x)*(x**4 + 1)**(5/4) + sqrt(x)*(x**4 + 1)**(1/4)*x**4 + sqrt(x) 
*(x**4 + 1)**(1/4) - 5*sqrt(x**4 + 1)*int((x**4 + 1)**(3/4)/(sqrt(x)*x**10 
 + sqrt(x)*x**8 + 2*sqrt(x)*x**6 + 2*sqrt(x)*x**4 + sqrt(x)*x**2 + sqrt(x) 
),x)*x**4 - 5*sqrt(x**4 + 1)*int((x**4 + 1)**(3/4)/(sqrt(x)*x**10 + sqrt(x 
)*x**8 + 2*sqrt(x)*x**6 + 2*sqrt(x)*x**4 + sqrt(x)*x**2 + sqrt(x)),x) + 3* 
sqrt(x**4 + 1)*int((sqrt(x)*(x**4 + 1)**(3/4)*x**5)/(x**10 + x**8 + 2*x**6 
 + 2*x**4 + x**2 + 1),x)*x**4 + 3*sqrt(x**4 + 1)*int((sqrt(x)*(x**4 + 1)** 
(3/4)*x**5)/(x**10 + x**8 + 2*x**6 + 2*x**4 + x**2 + 1),x) - 2*sqrt(x**4 + 
 1)*int((sqrt(x)*(x**4 + 1)**(3/4)*x)/(x**10 + x**8 + 2*x**6 + 2*x**4 + x* 
*2 + 1),x)*x**4 - 2*sqrt(x**4 + 1)*int((sqrt(x)*(x**4 + 1)**(3/4)*x)/(x**1 
0 + x**8 + 2*x**6 + 2*x**4 + x**2 + 1),x)))/(3*sqrt(x**4 + 1)*(x**4 + 1))