\(\int \frac {1+x^6}{\sqrt {1-x^2+x^4} (1-x^6)} \, dx\) [678]

Optimal result

Integrand size = 29, antiderivative size = 53 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{3} \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2+x^4}}\right )+\frac {1}{3} \text {arctanh}\left (\frac {x}{\sqrt {1-x^2+x^4}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{3} \left (\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt {1-x^2+x^4}}\right )\right ) \] Input:

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

Output:

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 2.22 (sec) , antiderivative size = 371, normalized size of antiderivative = 7.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2019, 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.

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

\(\Big \downarrow \) 2019

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

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (\frac {\sqrt {x^4-x^2+1}}{3 (1-x)}+\frac {\sqrt {x^4-x^2+1}}{3 (x+1)}+\frac {\left (1-\sqrt [3]{-1}\right ) \sqrt {x^4-x^2+1}}{6 \left (1-\sqrt [3]{-1} x\right )}+\frac {\left (1-\sqrt [3]{-1}\right ) \sqrt {x^4-x^2+1}}{6 \left (\sqrt [3]{-1} x+1\right )}+\frac {\left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}{6 \left (1-(-1)^{2/3} x\right )}+\frac {\left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}{6 \left ((-1)^{2/3} x+1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4-x^2+1}}\right )+\frac {\left (1+3 i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{12 \sqrt {x^4-x^2+1}}+\frac {\left (1-3 i \sqrt {3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{12 \sqrt {x^4-x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {3}{4}\right )}{6 \sqrt {x^4-x^2+1}}-\frac {\left (1-(-1)^{2/3}\right ) \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan (x),\frac {3}{4}\right )}{6 \left (1+(-1)^{2/3}\right ) \sqrt {x^4-x^2+1}}-\frac {i \left (x^2+1\right ) \sqrt {\frac {x^4-x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{4},2 \arctan (x),\frac {3}{4}\right )}{2 \sqrt {3} \sqrt {x^4-x^2+1}}+\frac {1}{3} \text {arctanh}\left (\frac {x}{\sqrt {x^4-x^2+1}}\right )\)

Input:

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

Output:

(Sqrt[2]*ArcTan[(Sqrt[2]*x)/Sqrt[1 - x^2 + x^4]])/3 + ArcTanh[x/Sqrt[1 - x 
^2 + x^4]]/3 - ((1 + x^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*Ar 
cTan[x], 3/4])/(6*Sqrt[1 - x^2 + x^4]) + ((1 - (3*I)*Sqrt[3])*(1 + x^2)*Sq 
rt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x], 3/4])/(12*Sqrt[1 - 
x^2 + x^4]) + ((1 + (3*I)*Sqrt[3])*(1 + x^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2 
)^2]*EllipticF[2*ArcTan[x], 3/4])/(12*Sqrt[1 - x^2 + x^4]) - ((I/2)*(1 + x 
^2)*Sqrt[(1 - x^2 + x^4)/(1 + x^2)^2]*EllipticPi[1/4, 2*ArcTan[x], 3/4])/( 
Sqrt[3]*Sqrt[1 - x^2 + x^4]) - ((1 - (-1)^(2/3))*(1 + x^2)*Sqrt[(1 - x^2 + 
 x^4)/(1 + x^2)^2]*EllipticPi[1/4, 2*ArcTan[x], 3/4])/(6*(1 + (-1)^(2/3))* 
Sqrt[1 - x^2 + x^4])
 

Defintions of rubi rules used

rule 2009

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

rule 2019

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px 
, Qx, x]^p*Qx^(p + q), x] /; FreeQ[q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && 
 EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]
 

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 = 2.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98

method	result	size
elliptic	\(\frac {\left (-\frac {2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}-x^{2}+1}}{2 x}\right )}{3}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {x^{4}-x^{2}+1}}{x}\right )}{3}\right ) \sqrt {2}}{2}\)	\(52\)
trager	\(\frac {\ln \left (-\frac {\sqrt {x^{4}-x^{2}+1}+x}{\left (-1+x \right ) \left (1+x \right )}\right )}{3}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{4}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 \sqrt {x^{4}-x^{2}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{\left (x^{2}+x +1\right ) \left (x^{2}-x +1\right )}\right )}{6}\)	\(106\)
default	\(\frac {\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}-3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}+3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}\)	\(116\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {2 x^{2}-3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}-3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (x^{2}+3 x +1\right ) \sqrt {2}}{2 \sqrt {x^{4}-x^{2}+1}}\right )}{6}-\frac {\operatorname {arctanh}\left (\frac {2 x^{2}+3 x +2}{\sqrt {x^{4}-x^{2}+1}}\right )}{6}\)	\(116\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int \frac {1+x^6}{\sqrt {1-x^2+x^4} \left (1-x^6\right )} \, dx=\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {x^{4} - x^{2} + 1} x}{x^{4} - 3 \, x^{2} + 1}\right ) + \frac {1}{3} \, \log \left (\frac {x + \sqrt {x^{4} - x^{2} + 1}}{x^{2} - 1}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*int(sqrt(x**4 - x**2 + 1)/(x**8 + x**4 + 1),x) - 2*int((sqrt(x**4 - x** 
2 + 1)*x**4)/(x**8 + x**4 + 1),x) - log(x**2 - 1) + log( - sqrt(x**4 - x** 
2 + 1) - x))/3